Explanation:
Okay, so I need to find the probability that a randomly chosen point inside rectangle ABCD is closer to the center O than to any of the four vertices. The rectangle has sides in the ratio AB/AD = k. Let me try to visualize this. The rectangle is centered at O, which is the intersection of the diagonals. The vertices are A, B, C, D. The Asymptote code draws a rectangle with coordinates from (-250,100) to (250,-100), so maybe it's symmetric? But the actual coordinates might not matter because we can consider a general rectangle with AB = k*AD.

First, I need to understand what region inside the rectangle is closer to O than to any of the vertices. So for each vertex, I can think about the set of points that are closer to O than to that vertex. Then, the region we want is the intersection of all these sets, right? Because we want points that are closer to O than to any of the four vertices.

Alternatively, maybe it's easier to think about the Voronoi diagram for the center and the four vertices? But I'm not sure. Maybe I can find the boundaries where the distance to O equals the distance to a vertex, and then figure out the region where the distance to O is less than to any vertex.

Let me recall that the set of points equidistant to two points is the perpendicular bisector of the segment joining them. So, if I want points closer to O than to vertex A, I can find the perpendicular bisector between O and A, and then take the region closer to O.

But since there are four vertices, I need to find regions closer to O than to any of A, B, C, D. So the intersection of the four regions defined by being closer to O than to each vertex.

Let me set up coordinates to make this easier. Let me assume that the rectangle has length AB = k * AD. Let me denote AD = a, so AB = k*a. Let me consider coordinates. Let me place the rectangle with its center at the origin O(0,0). Then, the coordinates of the vertices can be determined.

Since the rectangle has center at O, the coordinates would be:

A is (- (AB/2), (AD/2)) ?

Wait, actually, the rectangle's sides are AB and AD. Let me think.

Let me define coordinate system such that O is (0,0). Let me denote the rectangle with length AB = k and AD = 1 for simplicity? Wait, but AB and AD are adjacent sides. Wait, AB is the length, AD is the width? Let me think.

If AB/AD = k, then AB = k*AD. So let me denote the rectangle with sides of length 2a and 2b, so that the center O is at (0,0), and the vertices are at (-a, -b), (-a, b), (a, b), (a, -b). Wait, then AB would be the horizontal side from (-a, -b) to (a, -b)? Wait no, AB is adjacent to A and B. Let me check the Asymptote code. The Asymptote code has A at (-250,100), B at (250,100), C at (250,-100), D at (-250,-100). So AB is horizontal, AD is vertical. So AB is the top side, AD is the left side. So if AB is horizontal, then AB is length 2*250? But in the Asymptote code it's from -250 to 250 on x-axis, so AB is length 500 and AD is length 200? But since AB/AD = k, so k = 500 / 200 = 2.5? So in the general case, AB = 2a and AD = 2b, so that AB / AD = (2a)/(2b) = a/b = k. So a = k*b. Therefore, if I take the rectangle with length AB = 2a and AD = 2b, then the coordinates of the vertices are:

A(-a, b), B(a, b), C(a, -b), D(-a, -b). Wait, but in the Asymptote code, A is (-250, 100), B is (250, 100), so if AB is horizontal, that's correct. So yes, if AB is horizontal, then A is (-a, b) and B is (a, b), etc.

But since AB/AD = k, and AB is horizontal length 2a, AD is vertical length 2b, so AB / AD = (2a)/(2b) = a/b = k. Therefore, a = k*b.

Therefore, the rectangle can be parametrized with variables a and b, where a = k*b. So maybe to make it even simpler, I can set b = 1, then a = k. Therefore, the rectangle has width 2a = 2k and height 2b = 2. Then the coordinates of the vertices are A(-k, 1), B(k, 1), C(k, -1), D(-k, -1). The center O is at (0,0). That might be easier for calculations.

Alternatively, maybe using coordinates with sides of length 2 and 2k? Wait, but the ratio AB/AD is k, so AB is length 2k, AD is length 2? Then AB/AD = k. Maybe that's another way. But maybe it's better to normalize.

Let me just use coordinates where the rectangle has length 2 and width 2k? Wait, but then AB is length 2, AD is length 2k. Then AB/AD = 1/k. Hmm, but the problem says AB/AD = k. So perhaps if AB is length 2k, AD is length 2. Then AB/AD = k. So yes, that would work.

So let me define the rectangle with vertices at (-k, -1), (-k, 1), (k, 1), (k, -1). Wait, that would make AB from (-k,1) to (k,1), so AB length is 2k. AD from (-k,1) to (-k,-1), so AD length is 2. So AB/AD = (2k)/2 = k. Perfect. Then center O is (0,0). So we can work with coordinates from x=-k to x=k, y=-1 to y=1.

But maybe I should use variables. Let me think. Let me consider general rectangle with sides of length 2a and 2b, so AB = 2a, AD = 2b, then AB/AD = a/b = k. Therefore, a = k*b.

But perhaps to make it even more general, maybe I can set coordinates with O at (0,0), and the rectangle extending from (-a, -b) to (a, b). So AB is horizontal from (-a, b) to (a, b), so AB = 2a. AD is vertical from (-a, b) to (-a, -b), so AD = 2b. Therefore, AB/AD = a/b = k. So a = k*b.

But maybe I can set b = 1, then a = k. So coordinates are from (-k, -1) to (k, 1). Therefore, center at (0,0). So let me use this coordinate system.

So the rectangle is [ -k, k ] in x-axis and [ -1, 1 ] in y-axis. The center is O(0,0). Now, the problem is to find the probability that a random point inside this rectangle is closer to O than to any of the four vertices. Since the area of the rectangle is (2k)(2) = 4k. So the probability will be [Area of desired region] / (4k).

Therefore, my task reduces to finding the area of the region within the rectangle where for any point (x,y), distance to O is less than distance to A, B, C, D.

Alternatively, the region where distance to O is less than the minimum distance to the four vertices.

But since the rectangle is symmetric, maybe I can focus on one quadrant and multiply by 4?

Yes! Because of symmetry, the region should be symmetric across both axes, so I can compute the area in one quadrant and multiply by 4. Let me think.

Let me consider the first quadrant where x ≥ 0 and y ≥ 0. In this quadrant, the vertices are B(k, 1) and A(-k, 1) is in the second quadrant, but in the first quadrant, the closest vertex is B(k,1) and D(-k,-1) is not in the quadrant. Wait, but in the first quadrant, the vertices are B(k,1) and maybe also... Wait, the rectangle has four vertices: A(-k,1), B(k,1), C(k,-1), D(-k,-1). So in the first quadrant, the only vertex is B(k,1). So in the first quadrant, the region closer to O than to any vertex would be the region closer to O than to B, and also closer to O than to A, C, D? Wait, but in the first quadrant, points can't be closer to A, C, or D than to O? Wait, actually, points in the first quadrant can be closer to A, which is (-k,1). Wait, but A is in the second quadrant. Wait, but a point in the first quadrant might be closer to A than to O?

Wait, for example, take the point (k/2, 1). The distance to O is sqrt( (k/2)^2 + 1^2 ). The distance to A(-k,1) is sqrt( (k/2 + k)^2 + (1 - 1)^2 ) = sqrt( (3k/2)^2 ) = 3k/2. Compare to distance to O: sqrt( (k^2)/4 + 1 ). For k > 0, this is less than 3k/2, since sqrt(k²/4 + 1) < sqrt(k²/4 + k²/4) = sqrt(k²/2) = k / sqrt(2) ≈ 0.707k, which is less than 1.5k. Wait, so for k > 0, this point is closer to O than to A? Wait, but if k is very large, then maybe not?

Wait, suppose k is very large. Let me take k approaching infinity. Then, the distance to O for point (k/2,1) is sqrt( (k/2)^2 + 1 ) ≈ k/2. The distance to A(-k,1) is sqrt( (3k/2)^2 + 0 ) = 3k/2. So yes, distance to O is half of distance to A. So even for large k, (k/2,1) is closer to O. So maybe points in the first quadrant are always closer to O than to A? Wait, but is that true?

Wait, suppose we take a point very close to A, but in the first quadrant? Wait, but A is at (-k,1). So the first quadrant is x ≥ 0, y ≥ 0. To be close to A, a point would have to be near (-k,1), but that's in the second quadrant. So in the first quadrant, points cannot be close to A. Similarly, points in the first quadrant can't be close to D or C? Because D is (-k,-1) and C is (k,-1). So in the first quadrant, the only vertex that is "nearby" is B(k,1). So maybe in the first quadrant, the only vertex we need to compare with O is B(k,1). Therefore, the region where the point is closer to O than to any vertex is the region in the first quadrant where it's closer to O than to B, and also closer to O than to A, C, D? But since points in the first quadrant are not close to A, C, D, except for B?

Wait, maybe not. For example, take a point near the center. It's closer to O than to any vertex. Take a point near the side of the rectangle. Let me check. Suppose the point is (x,y) in the first quadrant. What's the distance to O: sqrt(x² + y²). What's the distance to B(k,1): sqrt( (x - k)² + (y - 1)² ). The distance to A(-k,1) is sqrt( (x + k)² + (y - 1)² ). Similarly for others. But since the point is in first quadrant, x ≥ 0, y ≥ 0. So the distances to A, C, D would be:

To A(-k,1): distance squared is (x + k)^2 + (y - 1)^2.

To C(k,-1): distance squared is (x - k)^2 + (y + 1)^2.

To D(-k,-1): distance squared is (x + k)^2 + (y + 1)^2.

So comparing to O: distance squared is x² + y².

So for the point to be closer to O than to A, we need:

x² + y² < (x + k)^2 + (y - 1)^2

Similarly for other vertices. Let me check this inequality:

x² + y² < (x + k)^2 + (y - 1)^2

Expanding the right-hand side: x² + 2kx + k² + y² - 2y + 1

Subtract left-hand side:

0 < 2kx + k² - 2y + 1

Which simplifies to:

2kx - 2y + (k² + 1) > 0

Similarly, for distance to O being less than distance to C:

x² + y² < (x - k)^2 + (y + 1)^2

Expanding: x² - 2kx + k² + y² + 2y + 1

Subtract left-hand side:

0 < -2kx + k² + 2y + 1

=> -2kx + 2y + (k² + 1) > 0

Similarly, for distance to D:

x² + y² < (x + k)^2 + (y + 1)^2

Right-hand side: x² + 2kx + k² + y² + 2y + 1

Subtract left-hand side:

0 < 2kx + 2y + k² + 1

So 2kx + 2y + (k² + 1) > 0.

But since x ≥ 0, y ≥ 0, and k > 0, these inequalities are all true? Wait, for distance to O being less than distance to A, the inequality is 2kx - 2y + (k² + 1) > 0. Since x ≥ 0, y ≥ 0, and k is positive, this would be:

Is 2kx - 2y + (k² + 1) > 0?

But since x and y can vary, this inequality may or may not hold. For example, if x is very small and y is large.

Wait, but in the first quadrant, points can be closer to A? Let me take a point near (0,1). Let me compute distance to O: sqrt(0² + 1²) = 1. Distance to A(-k,1): sqrt( (0 + k)^2 + (1 - 1)^2 ) = k. So 1 < k? That is true if k > 1. Wait, but k is AB / AD. If the rectangle is very long horizontally, then k is large. If the rectangle is very tall, k is small. So depending on k, the point (0,1) could be closer to O or to A.

Wait, for example, if k = 1, then distance to A is 1, same as distance to O. If k > 1, then distance to A is larger, so O is closer. If k < 1, distance to A is smaller, so point (0,1) is closer to A.

Therefore, the condition that the point is closer to O than to A is not automatically satisfied in the first quadrant. Therefore, I need to check all four inequalities.

But this is getting complicated. Maybe instead of considering each vertex, I can find the region where distance to O is less than distance to any of the vertices. Which would be the intersection of all regions where distance to O is less than distance to each vertex.

Alternatively, since the rectangle is convex and the set of points closer to O than to any vertex is the intersection of the regions closer to O than to each vertex. Therefore, the boundaries are the perpendicular bisectors between O and each vertex. Then the region we want is the intersection of all four regions.

So maybe to find the boundaries, let's find the perpendicular bisector between O and each vertex. For example, the set of points equidistant to O and vertex A.

Let me take vertex A(-k,1). The set of points equidistant to O(0,0) and A(-k,1) is the perpendicular bisector of OA.

The midpoint of OA is (-k/2, 1/2). The line OA has slope (1 - 0)/(-k - 0) = -1/k. Therefore, the perpendicular bisector will have slope k.

So equation of the perpendicular bisector: passes through (-k/2, 1/2) with slope k.

So equation: y - 1/2 = k(x + k/2)

Similarly, for the other vertices.

But since we are in the first quadrant, maybe I can find the intersection of these bisectors and find the region.

Alternatively, since the problem is symmetric, maybe the region closer to O than any vertex is a smaller rectangle or a polygon whose area we can compute.

Alternatively, perhaps the region is the set of points closer to O than to any vertex, which is the intersection of four regions. Each region is defined by being closer to O than to one vertex.

But since we have symmetry, maybe it's enough to consider one region and multiply.

But perhaps it's easier to find the locus of points equidistant to O and to each vertex.

Let me try to find equations for these bisectors.

First, let's take vertex A(-k, 1). As above, the perpendicular bisector is y - 1/2 = k(x + k/2).

Similarly, for vertex B(k,1), the perpendicular bisector would be symmetric. The midpoint of OB is (k/2, 1/2), and the line OB has slope (1 - 0)/(k - 0) = 1/k, so the perpendicular bisector has slope -k. Equation: y - 1/2 = -k(x - k/2).

Similarly, for vertex C(k, -1), the perpendicular bisector is y + 1/2 = -k(x - k/2) (since midpoint is (k/2, -1/2), and line OC has slope (-1)/k, so perpendicular bisector has slope k). Wait:

Wait, for vertex C(k, -1): midpoint between O(0,0) and C(k, -1) is (k/2, -1/2). The line OC has slope (-1 - 0)/(k - 0) = -1/k, so the perpendicular bisector has slope k. Therefore, equation is y - (-1/2) = k(x - k/2) => y + 1/2 = k(x - k/2).

Similarly for vertex D(-k, -1): midpoint is (-k/2, -1/2), line OD has slope (-1)/(-k) = 1/k, so perpendicular bisector has slope -k. Equation: y - (-1/2) = -k(x - (-k/2)) => y + 1/2 = -k(x + k/2).

Therefore, these four bisectors divide the plane into regions. The region closer to O than any vertex is the intersection of the regions closer to O than each vertex. Therefore, this region is bounded by these four bisectors.

Therefore, in each quadrant, the region is bounded by two bisectors. For example, in the first quadrant, the boundaries are the bisectors between O and B, and between O and A? Wait, no. Wait in the first quadrant, which bisectors are relevant?

In the first quadrant, the closest vertices are B and... Well, A is in the second quadrant, but maybe the bisector between O and A will intersect the first quadrant?

Let me check the equation for the bisector between O and A: y - 1/2 = k(x + k/2). Let me see where this line intersects the first quadrant. Since in the first quadrant, x ≥ 0, y ≥ 0.

Let me solve for y: y = kx + k^2/2 + 1/2.

But when x = 0, y = k*(0) + k²/2 + 1/2 = (k² + 1)/2. However, since the rectangle has y from -1 to 1, so if (k² + 1)/2 > 1, which would be when k² +1 > 2 => k² >1 => |k|>1. So if k >1, then the bisector between O and A intersects the first quadrant above y=1, which is outside the rectangle. Therefore, within the rectangle, the bisector between O and A would intersect the top side of the rectangle at y=1. Let me find that intersection point.

Set y = 1 in the equation of bisector between O and A: 1 - 1/2 = k(x + k/2) => 1/2 = kx + k²/2 => Multiply both sides by 2: 1 = 2kx + k² => 2kx = 1 - k² => x = (1 - k²)/(2k). However, since we are in the first quadrant, x must be ≥0. Therefore, (1 - k²)/(2k) ≥ 0. Since k > 0 (as AB/AD =k, AB and AD are positive lengths), this requires 1 -k² ≥ 0 => k² ≤1 => k ≤1. Therefore, if k ≤1, the bisector between O and A intersects the top edge at x = (1 - k²)/(2k). If k >1, then (1 -k²)/ (2k) is negative, meaning the bisector between O and A does not intersect the top edge in the first quadrant. Instead, within the rectangle, the bisector between O and A would intersect the left edge or something else?

Wait, but the bisector between O and A is y = kx + k²/2 + 1/2. If we set x = -k (the left side of the rectangle), but x is in the first quadrant (x ≥ 0). So maybe the bisector doesn't intersect the rectangle in the first quadrant when k >1. Therefore, in that case, within the rectangle, the region closer to O than A would be the entire first quadrant?

Wait, maybe not. Let me think. For example, when k >1, take a point near the top left corner of the rectangle. Wait, but in the first quadrant, the left side is x=0, y from 0 to 1. Let me take the point (0,1). Its distance to O is 1. Its distance to A(-k,1) is sqrt( (k)^2 + 0 ) = k. Since k >1, so distance to O is less than distance to A. Therefore, the entire first quadrant is closer to O than to A? Wait, but if k is very large, say k approaches infinity, then for any point (x, y) in the first quadrant, distance to O is sqrt(x² + y²), and distance to A is sqrt( (x + k)^2 + (y - 1)^2 ). For large k, distance to A is approximately k + x, since (x +k)^2 dominates. Then distance to O is sqrt(x² + y²) ≈ sqrt(x² + y²) << k + x. Therefore, for large k, the entire first quadrant is closer to O than to A. Therefore, for k ≥1, the region closer to O than A in the first quadrant is the entire first quadrant? Wait, but maybe not entirely?

Wait, suppose k = 2. Take a point (x,y) in the first quadrant. For it to be closer to O than to A, we need sqrt(x² + y²) < sqrt( (x + 2)^2 + (y -1)^2 ). Squaring both sides:

x² + y² < (x + 2)^2 + (y -1)^2

Expand RHS: x² + 4x + 4 + y² - 2y + 1 = x² + y² + 4x - 2y + 5

Subtract LHS: 0 < 4x - 2y + 5 => 4x - 2y + 5 > 0 => 2x - y + 5/2 > 0 => y < 2x + 5/2. But since y ≤ 1 and x ≥ 0, this inequality is always true because 2x + 5/2 ≥ 5/2 > 1. Therefore, for k = 2, the entire first quadrant is closer to O than to A. Therefore, when k ≥1, the region closer to O than A in the first quadrant is the entire first quadrant.

Similarly, for the region closer to O than C or D? Let me check. For point (x,y) in first quadrant, distance to C(k, -1) is sqrt( (x - k)^2 + (y + 1)^2 ). Compare to distance to O: sqrt(x² + y²). Squaring both sides:

x² + y² < (x - k)^2 + (y + 1)^2

Expand RHS: x² - 2kx + k² + y² + 2y + 1

Subtract LHS: 0 < -2kx + k² + 2y + 1 => 2kx - 2y + ( -k² -1 ) < 0 ?

Wait, wait:

Original inequality: x² + y² < (x - k)^2 + (y + 1)^2 => 0 < -2kx + k² + 2y + 1.

So 2kx - 2y + (-k² -1) < 0? Wait, no:

Wait, 0 < -2kx + k² + 2y + 1 is equivalent to 2kx - 2y + (-k² -1) < 0? Let me check:

Original inequality: 0 < -2kx + k² + 2y + 1 => rearranged: 2kx - 2y + (-k² -1) < 0? Wait, no. Let me rearrange:

Starting from 0 < -2kx + k² + 2y + 1, subtract both sides by (-2kx + k² + 2y + 1):

Wait maybe better to rearrange:

-2kx + 2y + k² + 1 > 0 => 2y > 2kx - k² - 1 => y > kx - (k² +1)/2.

But for points in the first quadrant, y ≥ 0. So depending on k and x, this inequality may or may not hold.

For example, take k = 2, and check if y > 2x - (4 + 1)/2 = 2x - 2.5. In the first quadrant, y ≥ 0. Let me see when 0 > 2x - 2.5 => 2x < 2.5 => x < 1.25. So for x < 1.25, y ≥ 0 satisfies the inequality y > 2x - 2.5. However, for x ≥1.25, the inequality y > 2x - 2.5 may not hold. For example, take x = 1.5, then 2x - 2.5 = 3 - 2.5 = 0.5. So y needs to be greater than 0.5. Therefore, for x ≥1.25, the region where y > 0.5 would satisfy the inequality. However, if y is less than 0.5, it would not.

Therefore, for k=2, the region closer to O than to C in the first quadrant is y > 2x - 2.5. So this is a line cutting through the first quadrant. Similarly, for vertex D(-k, -1), the bisector is y + 1/2 = -k(x + k/2). In the first quadrant, this would correspond to something else? But maybe we don't have to consider D in the first quadrant? For points in the first quadrant, distance to D is sqrt( (x + k)^2 + (y + 1)^2 ). Since x ≥0, y ≥0, this is larger than sqrt( (x +k)^2 + 1 ), which is larger than sqrt( k^2 + 1 ) if x=0. So comparing to distance to O, sqrt(x² + y²). For points in first quadrant, is it possible to be closer to D than to O?

Yes, for example, take the point (0, 0). Distance to O: 0, which is less than distance to D. But take a point near (0, -something)? Wait, no, in first quadrant. Take point (0, 1). Distance to O: 1, distance to D(-k, -1): sqrt( (k)^2 + (2)^2 ). If k is 2, this is sqrt(4 + 4) = sqrt(8) ≈ 2.828. So distance to O is less. If we take a point (x, y) where y is very small. For example, take y = 0, x approaching k. Then distance to O is sqrt(x² + 0) = x. Distance to D is sqrt( (x + k)^2 + (0 + 1)^2 ) ≈ sqrt( (2k)^2 + 1 ). For large k, this is about 2k, so distance to O is x, which is less than 2k. Therefore, even points near the right side are closer to O than to D. So maybe in the first quadrant, the only vertex we need to compare with O is B and C? Wait, but we had an inequality for C and D? Wait, but we considered distance to C and D. However, for the first quadrant, maybe the critical vertices are B and C. Wait, but D is at (-k, -1), so in the first quadrant, points can't be closer to D than to O? Wait, let's check.

Take a point (x, y) in first quadrant. Distance to D is sqrt( (x +k)^2 + (y +1)^2 ). Distance to O is sqrt(x² + y²). Let me compare:

sqrt( (x +k)^2 + (y +1)^2 ) < sqrt(x² + y² )?

Squaring both sides:

(x +k)^2 + (y +1)^2 < x² + y²

Expanding:

x² + 2kx + k² + y² + 2y +1 < x² + y²

Simplify:

2kx + k² + 2y + 1 < 0

But 2kx + 2y + k² + 1 < 0. Since x ≥0, y ≥0, k >0, this is impossible. Therefore, in the first quadrant, no point is closer to D than to O. Similarly, for distance to A, we saw that if k ≥1, then all points are closer to O than to A. Therefore, in the first quadrant, the only constraint is being closer to O than to B or C.

Therefore, in the first quadrant, the region where points are closer to O than to any vertex is the intersection of regions closer to O than B and C. Therefore, we need to find the area in the first quadrant where distance to O < distance to B and distance to O < distance to C. Then multiply by 4 to get total area.

So let me focus on first quadrant. Let me write down the inequalities:

For distance to O < distance to B(k,1):

sqrt(x² + y²) < sqrt( (x - k)^2 + (y - 1)^2 )

Squaring both sides:

x² + y² < (x - k)^2 + (y - 1)^2

Which simplifies to:

x² + y² < x² - 2kx + k² + y² - 2y + 1

Cancelling x² and y²:

0 < -2kx + k² - 2y + 1

Which rearranged is:

2kx + 2y < k² + 1

Similarly, for distance to O < distance to C(k, -1):

sqrt(x² + y²) < sqrt( (x - k)^2 + (y + 1)^2 )

Squaring both sides:

x² + y² < (x - k)^2 + (y + 1)^2

Expanding:

x² + y² < x² - 2kx + k² + y² + 2y + 1

Simplify:

0 < -2kx + k² + 2y + 1

Which rearranged is:

2kx - 2y < -k² - 1

Wait, let me do it again:

Original inequality:

0 < -2kx + k² + 2y + 1 => rearranged:

2kx - 2y < -k² -1 ?

Wait:

Starting from 0 < -2kx + k² + 2y + 1,

Moving terms:

2kx - 2y < -k² -1 ?

Wait, if we move 2kx to the left and 0 to the right? Wait, perhaps better to rearrange:

Let me write:

From 0 < -2kx + k² + 2y + 1,

Bring 2kx to the right and others to the left:

-2kx + k² + 2y + 1 > 0 => 2y > 2kx - k² -1 => y > kx - (k² +1)/2.

But in the previous inequality for distance to B, we had 2kx + 2y < k² + 1 => y < (k² +1)/2 -kx.

So combining both inequalities:

In the first quadrant, we have:

y < (k² +1)/2 -kx (from distance to B)

and

y > kx - (k² +1)/2 (from distance to C)

So these two inequalities define a region between two lines.

Therefore, in the first quadrant, the region closer to O than to B or C is between these two lines. Therefore, to find the area in the first quadrant, we need to find the area between these two lines, within the rectangle [0 ≤ x ≤k, 0 ≤ y ≤1].

Therefore, the bounds for y are:

Lower bound: y_lower = kx - (k² +1)/2

Upper bound: y_upper = (k² +1)/2 -kx

But we need to ensure that these bounds are within the rectangle, i.e., between 0 and 1.

Therefore, we need to find the range of x where y_lower ≥0 and y_upper ≤1. However, since the region is between these two lines, and in the first quadrant, we need to find the intersection of these inequalities with the rectangle.

Alternatively, maybe we can find the region where both inequalities hold, which is the area between the two lines.

But to find that area, perhaps we need to find where these lines intersect the rectangle.

First, let's find where y_lower = kx - (k² +1)/2 intersects the rectangle.

Similarly for y_upper.

Let me analyze y_lower = kx - (k² +1)/2.

Since in the first quadrant, y ≥ 0. So solving for when y_lower = 0:

kx - (k² +1)/2 = 0 => x = (k² +1)/(2k)

Similarly, y_upper = (k² +1)/2 -kx. Setting y_upper = 1:

(k² +1)/2 -kx = 1 => -kx = 1 - (k² +1)/2 = (2 - k² -1)/2 = (1 - k²)/2 => x = (k² -1)/(2k)

Therefore, the line y_upper intersects the top edge y=1 at x=(k² -1)/(2k), and the line y_lower intersects the bottom edge y=0 at x=(k² +1)/(2k).

But depending on k, these x-values may be within the rectangle [0, k].

Let me analyze for different values of k.

First, consider k >1.

If k >1, then (k² -1)/(2k) is positive because k² -1 >0. Also, (k² +1)/(2k) is greater than (k²)/(2k) = k/2. Since k >1, let's see if (k² +1)/(2k) is less than k?

(k² +1)/(2k) < k ?

Multiply both sides by 2k (positive):

k² +1 < 2k² => 1 < k² => which is true since k >1.

Therefore, for k >1, the line y_upper intersects y=1 at x=(k² -1)/(2k), and lies between 0 and k. Similarly, the line y_lower intersects y=0 at x=(k² +1)/(2k), which is between k/2 and k (since (k² +1)/(2k) = k/2 + 1/(2k), which is greater than k/2).

Therefore, in the first quadrant, the region between the lines y_lower and y_upper is bounded between x values from x_low to x_high, where x_low is where y_upper meets y=1, and x_high is where y_lower meets y=0. Wait, but how exactly?

Wait, perhaps the region between the two lines is a parallelogram or some polygon. Let me sketch mentally.

For k >1:

The line y_upper = (k² +1)/2 -kx starts at the top edge at x=(k² -1)/(2k), y=1, and goes down to where?

At x=0, y_upper would be (k² +1)/2 - 0 = (k² +1)/2 >1, since for k >1, (k² +1)/2 > (1 +1)/2 =1. Therefore, the line y_upper intersects the left edge x=0 above y=1, which is outside the rectangle. Therefore, within the rectangle, the line y_upper goes from ( (k² -1)/(2k), 1 ) down to some point where it meets y_lower?

Wait, no. Wait, the two lines y_upper and y_lower intersect each other?

Let me find where y_lower = y_upper:

kx - (k² +1)/2 = (k² +1)/2 -kx

=> kx +kx = (k² +1)/2 + (k² +1)/2

=> 2kx = k² +1 => x = (k² +1)/(2k)

Which is the same x as where y_lower meets y=0. So the two lines intersect at x=(k² +1)/(2k), y=0?

Wait, plug x=(k² +1)/(2k) into y_lower:

y_lower = k*( (k² +1)/(2k) ) - (k² +1)/2 = (k² +1)/2 - (k² +1)/2 = 0. So yes, they intersect at point ((k² +1)/(2k), 0). Similarly, if we plug x=(k² -1)/(2k) into y_upper:

y_upper = (k² +1)/2 -k*( (k² -1)/(2k) ) = (k² +1)/2 - (k² -1)/2 = [k² +1 -k² +1]/2 = 2/2 = 1. So that's the point.

Therefore, in the first quadrant, the region between the two lines is a quadrilateral with vertices at:

1. Intersection of y_upper and y=1: ((k² -1)/(2k), 1)

2. Intersection of y_upper and y_lower: ((k² +1)/(2k), 0)

3. Intersection of y_lower and y=0: ((k² +1)/(2k), 0) – same as above?

Wait, no, y_lower is the lower bound, so it goes from ((k² +1)/(2k), 0) to... Wait, but if we think of the region between y_upper and y_lower, it's bounded between these two lines. But since for x between 0 and (k² -1)/(2k), y_upper is above y=1, which is outside the rectangle. Therefore, within the rectangle, the region between the two lines is from x = (k² -1)/(2k) to x = (k² +1)/(2k), between the two lines.

Wait, let me think again.

When x is less than (k² -1)/(2k), the upper bound y_upper would be greater than 1, which is outside the rectangle. Therefore, within the rectangle, the upper bound is y=1. Similarly, for x greater than (k² +1)/(2k), the lower bound y_lower would be negative, which is outside the rectangle, so the lower bound is y=0.

Therefore, within the rectangle, the region closer to O than to B or C is bounded between:

- For x between 0 and x1 = (k² -1)/(2k): between y=0 and y=1 (since the lines are outside the rectangle). Wait, no, because we have constraints from both B and C.

Wait, no. Wait, we need both distance to O < distance to B and distance to O < distance to C. So for x between 0 and x1, which is where y_upper meets y=1, we have that the upper bound is y_upper which is above y=1, so within the rectangle, the upper bound is y=1. However, the lower bound is y_lower which may be below y=0 or not?

Wait, for x between 0 and x1, the lower bound y_lower = kx - (k² +1)/2. For x=0, that is - (k² +1)/2 <0. Therefore, within the rectangle, the lower bound is y=0. Therefore, for x between 0 and x1, the region is between y=0 and y=1, but also must satisfy the inequalities from B and C. However, we already derived that the inequalities reduce to between the lines which for x in [0, x1], the region is between y=0 and y=1? Wait, no, that can't be.

Wait, no. The inequalities are:

For all points in the region, they must satisfy both inequalities:

y < (k² +1)/2 -kx (from B)

and

y > kx - (k² +1)/2 (from C)

But since in this region, for x between 0 and x1, (k² +1)/2 -kx is greater than 1, so y < (k² +1)/2 -kx is automatically satisfied since y ≤1. Similarly, y > kx - (k² +1)/2. For x between 0 and x1, kx - (k² +1)/2 is less than 0, since at x= x1, it's zero. Therefore, y > [negative number], which is automatically satisfied since y ≥0. Therefore, for x between 0 and x1, all points satisfy both inequalities. Hence, the entire region from x=0 to x=x1, y=0 to y=1, is within the desired region.

Then, for x between x1 and x2 = (k² +1)/(2k), the upper bound is y_upper = (k² +1)/2 -kx, which decreases from 1 to 0. The lower bound is y_lower = kx - (k² +1)/2, which increases from 0 to 0. Therefore, the region between these lines is a sort of "triangle" between x1 and x2.

But wait, when x is between x1 and x2, the bounds are between y_lower and y_upper, which meet at (x2, 0). Therefore, the area between x1 and x2 is bounded by the two lines and the x-axis?

Wait, actually, since y_lower is increasing from 0 to 0 (wait, at x=x2, y_lower is 0), while y_upper is decreasing from 1 to 0. Therefore, the region between x1 and x2 is bounded above by y_upper and below by y_lower, forming a region that is a triangle?

Wait, let me take specific values. Let me take k=2, which is >1.

For k=2, x1 = (k² -1)/(2k) = (4 -1)/4 = 3/4 = 0.75

x2 = (k² +1)/(2k) = (4 +1)/4 = 5/4 = 1.25

Therefore, between x=0.75 and x=1.25, the region is between the two lines y_upper = (4 +1)/2 - 2x = 2.5 - 2x, and y_lower = 2x - (4 +1)/2 = 2x - 2.5.

Therefore, for x between 0.75 and 1.25, the area between these two lines.

At x=0.75, y_upper = 2.5 - 2*(0.75) = 2.5 - 1.5 = 1.0, and y_lower = 2*(0.75) - 2.5 = 1.5 - 2.5 = -1.0. But since y ≥0, the lower bound is 0.

Wait, this seems conflicting with previous thought. Wait, earlier I thought that between x1 and x2, the region between the lines is bounded by those lines, but at x= x1, y_lower is negative, so within the rectangle, the lower bound is y=0. Therefore, the region between x1 and x2 is between y=0 and y_upper = 2.5 - 2x. Wait, but then why was there a y_lower?

Wait, perhaps my earlier analysis was incorrect. Let me re-express.

We had two inequalities:

1. y < (k² +1)/2 -kx

2. y > kx - (k² +1)/2

But since in the rectangle, y ∈ [0,1], the effective region is:

y ∈ [max(0, kx - (k² +1)/2 ), min(1, (k² +1)/2 -kx ) ]

Therefore, for x between 0 and x1 = (k² -1)/(2k), where (k² +1)/2 -kx ≥1 and kx - (k² +1)/2 ≤0, the region is y ∈ [0,1].

For x between x1 and x2 = (k² +1)/(2k), where (k² +1)/2 -kx decreases from 1 to 0, and kx - (k² +1)/2 increases from 0 to 0, the region is y ∈ [kx - (k² +1)/2, (k² +1)/2 -kx ].

For x between x2 and k, where (k² +1)/2 -kx < 0, and kx - (k² +1)/2 >0, the region would be y ∈ [kx - (k² +1)/2, 0], but since y ≥0, that would be empty. Hence, the region is only between x=0 and x=x2.

Wait, but x2 is (k² +1)/(2k). For k >1, since (k² +1)/(2k) is less than k (as we saw earlier), so between x= x1 and x= x2, the region is between the two lines. Beyond x2, the region would be y ≥ [kx - (k² +1)/2 ] which is positive, but the upper bound would be negative, hence no region.

Therefore, in the first quadrant, the area closer to O than to B or C is:

- For x from 0 to x1: area = width * height = (x1 - 0) * 1 = x1

- For x from x1 to x2: area is the integral from x1 to x2 of [ (k² +1)/2 -kx - (kx - (k² +1)/2 ) ] dx = integral from x1 to x2 of [ (k² +1)/2 -kx -kx + (k² +1)/2 ] dx = integral of [k² +1 - 2kx ] dx

Wait:

Wait, the upper bound minus lower bound is [(k² +1)/2 -kx] - [kx - (k² +1)/2] = (k² +1)/2 -kx -kx + (k² +1)/2 = k² +1 - 2kx.

Therefore, the integral from x1 to x2 of (k² +1 - 2kx) dx.

So total area in first quadrant is:

A1 = x1 + ∫ from x1 to x2 (k² +1 - 2kx) dx

Compute this integral:

First, compute x1 = (k² -1)/(2k)

x2 = (k² +1)/(2k)

Compute integral ∫ (k² +1 - 2kx ) dx from x1 to x2.

Antiderivative is:

(k² +1)x - kx² evaluated from x1 to x2.

So compute at x2: (k² +1)x2 - kx2²

At x1: (k² +1)x1 - kx1²

Thus, the integral is [ (k² +1)x2 - kx2² ] - [ (k² +1)x1 - kx1² ]

Therefore, total area A1 is x1 + [ (k² +1)x2 - kx2² - (k² +1)x1 + kx1² ]

Simplify:

A1 = x1 + (k² +1)(x2 - x1) -k(x2² -x1² )

Note that x2² -x1² = (x2 -x1)(x2 +x1)

Therefore, A1 = x1 + (k² +1)(x2 -x1) -k(x2 -x1)(x2 +x1 )

Factor out (x2 -x1):

A1 = x1 + (x2 -x1)[(k² +1) -k(x2 +x1 ) ]

Now compute (x2 +x1):

x1 + x2 = [ (k² -1)/(2k) ] + [ (k² +1)/(2k) ] = [ (k² -1 +k² +1 ) / (2k) ] = (2k²)/ (2k) ) = k

Therefore, x2 +x1 = k

Therefore, A1 = x1 + (x2 -x1)[(k² +1) -k*k ] = x1 + (x2 -x1)[k² +1 -k²] = x1 + (x2 -x1)(1) = x1 + x2 -x1 = x2

Therefore, A1 = x2

Wait! That's unexpected. So the total area in first quadrant is x2? Which is (k² +1)/(2k). But let me check with k=2:

For k=2, x2 = (4 +1)/4 = 5/4 =1.25. Therefore, A1=1.25? But in the first quadrant, the area closer to O than to B or C is supposed to be... Let me check with k=2.

If k=2, then rectangle is from x=-2 to 2, y=-1 to 1. In first quadrant, area closer to O is x from 0 to 1.25, and... Wait, but according to the calculation, the area is x2 = 1.25? Which is a length? Wait, no, A1 is supposed to be an area. Wait, I think there was a miscalculation here.

Wait, going back, when I said A1 = x1 + ∫ ... which resulted in A1 = x2. But x1 and x2 are lengths along x-axis, so how can area be equal to x2? There must be a mistake in the calculation.

Let me retrace:

Original A1 = area from x=0 to x=x1 (which is a rectangle with width x1 and height 1) plus the integral from x=x1 to x=x2 of [k² +1 - 2kx] dx.

Then after computing, we found that A1 = x2. But when k=2, x2 is 5/4, but the first part x1 was 3/4, and then the integral? Let me compute it numerically for k=2.

For k=2, x1 = 3/4, x2 =5/4.

Compute A1 = x1 + ∫_{x1}^{x2} [k² +1 - 2kx] dx.

k² +1 = 4 +1 =5, 2k =4.

Thus, integral becomes ∫_{3/4}^{5/4} (5 -4x) dx.

Antiderivative is 5x - 2x² evaluated from 3/4 to 5/4.

At 5/4: 5*(5/4) - 2*(25/16) = 25/4 - 50/16 = 25/4 - 25/8 = 25/8.

At 3/4: 5*(3/4) - 2*(9/16) = 15/4 - 9/8 = 30/8 - 9/8 = 21/8.

Therefore, the integral is 25/8 -21/8 = 4/8 = 1/2.

Then A1 = x1 + integral = 3/4 + 1/2 = 5/4, which is equal to x2. Therefore, for k=2, A1=5/4. But the area in first quadrant is 5/4? Wait, but the entire first quadrant has area k*1 =2*1=2. Wait, no! Wait, the rectangle in first quadrant is [0, k] x [0,1], so area is k*1 =k. For k=2, area is 2. But according to this, A1 =5/4 which is less than 2. So that seems possible.

Wait, but let me check if this answer makes sense. For k=2, the region we found is 5/4. But let me think of another way. If we consider that the region closer to O than to B or C is the set of points between the two lines. For k=2, the area is 5/4. But if we consider the entire first quadrant area is 2, so probability would be 5/4 divided by 2 = 5/8. But let me check if that's correct.

Alternatively, let's think of the area where distance to O is less than distance to B and C. For k=2, it's the area between the two lines which we found to be 5/4. But maybe I can compute it another way.

Alternatively, the region closer to O than to B and C is the intersection of two regions. Each region is a half-plane. The intersection is the area between the two lines we found. But according to the calculation, that area is x2 = (k² +1)/(2k). Wait, for k=2, that is 5/4, which seems correct. But how does that relate to the rectangle?

Wait, maybe it's better to accept that the area in first quadrant is x2 = (k² +1)/(2k). Therefore, total area over all quadrants would be 4*(k² +1)/(2k) = 2(k² +1)/k. But wait, no. Wait, in the first quadrant, we found A1 = x2 = (k² +1)/(2k). Therefore, total area is 4*A1? Wait, no. Wait, the region is symmetric in all four quadrants? Wait, no. Wait, we did the analysis for the first quadrant, and the same applies to other quadrants. Therefore, total area is 4*A1. However, A1 = (k² +1)/(2k). Therefore, total area is 4*(k² +1)/(2k) = 2(k² +1)/k. However, the total area of the rectangle is 2k*2 = 4k. Therefore, probability is [2(k² +1)/k ] / (4k) ) = [ (k² +1)/k ] / (2k ) = (k² +1)/(2k² )

Wait, but for k=2, probability is (4 +1)/ (2*4) = 5/8, which matches our earlier calculation. But let me check if that answer makes sense. For k=1, which makes the rectangle a square. For k=1, probability is (1 +1)/2 =1, which is 1. Which doesn't make sense? Wait, wait. If k=1, the rectangle is a square with sides 2 and 2, area 4. The region closer to O than any vertex... Wait, in a square, the set of points closer to center than to any vertex is known to be a smaller square or another shape?

Wait, for a square, the region closer to center than to any vertex is the set of points where distance to center < distance to vertices. For a square, the set of points equidistant to center and a vertex is the perpendicular bisector between center and vertex, which in case of square would be a line. For example, for the vertex (1,1), the perpendicular bisector would be the line y = x, but wait:

Wait, let's take square with center at (0,0), vertices at (1,1), (-1,1), etc. For the vertex (1,1), the perpendicular bisector between (0,0) and (1,1) is the line y = -x + 1? Wait, midpoint is (0.5, 0.5). The line from (0,0) to (1,1) has slope 1, so perpendicular bisector has slope -1. Equation: y - 0.5 = -1(x -0.5) => y = -x +1. Similarly for other vertices. So the region closer to center than to any vertex is the intersection of four regions, bounded by these lines. The intersection would be a smaller square rotated by 45 degrees, whose vertices are at the intersections of these bisectors.

For example, the intersection of y = -x +1 and y = x +1 (from the other bisector) is at x=0, y=1. Similarly, other intersections would form a square. But the area of this inner square can be calculated.

For the square case (k=1), our formula gives probability = (1 +1)/(2*1² )= 2/2 =1, which can't be correct. Because the region closer to center than to any vertex can't be the entire square. For example, the corner points are exactly at distance sqrt(2)/2 from the center and distance 0 from the vertex, so those points are closer to the vertex. Therefore, the region closer to center is definitely smaller than the whole square. Hence, my calculation must be wrong.

Wait, so where is the mistake?

Earlier, I concluded that total area is 4*A1, where A1 = x2 = (k² +1)/(2k). But for k=1, this would be (1 +1)/2*1=1, so total area 4*1=4, which is equal to the area of the rectangle. Which is wrong. Hence, there's a mistake in reasoning.

Wait, when I considered the first quadrant area A1, I calculated it as x2. But when k=1, x2 = (1 +1)/(2*1)=1. Therefore, A1=1. But in the first quadrant, which is a square of area 1*1=1, so A1=1 would mean the entire quadrant is included, which is not correct. Hence, my mistake must be in the calculation of A1.

Going back, when I calculated A1 for general k, I had:

A1 = x1 + integral ... which became A1 = x2. But when k=1, this would be x2 = (1 +1)/2*1=1, but in reality, the area should be less than 1. Hence, error in the derivation. Let me check again.

Earlier steps:

I had A1 = x1 + [ (k² +1)(x2 -x1) -k(x2² -x1² ) ]

Then I factored out (x2 -x1):

A1 = x1 + (x2 -x1)[(k² +1) -k(x2 +x1 ) ]

Then I used x2 +x1 =k, so:

A1 = x1 + (x2 -x1)[k² +1 -k²] = x1 + (x2 -x1)(1) = x1 +x2 -x1 =x2.

But for k=1, x1 = (1 -1)/(2*1) =0, x2 = (1 +1)/ (2*1)=1. Hence, A1 = x2 =1, which is the entire first quadrant area. But this contradicts the square case.

But earlier, for k=1, the region closer to O than to any vertex is not the entire square. Therefore, there is a miscalculation in the steps. Where?

Wait, the error comes from the assumption that the area in the first quadrant is x2. But for k=1, this suggests that the entire first quadrant is included, which is not true. Therefore, my mistake is in the integral calculation.

Let me redo the integral calculation for general k. Let me go back.

First quadrant area A1 is:

Integral from x=0 to x=x1 of 1 dx + Integral from x=x1 to x=x2 of [ upper - lower ] dx

Which is:

x1 + ∫_{x1}^{x2} [ (k² +1)/2 -kx - (kx - (k² +1)/2 ) ] dx = x1 + ∫_{x1}^{x2} [k² +1 -2kx ] dx.

Yes, which we did. Then computed that as x2. But in reality, when k=1, this integral would be:

x1 = (1 -1)/2 =0

x2 = (1 +1)/2 =1

Integral from 0 to1 of [1 +1 - 2*1*x] dx = ∫₀¹ (2 - 2x) dx = [2x -x²]₀¹ = (2*1 -1) -0 =1. Hence, A1=0 +1=1. However, in reality, when k=1, the region closer to O than to any vertex should be the set of points where distance to O < distance to vertices. For k=1, the square has vertices at (±1, ±1). The region closer to O than to any vertex is the set of points where sqrt(x² + y²) < sqrt( (x -1)^2 + (y -1)^2 ), and similarly for other vertices. Which simplifies to the inequalities we had before.

But when k=1, the lines y_upper = (k² +1)/2 -kx = (1 +1)/2 -1*x =1 -x, and y_lower =kx - (k² +1)/2 =x -1.

In the first quadrant, between x=0 and x=1, the region between y_lower =x -1 and y_upper=1 -x. However, y_lower =x -1 is negative for x <1, and y_upper =1 -x goes from 1 to 0. Hence, the area between them in the first quadrant is the area between y=0 and y=1 -x from x=0 to x=1. Therefore, the area should be ∫₀¹ (1 -x) dx = [x - x²/2]₀¹ = 1 - 1/2 =1/2. But according to my previous calculation, A1=1, which is wrong. Hence, the error was in the previous conclusion that A1 = x2. Where is the mistake?

Ah! The mistake was in the final step where I said A1 =x1 + ... =x2. But when I substituted the numbers for k=1, I saw that A1=0 +1=1, but the actual area is 1/2. Hence, there is a miscalculation in the algebra.

Let me re-express the integral.

Earlier, I had:

The integral ∫_{x1}^{x2} [k² +1 -2kx ] dx = [ (k² +1)x -k x² ] from x1 to x2.

For k=1, that integral becomes:

[ (1 +1)x -1*x² ] from x=0 to x=1: [2x -x²]₀¹ = (2*1 -1) -0 =1, which matches. But earlier I thought that the area should be ∫₀¹ (1 -x) dx =1/2. Why the discrepancy?

Because the integrand [k² +1 -2kx ] for k=1 is [2 -2x], and I integrated it from 0 to1, which gives 1. However, the region between y_lower and y_upper is between y_lower =x -1 and y_upper=1 -x. But in reality, in the first quadrant, the lower bound is y=0, not y_lower =x -1. Therefore, my earlier assumption that the area is ∫ [upper - lower] dx from x1 to x2 is only valid when upper and lower bounds are within the rectangle. However, when k=1, for x between x1=0 and x2=1, the lower bound is not y_lower =x -1 but y=0, because y_lower is negative. Therefore, the actual region is between y=0 and y_upper=1 -x, hence area ∫₀¹ (1 -x) dx =1/2, but according to the formula [upper - lower] = [1 -x - (x -1)] = 2 -2x, which is double the correct integrand. Hence, my mistake was in the expression of the integrand.

Wait, how did we get [upper - lower] as k² +1 -2kx? Let's recalculate.

We had upper bound y_upper = (k² +1)/2 -kx, lower bound y_lower =kx - (k² +1)/2. Therefore, upper - lower = [(k² +1)/2 -kx] - [kx - (k² +1)/2] = (k² +1)/2 -kx -kx + (k² +1)/2 = k² +1 -2kx.

But in the case of k=1, this is 2 -2x. However, in reality, the actual upper bound is y_upper = 1 -x and lower bound is y=0, so upper - lower = 1 -x -0 =1 -x. But according to the formula, it is 2 -2x. Hence, discrepancy. Why?

Because for x in [x1, x2], the lower bound is not y_lower =kx - (k² +1)/2, but y=0 when y_lower <0. Hence, my previous assumption that upper - lower =k² +1 -2kx is only valid when y_lower ≥0, i.e., when x ≥x2? Wait, no. Wait, for x between x1 and x2, since x1 is where y_upper meets y=1, and x2 is where y_lower meets y=0.

Wait, for k=1, x1=0, x2=1. For x between 0 and 1, y_lower =x -1 is negative, so effective lower bound is 0. Upper bound is y_upper =1 -x. Therefore, the integrand should be y_upper - 0 =1 -x. However, according to the formula, upper - lower =2 -2x, which is double of 1 -x. Hence, the mistake was in the previous derivation where I assumed that both upper and lower bounds are active, but actually for x between x1 and x2, only the upper bound is active (since lower bound is y=0). Hence, my earlier approach is flawed.

This suggests that my earlier analysis was incorrect. Let me correct it.

The correct region in the first quadrant is:

For x between 0 and x1, the region is between y=0 and y=1 (since both inequalities are automatically satisfied).

For x between x1 and x2, the region is between y=0 and y_upper = (k² +1)/2 -kx.

Wait, why? Because in this range, the lower bound y_lower is negative, so the effective lower bound is y=0. The upper bound is still y_upper = (k² +1)/2 -kx.

Therefore, the area in first quadrant is:

A1 = ∫₀^{x1} 1 dx + ∫_{x1}^{x2} [ (k² +1)/2 -kx - 0 ] dx

= x1 + ∫_{x1}^{x2} [ (k² +1)/2 -kx ] dx

Let me compute this correctly.

For k=1, this becomes:

x1 = 0, x2 =1

Integral from 0 to1 of [ (1 +1)/2 -1*x ] dx = ∫₀¹ [1 -x] dx = [x - x²/2]₀¹ = 1 - 1/2 =1/2, which matches the correct area.

Previously, I incorrectly took upper - lower as k² +1 -2kx, but it should be upper - 0 = (k² +1)/2 -kx. Hence, the previous error.

Therefore, let's recalculate the general case.

A1 = x1 + ∫_{x1}^{x2} [ (k² +1)/2 -kx ] dx

Compute this integral.

First, compute x1 = (k² -1)/(2k)

x2 = (k² +1)/(2k)

Integral ∫ [ (k² +1)/2 -kx ] dx from x1 to x2

Antiderivative is (k² +1)/2 * x - (k/2)x²

Evaluate at x2:

(k² +1)/2 * x2 - (k/2)x2²

At x1:

(k² +1)/2 * x1 - (k/2)x1²

Thus, integral is:

[(k² +1)/2 * x2 - (k/2)x2²] - [(k² +1)/2 * x1 - (k/2)x1²]

Factor out (k/2):

= (k/2)[ ( (k² +1)/k * x2 - x2² ) - ( (k² +1)/k * x1 - x1² ) ]

Wait, perhaps it's easier to compute numerically.

Let me compute the difference:

Let me denote C = (k² +1)/2

Then integral = [C*x2 - (k/2)x2²] - [C*x1 - (k/2)x1²] = C(x2 -x1) - (k/2)(x2² -x1² )

As before, x2 -x1 = [(k² +1)/(2k) - (k² -1)/(2k)] = [ (k² +1 -k² +1 ) / (2k) ] = 2/(2k) = 1/k

Also, x2² -x1² = (x2 -x1)(x2 +x1) = (1/k)*k =1

Therefore, integral = C*(1/k) - (k/2)(1) = [(k² +1)/2]*(1/k) - k/2 = (k² +1)/(2k) - k/2 = [k² +1 -k²]/(2k) = 1/(2k)

Therefore, A1 = x1 + 1/(2k)

But x1 = (k² -1)/(2k)

So A1 = (k² -1)/(2k) + 1/(2k) = (k² -1 +1)/ (2k) = k² / (2k) = k/2

Therefore, the area in the first quadrant is k/2. Hence, total area over all four quadrants is 4*(k/2) = 2k.

Wait, for k=1, this gives total area 2*1=2. The area of the rectangle is 4*1=4. So probability is 2/4=1/2. But earlier analysis for k=1 suggested the region is smaller. Let me check.

For k=1, the total area closer to O than any vertex is 2. The rectangle has area 4. So probability 1/2. Let me check this with the square.

In a square, the region closer to center than to any vertex is a smaller square rotated by 45 degrees, whose vertices are at the midpoints of the sides of the original square. Wait, no. Let me think.

For a square with side length 2 (from -1 to 1), the set of points closer to center than to any vertex is the set of points where distance to center < distance to any vertex. The region is the intersection of four regions, each bounded by perpendicular bisectors. These bisectors are lines like y = -x +1, etc. The intersection of these regions is a square rotated by 45 degrees, with vertices at (0,1), (1,0), (0,-1), (-1,0). Wait, no. Let me compute the intersection of the regions.

For example, the region closer to O than to vertex (1,1) is the region below y = -x +1. Similarly, closer to O than to (-1,1) is y < x +1, closer to O than to (-1,-1) is y > -x -1, closer to O than to (1,-1) is y > x -1. The intersection of these is the region where -x -1 < y < -x +1 and x -1 < y < x +1, which is the region between the lines y = |x| -1 and y = -|x| +1. Wait, this is a square with vertices at (1,0), (0,1), (-1,0), (0,-1). The area of this square is 2 (since from -1 to 1 in x and y, but it's a diamond). The area is 2*1*1 =2? Wait, the area of this diamond is 2*(base*height)/2 = 2*1*1 =2? Wait, actually, the area can be calculated as follows: it's a square with diagonals of length 2 (from -1 to 1 on both axes). The area of a square with diagonal d is d²/2. So here, d=2, area is 4/2=2. Yes, which matches. So the area is 2, and the total area of the square is 4, so probability is 2/4=1/2. Which matches our result of k=1, so A1= k/2 =1/2, total area 2k=2*1=2, probability 2/4=1/2. Okay, this makes sense.

For k=2, total area is 2k=4, while the rectangle has area 4k=8, so probability is 4/8=1/2. Wait, but earlier calculation with k=2 suggested probability 5/8? Wait, no, when I thought I found probability 5/8, that was based on a wrong calculation. Now, with corrected calculation, probability is 2k / (4k) = 1/2. Wait, this suggests the probability is always 1/2, regardless of k? But that seems counterintuitive. For example, if the rectangle is very long and skinny (k very large), intuitively, the region closer to center might be smaller. But according to this, it's always 1/2.

Wait, but let's test with k approaching infinity. Suppose k is very large, so the rectangle is very long horizontally. What is the region closer to O than to any vertex? The vertices are at (k,1), (-k,1), etc. For a point in the middle of the rectangle, say (0,0), it's closer to O. For a point near the center, it's closer to O. For a point near the edge, say (x, y) near (k, 1), it's closer to vertex B(k,1) than to O. But the region closer to O would be the set of points where distance to O < distance to any vertex. For large k, what does this region look like?

If k is large, then for a point (x,y), distance to O is sqrt(x² + y²). Distance to nearest vertex is min(distance to (k,1), distance to (-k,1), distance to (k,-1), distance to (-k,-1)).

For a point in the middle, it's closer to O. For a point near the side, like near (k,0), distance to O is sqrt(k² + 0) =k. Distance to vertex (k,1) is sqrt(0 +1) =1, which is less. So the region closer to O would be bounded away from the sides.

But according to our result, the probability is always 1/2. Let's see: if the area of the region closer to O is 2k, and the total area is 4k, then probability is 1/2. But is that true?

Wait, in our calculation, we derived that the area closer to O is 2k for any k, but let's check with k approaching infinity. If k approaches infinity, then the region closer to O should be a smaller and smaller fraction of the rectangle. But according to our result, it's always 1/2. Contradiction! Hence, there must be a mistake.

Wait, let's re-examine the calculation. For general k, we found that the area in first quadrant is k/2. So total area is 4*(k/2) =2k. Total area of rectangle is 4k. Therefore, probability is 2k / 4k =1/2. But for k approaching infinity, this suggests the probability is 1/2, which doesn't make sense. What's wrong?

Wait, but if k approaches infinity, the rectangle becomes very long and thin. The region closer to O than to any vertex should be a horizontal strip around the center. Let's calculate its area.

For a very large k, the region closer to O than to any vertex is the set of points where distance to O < distance to any vertex. Let's take a point (x,y). Distance to O: sqrt(x² + y²). Distance to nearest vertex: the closest vertex would be either (k,1), (-k,1), (k,-1), or (-k,-1). Let's assume the point is in the right half, x ≥0. The closest vertex would be (k,1) if the point is near the right side, or perhaps another.

For a point near the middle, say x=0, y=0, it's closer to O. For a point near (k,1), it's closer to that vertex. The boundary between being closer to O or to (k,1) is the perpendicular bisector between O and (k,1). Which we already derived as y = kx + (k² +1)/2. Wait, no, earlier we had for the bisector between O and A(-k,1) was y = kx + (k² +1)/2. But for the bisector between O and B(k,1), it's y = -kx + (k² +1)/2.

Wait, in the first quadrant, the bisector between O and B(k,1) is y = -kx + (k² +1)/2. For large k, this line is approximately y = -kx + (k²)/2, which simplifies to y ≈ -kx + k²/2. For x =0, y≈ k²/2, which is way outside the rectangle (which has y up to 1). For y=1, solving for x: 1 = -kx +k²/2 => kx =k²/2 -1 => x ≈k/2 - 1/k. For large k, x≈k/2. So the bisector intersects the top edge y=1 at x≈k/2. Hence, in the first quadrant, the region closer to O than to B is to the left of this line, i.e., x < (k²/2 -1)/k ≈k/2.

Similarly, the region closer to O than to C(k,-1) in first quadrant is above the line y = kx - (k² +1)/2. For large k, this is approximately y ≈kx -k²/2. Setting y=0: 0 ≈kx -k²/2 => x≈k/2. Hence, the region closer to O than to C is above y ≈kx -k²/2, which for x near k/2 is y≈0.

Thus, the region closer to O than to any vertex in first quadrant is approximately the region where x <k/2 and y > something, but for large k, this region is a rectangle from x=0 to x=k/2, y= -1 to y=1? No, because it's bounded by the bisectors.

Wait, actually, for large k, the region closer to O than to any vertex is approximately the set of points where |x| <k/2 and |y| < something, but I need to think carefully.

Actually, according to our previous calculation, the area closer to O is 2k, which for large k would be a strip from x=-k/2 to x=k/2, y=... but the height?

Wait, the area is 2k, while the total area is 4k. Hence, the region is 1/2 of the rectangle. For large k, the region closer to O is a central stripe of width k (from x=-k/2 to x=k/2), but height? If the width is k, then the height would have to be 2 to maintain area 2k. But the rectangle has height 2, so it would be the central stripe from x=-k/2 to x=k/2, full height. But this is impossible because for x=k/2, the distance to O is sqrt( (k/2)^2 + y^2 ), while distance to vertex B(k,1) is sqrt( (k/2)^2 + (1 - y)^2 ). Setting them equal: (k/2)^2 + y^2 = (k/2)^2 + (1 - y)^2 => y² = (1 - y)^2 => y² =1 -2y + y² => 0 =1 -2y => y=1/2. Hence, the boundary is at y=1/2 for x=k/2. Similarly on other sides. Hence, the region closer to O is a central square from x=-k/2 to x=k/2 and y=-1/2 to y=1/2? Wait, if that's the case, the area would be k * 1 =k, which is half of the total area 2k? No, total area is 4k. Wait, if it's a central rectangle of width k and height 1, area is k*1=k, which is k/4k =1/4. Doesn't match.

Wait, this suggests that my previous conclusion that the area is 2k is incorrect for large k. Hence, there is a consistent error in my calculation.

Let me re-examine the calculation for general k. We had:

A1 = k/2, total area = 2k, total rectangle area =4k, probability=1/2. But for k approaching infinity, this suggests the probability is always 1/2, which seems incorrect.

However, for k=1, it's correct (probability 1/2). For k approaching infinity, let me think of another approach.

Consider the rectangle with AB/AD =k, very large k. Let's assume k is very large, so the rectangle is long horizontally. The set of points closer to O than to any vertex is the set of points where distance to O < distance to any vertex. For a point (x,y), this means:

sqrt(x² + y²) < sqrt( (x -k)^2 + (y -1)^2 )

sqrt(x² + y²) < sqrt( (x +k)^2 + (y -1)^2 )

sqrt(x² + y²) < sqrt( (x -k)^2 + (y +1)^2 )

sqrt(x² + y²) < sqrt( (x +k)^2 + (y +1)^2 )

These are the four inequalities. Let's focus on the first one, which is for vertex B(k,1). Squaring:

x² + y² < (x -k)^2 + (y -1)^2 => 0 < -2kx + k² -2y +1 => 2kx + 2y < k² +1.

For large k, this becomes 2kx < k² => x < k/2.

Similarly, for the other vertices, we get similar bounds. For vertex C(k,-1), similar to B, we get 2kx + 2y < k² +1? No, let's check.

For vertex C(k,-1):

sqrt(x² + y²) < sqrt((x -k)^2 + (y +1)^2 )

Squaring: x² + y² < (x -k)^2 + (y +1)^2 => 0 < -2kx + k² + 2y +1 => 2kx - 2y < -k² -1 => 2kx - 2y < -k² -1.

For large k, this is 2kx ≈ -k² => x ≈ -k/2.

Similarly, for vertices A and D, we get x > -k/2 and x <k/2. Therefore, the region closer to O than to any vertex is the set of points where |x| <k/2 and |y| < something?

Wait, but what about y? Let's take the first inequality 2kx + 2y < k² +1. For x <k/2, this is satisfied for any y? No. For example, if x=0, it becomes 2y <k² +1, which is always true since y ≤1 and k is large. Similarly, for x=k/2, it becomes k² + 2y <k² +1 => 2y <1 => y <1/2. So the region closer to O than to B is x <k/2 and y < (k² +1 -2kx)/2. For large k, this is approximately y < (k²)/2 -kx. For x near k/2, y < (k²)/2 -k*(k/2) = (k²)/2 -k²/2 = 0. So the boundary is y=0 at x=k/2. Similarly on the other side, for x=-k/2, the boundary is y=1.

Wait, this suggests that the region closer to O than to any vertex is the central rectangle from x=-k/2 to x=k/2 and y from -1/2 to 1/2? No, because for x=0, the constraint is y < (k² +1)/2, which is always true since y ≤1. Hence, the region is x between -k/2 and k/2, and y between -1 and 1? But this would make the area k * 2 = 2k, which matches our previous calculation. But this would mean the probability is 2k / 4k =1/2, which seems to suggest that for any k, the region is the central rectangle of width k and height 2, which is half the area of the rectangle. But this seems to contradict the intuition for large k, but maybe it's correct?

Wait, if the region is the central rectangle from x=-k/2 to x=k/2 and y from -1 to 1, then for any k, this is indeed half the area of the rectangle (since total area is 2k*2 =4k, and central area is k*2=2k). But why is this the case?

Because the boundaries of the region closer to O are at x=±k/2 and y=±1/2? No, for x=0, the constraint is y < (k² +1)/2, which is always true. But for x=k/2, the constraint is y < 1/2. Similarly, for x=-k/2, y < -1/2. Hence, the region closer to O than to any vertex is the set of points where |x| <k/2 and |y| < 1/2? Wait, no, because for x=0, y can go up to 1. Wait, no, for x=0, the constraint from vertex B is 2*0 + 2y < k² +1 => y < (k² +1)/2, which is always true for y ≤1. Similarly for other vertices. Hence, the only constraints are from the vertices at the ends. For example, for x=k/2, the constraint from vertex B is y < 1/2, and from vertex C is y > -1/2. Hence, the region is x between -k/2 and k/2, and y between -1/2 and 1/2? No, because for x=0, y can be from -1 to 1. Wait, no. Let me think.

For a point (x, y) to be closer to O than to any vertex, it must satisfy all four inequalities. For example, the inequality from vertex B: 2kx + 2y < k² +1, from vertex C: 2kx - 2y < -k² -1, from vertex A: -2kx + 2y < -k² -1, and from vertex D: -2kx - 2y < -k² -1. Let me rewrite these inequalities:

1. 2kx + 2y < k² +1

2. 2kx - 2y < -k² -1

3. -2kx + 2y < -k² -1

4. -2kx - 2y < -k² -1

Let me analyze these inequalities. For large k, let's divide by 2k:

1. x + y/k < k/2 + 1/(2k)

Since y/k is small, this approximates to x < k/2

2. x - y/k < -k/2 - 1/(2k) ≈ -k/2

Which implies x < -k/2 - y/k - 1/(2k) ≈ -k/2 (since y/k is small)

Similarly, the other inequalities:

3. -x + y/k < -k/2 - 1/(2k) ≈ -k/2 => -x < -k/2 => x > k/2

4. -x - y/k < -k/2 - 1/(2k) ≈ -k/2 => -x < -k/2 => x > k/2

Wait, this seems conflicting. Wait, let me do it properly.

Let me take inequality 2: 2kx - 2y < -k² -1 => 2kx < 2y -k² -1. For large k, this implies 2kx is negative, so x is negative. Let me divide by 2k: x - (y)/k < -k/2 - 1/(2k). For large k, x < -k/2 - (small term), so x < -k/2.

Similarly, inequality 3: -2kx + 2y < -k² -1 => -2kx < -k² -1 -2y => 2kx > k² +1 +2y. For large k, this implies x > k/2 + (small term), so x > k/2.

Therefore, combining inequalities 2 and 3, we have x < -k/2 and x > k/2, which is impossible. This suggests a contradiction, which means my earlier conclusion was wrong.

Wait, this indicates that for large k, there is no region satisfying all inequalities, which is impossible because the center (0,0) should satisfy them. Hence, there must be a mistake in interpreting the inequalities.

Let me re-derive the four inequalities for distance to O < distance to each vertex.

For vertex B(k,1): sqrt(x² + y²) < sqrt((x -k)^2 + (y -1)^2)

Squared: x² + y² < x² - 2kx + k² + y² - 2y + 1

Simplifies to 0 < -2kx + k² - 2y + 1 => 2kx + 2y < k² +1 (1)

Similarly for vertex C(k,-1): sqrt(x² + y²) < sqrt((x -k)^2 + (y +1)^2)

Squared: x² + y² < x² - 2kx + k² + y² + 2y + 1

Simplifies to 0 < -2kx + k² + 2y +1 => 2kx - 2y < -k² -1 (2)

For vertex A(-k,1): sqrt(x² + y²) < sqrt((x +k)^2 + (y -1)^2)

Squared: x² + y² < x² + 2kx + k² + y² - 2y + 1

Simplifies to 0 < 2kx + k² - 2y +1 => -2kx - 2y < -k² -1 => 2kx + 2y > k² +1 (3)

For vertex D(-k,-1): sqrt(x² + y²) < sqrt((x +k)^2 + (y +1)^2)

Squared: x² + y² < x² + 2kx + k² + y² + 2y + 1

Simplifies to 0 < 2kx + k² + 2y +1 => -2kx - 2y < -k² -1 => 2kx + 2y > -k² -1 (4)

Wait, this is different from before. Earlier I had inequality (3) as 2kx + 2y > k² +1 and (4) as 2kx + 2y > -k² -1.

So for large k, the inequalities are:

1. 2kx + 2y < k² +1

2. 2kx - 2y < -k² -1

3. 2kx + 2y > k² +1

4. 2kx + 2y > -k² -1

Wait, but this is a contradiction: (1) says 2kx + 2y < k² +1 and (3) says 2kx + 2y > k² +1, which can only be true if 2kx + 2y = k² +1. Similarly for others. This suggests that the only points satisfying all inequalities are on the boundaries, which is impossible. Clearly, there is a mistake here.

Wait, let me re-derivate the inequalities for vertex A.

For vertex A(-k,1):

Distance to O: sqrt(x² + y²)

Distance to A: sqrt((x + k)^2 + (y - 1)^2)

Setting distance to O < distance to A:

x² + y² < (x + k)^2 + (y - 1)^2

Expanding right side: x² + 2kx + k² + y² - 2y + 1

Subtract left side: 0 < 2kx + k² - 2y + 1

Which gives 2kx - 2y + k² + 1 > 0 => 2kx - 2y > -k² -1

Wait, earlier I had 2kx + 2y > k² +1, which was incorrect. Let me correct this.

Original inequality: 0 < 2kx + k² - 2y + 1 => 2kx - 2y > -k² -1. So inequality (3) is 2kx - 2y > -k² -1.

Similarly, for vertex D(-k,-1):

Distance to O < distance to D:

x² + y² < (x + k)^2 + (y + 1)^2

Expanding: x² + y² < x² + 2kx + k² + y² + 2y + 1

Subtract: 0 < 2kx + k² + 2y + 1 => 2kx + 2y > -k² -1. So inequality (4) is 2kx + 2y > -k² -1.

Therefore, the four inequalities are:

1. 2kx + 2y < k² +1

2. 2kx - 2y < -k² -1

3. 2kx - 2y > -k² -1

4. 2kx + 2y > -k² -1

Wait, now this makes more sense. Let me re-express:

Inequality (1): 2kx + 2y < k² +1

Inequality (2): 2kx - 2y < -k² -1

Inequality (3): 2kx - 2y > -k² -1

Inequality (4): 2kx + 2y > -k² -1

Wait, but (2) and (3) are inequalities in the same expression: 2kx - 2y is between -k² -1 and -k² -1? No, (2) is 2kx - 2y < -k² -1 and (3) is 2kx - 2y > -k² -1? That would mean 2kx - 2y = -k² -1, which is a line, not a region. This suggests that there is a mistake in the derivation.

Wait, let me re-derive inequality (3). For vertex A(-k,1):

0 < 2kx + k² - 2y + 1 => 2kx + k² - 2y + 1 > 0 => 2kx - 2y > -k² -1. So this is inequality (3): 2kx - 2y > -k² -1

Similarly, for vertex D(-k,-1), we had 2kx + 2y > -k² -1.

Therefore, the four inequalities are:

1. 2kx + 2y < k² +1

2. 2kx - 2y < -k² -1

3. 2kx - 2y > -k² -1

4. 2kx + 2y > -k² -1

Wait, this seems contradictory. inequalities (2) and (3) are 2kx - 2y < -k² -1 and 2kx - 2y > -k² -1, which can only be true if 2kx - 2y = -k² -1, which is a line. Similarly for (1) and (4): 2kx + 2y < k² +1 and 2kx + 2y > -k² -1. This suggests that the region is between two lines for each pair, but how?

Actually, it seems that the correct inequalities are:

For each vertex, the inequality is a half-plane. For example, for vertex B: 2kx + 2y < k² +1, for vertex A: 2kx - 2y > -k² -1, for vertex C: 2kx - 2y < -k² -1, and for vertex D: 2kx + 2y > -k² -1. Wait, but this would mean:

- From vertex B: 2kx + 2y < k² +1

- From vertex A: 2kx - 2y > -k² -1

- From vertex C: 2kx - 2y < -k² -1

- From vertex D: 2kx + 2y > -k² -1

This is a contradiction unless the inequalities for A and C are swapped. Wait, let me recheck.

For vertex C(k, -1), the inequality was 2kx - 2y < -k² -1. For vertex A(-k,1), the inequality was 2kx - 2y > -k² -1. So these are two inequalities: 2kx - 2y > -k² -1 and 2kx - 2y < -k² -1, which is impossible unless 2kx - 2y = -k² -1. This suggests that there is an error in the derivation.

Wait, let me re-derive the inequality for vertex A(-k,1):

Original inequality: 0 < 2kx + k² - 2y +1 => 2kx - 2y + k² +1 > 0 => 2kx - 2y > -k² -1. Correct.

For vertex C(k, -1): original inequality was 0 < -2kx + k² + 2y +1 => -2kx + 2y > -k² -1 => 2kx - 2y < k² +1. Wait, wait, let me redo this.

For vertex C(k, -1):

0 < -2kx + k² + 2y +1 => -2kx + 2y > -k² -1 => 2kx - 2y < k² +1. Wait, this is different from what I had before. Previously, I had for vertex C: 2kx - 2y < -k² -1, which was incorrect. Let me re-derive:

For vertex C(k, -1):

Distance to O < distance to C:

x² + y² < (x -k)^2 + (y + 1)^2

Expand: x² + y² < x² - 2kx + k² + y² + 2y + 1

Simplify: 0 < -2kx + k² + 2y + 1 => -2kx + 2y > -k² -1 => 2kx - 2y < k² +1.

Yes, that's correct. So for vertex C, the inequality is 2kx - 2y < k² +1. Similarly, for vertex A, it's 2kx - 2y > -k² -1. Hence, the inequalities for the two vertices A and C are:

-2kx - 2y > -k² -1 (from A) and 2kx - 2y < k² +1 (from C). Wait, no, for vertex A, it's 2kx - 2y > -k² -1, and for vertex C, it's 2kx - 2y < k² +1.

Similarly, for vertex B, it's 2kx + 2y < k² +1, and for vertex D, it's 2kx + 2y > -k² -1.

Therefore, the four inequalities are:

1. 2kx + 2y < k² +1 (from B)

2. 2kx + 2y > -k² -1 (from D)

3. 2kx - 2y > -k² -1 (from A)

4. 2kx - 2y < k² +1 (from C)

Therefore, combining these, the region is defined by:

- For 2kx + 2y: between -k² -1 and k² +1

- For 2kx - 2y: between -k² -1 and k² +1

Which, for large k, translates to:

- 2kx + 2y is between -k² -1 and k² +1

- 2kx - 2y is between -k² -1 and k² +1

Dividing by 2k:

For the first pair: x + y/k is between -k/2 - 1/(2k) and k/2 + 1/(2k)

For the second pair: x - y/k is between -k/2 - 1/(2k) and k/2 + 1/(2k)

For large k, this approximates to:

x is between -k/2 and k/2, and y is between -k/2 and k/2. But since the original rectangle has y between -1 and 1, this suggests that for large k, the region is x between -k/2 and k/2, y between -1 and 1, but this contradicts the earlier conclusion.

Wait, no. Let me think of it this way: the inequalities 2kx + 2y < k² +1 and 2kx + 2y > -k² -1 can be rewritten as:

- (k² +1)/2 < kx + y < (k² +1)/2

Similarly for the other pair:

- (k² +1)/2 < kx - y < (k² +1)/2

For large k, this becomes:

- (k²)/2 < kx + y < k²/2

- (k²)/2 < kx - y < k²/2

Let me solve these inequalities for y.

From the first pair:

- k²/2 < kx + y < k²/2 => -k²/2 - kx < y < k²/2 - kx

From the second pair:

- k²/2 < kx - y < k²/2 => -k²/2 - kx < -y < k²/2 - kx => -k²/2 + kx < y < k²/2 + kx

Combining both:

y must satisfy both:

- k²/2 - kx < y < k²/2 - kx

and

- k²/2 + kx < y < k²/2 + kx

For large k, the intersection of these regions is y between max(-k²/2 - kx, -k²/2 + kx) and min(k²/2 - kx, k²/2 + kx).

Since k is positive, -k²/2 - kx < -k²/2 + kx if x > 0, and vice versa.

For x > 0, the lower bound is -k²/2 + kx and the upper bound is k²/2 - kx. For y to satisfy both, we need:

- k²/2 + kx < y < k²/2 - kx

Which requires that -k²/2 + kx < k²/2 - kx => -k²/2 - k²/2 < -kx -kx => -k² < -2kx => k² > 2kx => x < k/2

Similarly for x < 0, the lower bound is -k²/2 - kx and the upper bound is k²/2 + kx, and this requires x > -k/2.

Therefore, the region is x between -k/2 and k/2, and for each x, y is between -k²/2 + kx and k²/2 - kx. For large k, this is approximately y between -k²/2 + kx and k²/2 - kx. But since the rectangle has y between -1 and 1, the actual bounds are y between -1 and 1, but only where -k²/2 + kx < 1 and k²/2 - kx > -1.

For x between -k/2 and k/2, let's find the range of y:

For the upper bound: k²/2 - kx > -1 => k²/2 > kx -1. For x < k/2, kx < k²/2, so kx -1 < k²/2 -1. Hence, k²/2 - kx > k²/2 - (k²/2 -1) =1. So the upper bound is always >1, but since y ≤1, it is effectively 1.

Similarly, the lower bound: -k²/2 + kx > -1 => kx > -1 + k²/2. For x > -k/2, kx > -k²/2. So -k²/2 + kx > -k²/2 + (-k²/2) = -k², but this is always greater than -1 for large k. Hence, the lower bound is effectively -1.

Therefore, for x between -k/2 and k/2, the region is y between -1 and 1, which is the entire central strip. Hence, the area is 2k (width from -k/2 to k/2) * 2 (height from -1 to 1) = 4k._WAIT, but this is the entire area of the rectangle! This is impossible. Clearly, there is a mistake.

Wait, no. The region defined by the inequalities is the entire rectangle? That can't be. For example, take a point near the corner (k,1). It should not be in the region. But according to this, it is. Hence, there is a fundamental error in the derivation.

This suggests that my approach is incorrect, and I need to find another way.

Let me return to the original correct calculation for k=1, where the area is 2, and the probability is 1/2. For k=1, it's correct. For k=2, if I calculate the area using the correct method, what do I get?

For k=2, the area in first quadrant is k/2 =1. Total area is 2k =4. Total rectangle area is 4k=8. Probability is 4/8=1/2. But let's check by integrating.

For k=2, the region in first quadrant is the set of points where distance to O < distance to B and C.

We had earlier, for k=2, the area in first quadrant is 1. Let's verify by integrating.

The region in first quadrant is:

For x from 0 to x1 = (k² -1)/(2k) = (4 -1)/4 = 3/4, y from 0 to 1.

For x from 3/4 to x2 = (k² +1)/(2k) = 5/4, y from 0 to y_upper = (k² +1)/2 -kx = (5/2) - 2x.

So area is:

A1 = ∫₀³/4 1 dx + ∫³/4^5/4 (5/2 - 2x) dx

First integral: 3/4

Second integral: [ (5/2)x - x² ] from 3/4 to 5/4

At 5/4: (5/2)*(5/4) - (5/4)^2 = 25/8 - 25/16 = 25/16

At 3/4: (5/2)*(3/4) - (3/4)^2 = 15/8 - 9/16 = 30/16 - 9/16 = 21/16

Difference: 25/16 - 21/16 = 4/16 = 1/4

Total A1 = 3/4 + 1/4 = 1. Which matches k/2 = 2/2 =1. Hence, for k=2, the area in first quadrant is 1, total area 4, probability 4/8=1/2.

But for k=2, the region is the central rectangle from x=-1 to x=1 and y=-1 to 1? No, because for k=2, the rectangle is from x=-2 to 2, y=-1 to 1. The area closer to O is 4, which is half of 8. This suggests that for any k, the region is half the area of the rectangle. This is surprising but mathematically consistent with the calculations for k=1 and k=2. Hence, perhaps the probability is always 1/2, regardless of k.

But why? Intuitively, for a rectangle, the set of points closer to the center than to any vertex is always half the area? This seems counterintuitive for very long rectangles, but mathematically, it seems to hold.

Let me think about symmetry. The set of points closer to O than to any vertex is symmetric with respect to the center. If we consider the rectangle and its complement, perhaps there is some symmetry.

Alternatively, consider that the region closer to O than to any vertex is the intersection of four regions, each defined by being closer to O than to a vertex. For each vertex, the region closer to O is a half-plane, and the intersection is a convex region. But why its area is always half the rectangle?

Perhaps there is a general theorem here, but I'm not sure. However, given that it works for k=1 and k=2, and the calculations are consistent, I might have to accept that the probability is 1/2 for any k.

But the problem states AB/AD =k, and asks for the probability. If the answer is always 1/2, then the probability is 1/2, and the answer is \boxed{\dfrac{1}{2}}.

But I need to confirm with another value. Let's take k=0.5, which is a rectangle width 1 and height 2, so it's taller than it is wide.

For k=0.5, the rectangle is from x=-0.5 to 0.5, y=-1 to 1. The area is 1*2=2. According to our formula, the region closer to O is 2k=1, hence probability 1/2. Let's verify.

For k=0.5, the region closer to O than to any vertex is the set of points where distance to O < distance to any vertex.

For example, take a point (0,0). It's closer to O. Take a point (0,1). Distance to O is 1, distance to A(-0.5,1) is sqrt((0 +0.5)^2 + (1 -1)^2) = 0.5 < 1. So (0,1) is closer to A, hence not in the region. Hence, the region is not the entire rectangle. But according to our calculation, the area is 1, which is half of 2. Let's calculate it.

For k=0.5, the area in first quadrant is k/2 =0.5/2=0.25. Total area 4*0.25=1. Probability 1/2.

To verify, let's calculate the area in first quadrant.

For k=0.5, x1 = (k² -1)/(2k) = (0.25 -1)/1 = -0.75. But since we are in first quadrant x ≥0, x1 is negative, so the first part is from x=0 to x=x1 (which is 0 to -0.75, but x ≥0, so no area. Hence, A1 = ∫_{0}^{x2} [ (k² +1)/2 -kx ] dx, where x2 = (k² +1)/(2k) = (0.25 +1)/1 = 1.25. But the rectangle in first quadrant is x from 0 to 0.5, y from 0 to 1. Hence, the upper limit is min(x2, 0.5) = 0.5.

Therefore, A1 = ∫₀^0.5 [ (0.25 +1)/2 - 0.5x ] dx = ∫₀^0.5 [ 1.25/2 - 0.5x ] dx = ∫₀^0.5 (0.625 - 0.5x) dx = [0.625x - 0.25x²] from 0 to 0.5 = 0.625*0.5 - 0.25*(0.25) = 0.3125 - 0.0625 = 0.25, which matches k/2 =0.5/2=0.25. Hence, total area is 4*0.25=1, probability 1/2. This seems to hold.

Thus, despite the initial intuition, it appears the probability is always 1/2, regardless of k. This is surprising, but the calculations for specific cases support it. Hence, the answer is 1/2.

**Final Answer**
\boxed{\dfrac{1}{2}}

To determine the probability that a randomly chosen point inside rectangle $ABCD$ is closer to the center $O$ than to any of the four vertices, we analyze the geometric condition that defines this region.

---

### Setup and Geometry

- **Rectangle Setup**: Place rectangle $ABCD$ with center at the origin $O(0,0)$. Let the rectangle have sides of length $2k$ (horizontal) and $2$ (vertical), resulting in:
  - Vertices: $A = (-k, 1)$, $B = (k, 1)$, $C = (k, -1)$, $D = (-k, -1)$
  - Total area of the rectangle: $A = 2k \cdot 2 = 4k$

- We are to find the area of the region within the rectangle where the distance from a point $(x, y)$ to the center $O$ is *less than* the distance from that point to any of the four vertices $A$, $B$, $C$, and $D$.

This condition defines a set that is the **intersection** of four regions, each region being the set of points closer to $O$ than to a specific vertex. So the desired region is the **intersection** of four half-planes defined by perpendicular bisectors between $O$ and each vertex.

---

### Key Insight: Symmetry and Area

Due to the symmetry of the rectangle about the center and the equal contribution of each quadrant, we can compute the area of the region in one quadrant (say, the first quadrant where $x \geq 0$, $y \geq 0$) and multiply by 4.

Using symmetry and analytical geometry, we determine that in each quadrant, the region of points closer to the center than to any vertex is a shape bounded by two perpendicular bisectors. Integrating this region in each quadrant leads to the calculation of the total area of the region closer to the center than any vertex.

After detailed calculation (and verified for several cases including $k = 1, 2, 0.5$), it turns out that the **total area of this region is**:

$$
\text{Desired Region Area} = 2k
$$

Since the **total area of the rectangle** is $4k$, the probability that a randomly chosen point is closer to the center than to any vertex is:

$$
P = \frac{\text{Desired Region Area}}{\text{Total Area}} = \frac{2k}{4k} = \frac{1}{2}
$$

---

### Final Conclusion

Regardless of the value of $k$, the probability that a point randomly chosen from the interior of rectangle $ABCD$ is closer to its center than to any of its vertices is:

$$
\boxed{\dfrac{1}{2}}
$$

[Assistant]
Explanation:
Okay, so I need to find the probability that a randomly chosen point inside rectangle ABCD is closer to the center O than to any of the four vertices. The rectangle has sides in the ratio AB/AD = k. Let me try to visualize this. The rectangle is centered at O, which is the intersection of the diagonals. The vertices are A, B, C, D. The Asymptote code draws a rectangle with coordinates from (-250,100) to (250,-100), so maybe it's symmetric? But the actual coordinates might not matter because we can consider a general rectangle with AB = k*AD.

First, I need to understand what region inside the rectangle is closer to O than to any of the vertices. So for each vertex, I can think about the set of points that are closer to O than to that vertex. Then, the region we want is the intersection of all these sets, right? Because we want points that are closer to O than to any of the four vertices.

Alternatively, maybe it's easier to think about the Voronoi diagram for the center and the four vertices? But I'm not sure. Maybe I can find the boundaries where the distance to O equals the distance to a vertex, and then figure out the region where the distance to O is less than to any vertex.

Let me recall that the set of points equidistant to two points is the perpendicular bisector of the segment joining them. So, if I want points closer to O than to vertex A, I can find the perpendicular bisector between O and A, and then take the region closer to O.

But since there are four vertices, I need to find regions closer to O than to any of A, B, C, D. So the intersection of the four regions defined by being closer to O than to each vertex.

Let me set up coordinates to make this easier. Let me assume that the rectangle has length AB = k * AD. Let me denote AD = a, so AB = k*a. Let me consider coordinates. Let me place the rectangle with its center at the origin O(0,0). Then, the coordinates of the vertices can be determined.

Since the rectangle has center at O, the coordinates would be:

A is (- (AB/2), (AD/2)) ?

Wait, actually, the rectangle's sides are AB and AD. Let me think.

Let me define coordinate system such that O is (0,0). Let me denote the rectangle with length AB = k and AD = 1 for simplicity? Wait, but AB and AD are adjacent sides. Wait, AB is the length, AD is the width? Let me think.

If AB/AD = k, then AB = k*AD. So let me denote the rectangle with sides of length 2a and 2b, so that the center O is at (0,0), and the vertices are at (-a, -b), (-a, b), (a, b), (a, -b). Wait, then AB would be the horizontal side from (-a, -b) to (a, -b)? Wait no, AB is adjacent to A and B. Let me check the Asymptote code. The Asymptote code has A at (-250,100), B at (250,100), C at (250,-100), D at (-250,-100). So AB is horizontal, AD is vertical. So AB is the top side, AD is the left side. So if AB is horizontal, then AB is length 2*250? But in the Asymptote code it's from -250 to 250 on x-axis, so AB is length 500 and AD is length 200? But since AB/AD = k, so k = 500 / 200 = 2.5? So in the general case, AB = 2a and AD = 2b, so that AB / AD = (2a)/(2b) = a/b = k. So a = k*b. Therefore, if I take the rectangle with length AB = 2a and AD = 2b, then the coordinates of the vertices are:

A(-a, b), B(a, b), C(a, -b), D(-a, -b). Wait, but in the Asymptote code, A is (-250, 100), B is (250, 100), so if AB is horizontal, that's correct. So yes, if AB is horizontal, then A is (-a, b) and B is (a, b), etc.

But since AB/AD = k, and AB is horizontal length 2a, AD is vertical length 2b, so AB / AD = (2a)/(2b) = a/b = k. Therefore, a = k*b.

Therefore, the rectangle can be parametrized with variables a and b, where a = k*b. So maybe to make it even simpler, I can set b = 1, then a = k. Therefore, the rectangle has width 2a = 2k and height 2b = 2. Then the coordinates of the vertices are A(-k, 1), B(k, 1), C(k, -1), D(-k, -1). The center O is at (0,0). That might be easier for calculations.

Alternatively, maybe using coordinates with sides of length 2 and 2k? Wait, but the ratio AB/AD is k, so AB is length 2k, AD is length 2? Then AB/AD = k. Maybe that's another way. But maybe it's better to normalize.

Let me just use coordinates where the rectangle has length 2 and width 2k? Wait, but then AB is length 2, AD is length 2k. Then AB/AD = 1/k. Hmm, but the problem says AB/AD = k. So perhaps if AB is length 2k, AD is length 2. Then AB/AD = k. So yes, that would work.

So let me define the rectangle with vertices at (-k, -1), (-k, 1), (k, 1), (k, -1). Wait, that would make AB from (-k,1) to (k,1), so AB length is 2k. AD from (-k,1) to (-k,-1), so AD length is 2. So AB/AD = (2k)/2 = k. Perfect. Then center O is (0,0). So we can work with coordinates from x=-k to x=k, y=-1 to y=1.

But maybe I should use variables. Let me think. Let me consider general rectangle with sides of length 2a and 2b, so AB = 2a, AD = 2b, then AB/AD = a/b = k. Therefore, a = k*b.

But perhaps to make it even more general, maybe I can set coordinates with O at (0,0), and the rectangle extending from (-a, -b) to (a, b). So AB is horizontal from (-a, b) to (a, b), so AB = 2a. AD is vertical from (-a, b) to (-a, -b), so AD = 2b. Therefore, AB/AD = a/b = k. So a = k*b.

But maybe I can set b = 1, then a = k. So coordinates are from (-k, -1) to (k, 1). Therefore, center at (0,0). So let me use this coordinate system.

So the rectangle is [ -k, k ] in x-axis and [ -1, 1 ] in y-axis. The center is O(0,0). Now, the problem is to find the probability that a random point inside this rectangle is closer to O than to any of the four vertices. Since the area of the rectangle is (2k)(2) = 4k. So the probability will be [Area of desired region] / (4k).

Therefore, my task reduces to finding the area of the region within the rectangle where for any point (x,y), distance to O is less than distance to A, B, C, D.

Alternatively, the region where distance to O is less than the minimum distance to the four vertices.

But since the rectangle is symmetric, maybe I can focus on one quadrant and multiply by 4?

Yes! Because of symmetry, the region should be symmetric across both axes, so I can compute the area in one quadrant and multiply by 4. Let me think.

Let me consider the first quadrant where x ≥ 0 and y ≥ 0. In this quadrant, the vertices are B(k, 1) and A(-k, 1) is in the second quadrant, but in the first quadrant, the closest vertex is B(k,1) and D(-k,-1) is not in the quadrant. Wait, but in the first quadrant, the vertices are B(k,1) and maybe also... Wait, the rectangle has four vertices: A(-k,1), B(k,1), C(k,-1), D(-k,-1). So in the first quadrant, the only vertex is B(k,1). So in the first quadrant, the region closer to O than to any vertex would be the region closer to O than to B, and also closer to O than to A, C, D? Wait, but in the first quadrant, points can't be closer to A, C, or D than to O? Wait, actually, points in the first quadrant can be closer to A, which is (-k,1). Wait, but A is in the second quadrant. Wait, but a point in the first quadrant might be closer to A than to O?

Wait, for example, take the point (k/2, 1). The distance to O is sqrt( (k/2)^2 + 1^2 ). The distance to A(-k,1) is sqrt( (k/2 + k)^2 + (1 - 1)^2 ) = sqrt( (3k/2)^2 ) = 3k/2. Compare to distance to O: sqrt( (k^2)/4 + 1 ). For k > 0, this is less than 3k/2, since sqrt(k²/4 + 1) < sqrt(k²/4 + k²/4) = sqrt(k²/2) = k / sqrt(2) ≈ 0.707k, which is less than 1.5k. Wait, so for k > 0, this point is closer to O than to A? Wait, but if k is very large, then maybe not?

Wait, suppose k is very large. Let me take k approaching infinity. Then, the distance to O for point (k/2,1) is sqrt( (k/2)^2 + 1 ) ≈ k/2. The distance to A(-k,1) is sqrt( (3k/2)^2 + 0 ) = 3k/2. So yes, distance to O is half of distance to A. So even for large k, (k/2,1) is closer to O. So maybe points in the first quadrant are always closer to O than to A? Wait, but is that true?

Wait, suppose we take a point very close to A, but in the first quadrant? Wait, but A is at (-k,1). So the first quadrant is x ≥ 0, y ≥ 0. To be close to A, a point would have to be near (-k,1), but that's in the second quadrant. So in the first quadrant, points cannot be close to A. Similarly, points in the first quadrant can't be close to D or C? Because D is (-k,-1) and C is (k,-1). So in the first quadrant, the only vertex that is "nearby" is B(k,1). So maybe in the first quadrant, the only vertex we need to compare with O is B(k,1). Therefore, the region where the point is closer to O than to any vertex is the region in the first quadrant where it's closer to O than to B, and also closer to O than to A, C, D? But since points in the first quadrant are not close to A, C, D, except for B?

Wait, maybe not. For example, take a point near the center. It's closer to O than to any vertex. Take a point near the side of the rectangle. Let me check. Suppose the point is (x,y) in the first quadrant. What's the distance to O: sqrt(x² + y²). What's the distance to B(k,1): sqrt( (x - k)² + (y - 1)² ). The distance to A(-k,1) is sqrt( (x + k)² + (y - 1)² ). Similarly for others. But since the point is in first quadrant, x ≥ 0, y ≥ 0. So the distances to A, C, D would be:

To A(-k,1): distance squared is (x + k)^2 + (y - 1)^2.

To C(k,-1): distance squared is (x - k)^2 + (y + 1)^2.

To D(-k,-1): distance squared is (x + k)^2 + (y + 1)^2.

So comparing to O: distance squared is x² + y².

So for the point to be closer to O than to A, we need:

x² + y² < (x + k)^2 + (y - 1)^2

Similarly for other vertices. Let me check this inequality:

x² + y² < (x + k)^2 + (y - 1)^2

Expanding the right-hand side: x² + 2kx + k² + y² - 2y + 1

Subtract left-hand side:

0 < 2kx + k² - 2y + 1

Which simplifies to:

2kx - 2y + (k² + 1) > 0

Similarly, for distance to O being less than distance to C:

x² + y² < (x - k)^2 + (y + 1)^2

Expanding: x² - 2kx + k² + y² + 2y + 1

Subtract left-hand side:

0 < -2kx + k² + 2y + 1

=> -2kx + 2y + (k² + 1) > 0

Similarly, for distance to D:

x² + y² < (x + k)^2 + (y + 1)^2

Right-hand side: x² + 2kx + k² + y² + 2y + 1

Subtract left-hand side:

0 < 2kx + 2y + k² + 1

So 2kx + 2y + (k² + 1) > 0.

But since x ≥ 0, y ≥ 0, and k > 0, these inequalities are all true? Wait, for distance to O being less than distance to A, the inequality is 2kx - 2y + (k² + 1) > 0. Since x ≥ 0, y ≥ 0, and k is positive, this would be:

Is 2kx - 2y + (k² + 1) > 0?

But since x and y can vary, this inequality may or may not hold. For example, if x is very small and y is large.

Wait, but in the first quadrant, points can be closer to A? Let me take a point near (0,1). Let me compute distance to O: sqrt(0² + 1²) = 1. Distance to A(-k,1): sqrt( (0 + k)^2 + (1 - 1)^2 ) = k. So 1 < k? That is true if k > 1. Wait, but k is AB / AD. If the rectangle is very long horizontally, then k is large. If the rectangle is very tall, k is small. So depending on k, the point (0,1) could be closer to O or to A.

Wait, for example, if k = 1, then distance to A is 1, same as distance to O. If k > 1, then distance to A is larger, so O is closer. If k < 1, distance to A is smaller, so point (0,1) is closer to A.

Therefore, the condition that the point is closer to O than to A is not automatically satisfied in the first quadrant. Therefore, I need to check all four inequalities.

But this is getting complicated. Maybe instead of considering each vertex, I can find the region where distance to O is less than distance to any of the vertices. Which would be the intersection of all regions where distance to O is less than distance to each vertex.

Alternatively, since the rectangle is convex and the set of points closer to O than to any vertex is the intersection of the regions closer to O than to each vertex. Therefore, the boundaries are the perpendicular bisectors between O and each vertex. Then the region we want is the intersection of all four regions.

So maybe to find the boundaries, let's find the perpendicular bisector between O and each vertex. For example, the set of points equidistant to O and vertex A.

Let me take vertex A(-k,1). The set of points equidistant to O(0,0) and A(-k,1) is the perpendicular bisector of OA.

The midpoint of OA is (-k/2, 1/2). The line OA has slope (1 - 0)/(-k - 0) = -1/k. Therefore, the perpendicular bisector will have slope k.

So equation of the perpendicular bisector: passes through (-k/2, 1/2) with slope k.

So equation: y - 1/2 = k(x + k/2)

Similarly, for the other vertices.

But since we are in the first quadrant, maybe I can find the intersection of these bisectors and find the region.

Alternatively, since the problem is symmetric, maybe the region closer to O than any vertex is a smaller rectangle or a polygon whose area we can compute.

Alternatively, perhaps the region is the set of points closer to O than to any vertex, which is the intersection of four regions. Each region is defined by being closer to O than to one vertex.

But since we have symmetry, maybe it's enough to consider one region and multiply.

But perhaps it's easier to find the locus of points equidistant to O and to each vertex.

Let me try to find equations for these bisectors.

First, let's take vertex A(-k, 1). As above, the perpendicular bisector is y - 1/2 = k(x + k/2).

Similarly, for vertex B(k,1), the perpendicular bisector would be symmetric. The midpoint of OB is (k/2, 1/2), and the line OB has slope (1 - 0)/(k - 0) = 1/k, so the perpendicular bisector has slope -k. Equation: y - 1/2 = -k(x - k/2).

Similarly, for vertex C(k, -1), the perpendicular bisector is y + 1/2 = -k(x - k/2) (since midpoint is (k/2, -1/2), and line OC has slope (-1)/k, so perpendicular bisector has slope k). Wait:

Wait, for vertex C(k, -1): midpoint between O(0,0) and C(k, -1) is (k/2, -1/2). The line OC has slope (-1 - 0)/(k - 0) = -1/k, so the perpendicular bisector has slope k. Therefore, equation is y - (-1/2) = k(x - k/2) => y + 1/2 = k(x - k/2).

Similarly for vertex D(-k, -1): midpoint is (-k/2, -1/2), line OD has slope (-1)/(-k) = 1/k, so perpendicular bisector has slope -k. Equation: y - (-1/2) = -k(x - (-k/2)) => y + 1/2 = -k(x + k/2).

Therefore, these four bisectors divide the plane into regions. The region closer to O than any vertex is the intersection of the regions closer to O than each vertex. Therefore, this region is bounded by these four bisectors.

Therefore, in each quadrant, the region is bounded by two bisectors. For example, in the first quadrant, the boundaries are the bisectors between O and B, and between O and A? Wait, no. Wait in the first quadrant, which bisectors are relevant?

In the first quadrant, the closest vertices are B and... Well, A is in the second quadrant, but maybe the bisector between O and A will intersect the first quadrant?

Let me check the equation for the bisector between O and A: y - 1/2 = k(x + k/2). Let me see where this line intersects the first quadrant. Since in the first quadrant, x ≥ 0, y ≥ 0.

Let me solve for y: y = kx + k^2/2 + 1/2.

But when x = 0, y = k*(0) + k²/2 + 1/2 = (k² + 1)/2. However, since the rectangle has y from -1 to 1, so if (k² + 1)/2 > 1, which would be when k² +1 > 2 => k² >1 => |k|>1. So if k >1, then the bisector between O and A intersects the first quadrant above y=1, which is outside the rectangle. Therefore, within the rectangle, the bisector between O and A would intersect the top side of the rectangle at y=1. Let me find that intersection point.

Set y = 1 in the equation of bisector between O and A: 1 - 1/2 = k(x + k/2) => 1/2 = kx + k²/2 => Multiply both sides by 2: 1 = 2kx + k² => 2kx = 1 - k² => x = (1 - k²)/(2k). However, since we are in the first quadrant, x must be ≥0. Therefore, (1 - k²)/(2k) ≥ 0. Since k > 0 (as AB/AD =k, AB and AD are positive lengths), this requires 1 -k² ≥ 0 => k² ≤1 => k ≤1. Therefore, if k ≤1, the bisector between O and A intersects the top edge at x = (1 - k²)/(2k). If k >1, then (1 -k²)/ (2k) is negative, meaning the bisector between O and A does not intersect the top edge in the first quadrant. Instead, within the rectangle, the bisector between O and A would intersect the left edge or something else?

Wait, but the bisector between O and A is y = kx + k²/2 + 1/2. If we set x = -k (the left side of the rectangle), but x is in the first quadrant (x ≥ 0). So maybe the bisector doesn't intersect the rectangle in the first quadrant when k >1. Therefore, in that case, within the rectangle, the region closer to O than A would be the entire first quadrant?

Wait, maybe not. Let me think. For example, when k >1, take a point near the top left corner of the rectangle. Wait, but in the first quadrant, the left side is x=0, y from 0 to 1. Let me take the point (0,1). Its distance to O is 1. Its distance to A(-k,1) is sqrt( (k)^2 + 0 ) = k. Since k >1, so distance to O is less than distance to A. Therefore, the entire first quadrant is closer to O than to A? Wait, but if k is very large, say k approaches infinity, then for any point (x, y) in the first quadrant, distance to O is sqrt(x² + y²), and distance to A is sqrt( (x + k)^2 + (y - 1)^2 ). For large k, distance to A is approximately k + x, since (x +k)^2 dominates. Then distance to O is sqrt(x² + y²) ≈ sqrt(x² + y²) << k + x. Therefore, for large k, the entire first quadrant is closer to O than to A. Therefore, for k ≥1, the region closer to O than A in the first quadrant is the entire first quadrant? Wait, but maybe not entirely?

Wait, suppose k = 2. Take a point (x,y) in the first quadrant. For it to be closer to O than to A, we need sqrt(x² + y²) < sqrt( (x + 2)^2 + (y -1)^2 ). Squaring both sides:

x² + y² < (x + 2)^2 + (y -1)^2

Expand RHS: x² + 4x + 4 + y² - 2y + 1 = x² + y² + 4x - 2y + 5

Subtract LHS: 0 < 4x - 2y + 5 => 4x - 2y + 5 > 0 => 2x - y + 5/2 > 0 => y < 2x + 5/2. But since y ≤ 1 and x ≥ 0, this inequality is always true because 2x + 5/2 ≥ 5/2 > 1. Therefore, for k = 2, the entire first quadrant is closer to O than to A. Therefore, when k ≥1, the region closer to O than A in the first quadrant is the entire first quadrant.

Similarly, for the region closer to O than C or D? Let me check. For point (x,y) in first quadrant, distance to C(k, -1) is sqrt( (x - k)^2 + (y + 1)^2 ). Compare to distance to O: sqrt(x² + y²). Squaring both sides:

x² + y² < (x - k)^2 + (y + 1)^2

Expand RHS: x² - 2kx + k² + y² + 2y + 1

Subtract LHS: 0 < -2kx + k² + 2y + 1 => 2kx - 2y + ( -k² -1 ) < 0 ?

Wait, wait:

Original inequality: x² + y² < (x - k)^2 + (y + 1)^2 => 0 < -2kx + k² + 2y + 1.

So 2kx - 2y + (-k² -1) < 0? Wait, no:

Wait, 0 < -2kx + k² + 2y + 1 is equivalent to 2kx - 2y + (-k² -1) < 0? Let me check:

Original inequality: 0 < -2kx + k² + 2y + 1 => rearranged: 2kx - 2y + (-k² -1) < 0? Wait, no. Let me rearrange:

Starting from 0 < -2kx + k² + 2y + 1, subtract both sides by (-2kx + k² + 2y + 1):

Wait maybe better to rearrange:

-2kx + 2y + k² + 1 > 0 => 2y > 2kx - k² - 1 => y > kx - (k² +1)/2.

But for points in the first quadrant, y ≥ 0. So depending on k and x, this inequality may or may not hold.

For example, take k = 2, and check if y > 2x - (4 + 1)/2 = 2x - 2.5. In the first quadrant, y ≥ 0. Let me see when 0 > 2x - 2.5 => 2x < 2.5 => x < 1.25. So for x < 1.25, y ≥ 0 satisfies the inequality y > 2x - 2.5. However, for x ≥1.25, the inequality y > 2x - 2.5 may not hold. For example, take x = 1.5, then 2x - 2.5 = 3 - 2.5 = 0.5. So y needs to be greater than 0.5. Therefore, for x ≥1.25, the region where y > 0.5 would satisfy the inequality. However, if y is less than 0.5, it would not.

Therefore, for k=2, the region closer to O than to C in the first quadrant is y > 2x - 2.5. So this is a line cutting through the first quadrant. Similarly, for vertex D(-k, -1), the bisector is y + 1/2 = -k(x + k/2). In the first quadrant, this would correspond to something else? But maybe we don't have to consider D in the first quadrant? For points in the first quadrant, distance to D is sqrt( (x + k)^2 + (y + 1)^2 ). Since x ≥0, y ≥0, this is larger than sqrt( (x +k)^2 + 1 ), which is larger than sqrt( k^2 + 1 ) if x=0. So comparing to distance to O, sqrt(x² + y²). For points in first quadrant, is it possible to be closer to D than to O?

Yes, for example, take the point (0, 0). Distance to O: 0, which is less than distance to D. But take a point near (0, -something)? Wait, no, in first quadrant. Take point (0, 1). Distance to O: 1, distance to D(-k, -1): sqrt( (k)^2 + (2)^2 ). If k is 2, this is sqrt(4 + 4) = sqrt(8) ≈ 2.828. So distance to O is less. If we take a point (x, y) where y is very small. For example, take y = 0, x approaching k. Then distance to O is sqrt(x² + 0) = x. Distance to D is sqrt( (x + k)^2 + (0 + 1)^2 ) ≈ sqrt( (2k)^2 + 1 ). For large k, this is about 2k, so distance to O is x, which is less than 2k. Therefore, even points near the right side are closer to O than to D. So maybe in the first quadrant, the only vertex we need to compare with O is B and C? Wait, but we had an inequality for C and D? Wait, but we considered distance to C and D. However, for the first quadrant, maybe the critical vertices are B and C. Wait, but D is at (-k, -1), so in the first quadrant, points can't be closer to D than to O? Wait, let's check.

Take a point (x, y) in first quadrant. Distance to D is sqrt( (x +k)^2 + (y +1)^2 ). Distance to O is sqrt(x² + y²). Let me compare:

sqrt( (x +k)^2 + (y +1)^2 ) < sqrt(x² + y² )?

Squaring both sides:

(x +k)^2 + (y +1)^2 < x² + y²

Expanding:

x² + 2kx + k² + y² + 2y +1 < x² + y²

Simplify:

2kx + k² + 2y + 1 < 0

But 2kx + 2y + k² + 1 < 0. Since x ≥0, y ≥0, k >0, this is impossible. Therefore, in the first quadrant, no point is closer to D than to O. Similarly, for distance to A, we saw that if k ≥1, then all points are closer to O than to A. Therefore, in the first quadrant, the only constraint is being closer to O than to B or C.

Therefore, in the first quadrant, the region where points are closer to O than to any vertex is the intersection of regions closer to O than B and C. Therefore, we need to find the area in the first quadrant where distance to O < distance to B and distance to O < distance to C. Then multiply by 4 to get total area.

So let me focus on first quadrant. Let me write down the inequalities:

For distance to O < distance to B(k,1):

sqrt(x² + y²) < sqrt( (x - k)^2 + (y - 1)^2 )

Squaring both sides:

x² + y² < (x - k)^2 + (y - 1)^2

Which simplifies to:

x² + y² < x² - 2kx + k² + y² - 2y + 1

Cancelling x² and y²:

0 < -2kx + k² - 2y + 1

Which rearranged is:

2kx + 2y < k² + 1

Similarly, for distance to O < distance to C(k, -1):

sqrt(x² + y²) < sqrt( (x - k)^2 + (y + 1)^2 )

Squaring both sides:

x² + y² < (x - k)^2 + (y + 1)^2

Expanding:

x² + y² < x² - 2kx + k² + y² + 2y + 1

Simplify:

0 < -2kx + k² + 2y + 1

Which rearranged is:

2kx - 2y < -k² - 1

Wait, let me do it again:

Original inequality:

0 < -2kx + k² + 2y + 1 => rearranged:

2kx - 2y < -k² -1 ?

Wait:

Starting from 0 < -2kx + k² + 2y + 1,

Moving terms:

2kx - 2y < -k² -1 ?

Wait, if we move 2kx to the left and 0 to the right? Wait, perhaps better to rearrange:

Let me write:

From 0 < -2kx + k² + 2y + 1,

Bring 2kx to the right and others to the left:

-2kx + k² + 2y + 1 > 0 => 2y > 2kx - k² -1 => y > kx - (k² +1)/2.

But in the previous inequality for distance to B, we had 2kx + 2y < k² + 1 => y < (k² +1)/2 -kx.

So combining both inequalities:

In the first quadrant, we have:

y < (k² +1)/2 -kx (from distance to B)

and

y > kx - (k² +1)/2 (from distance to C)

So these two inequalities define a region between two lines.

Therefore, in the first quadrant, the region closer to O than to B or C is between these two lines. Therefore, to find the area in the first quadrant, we need to find the area between these two lines, within the rectangle [0 ≤ x ≤k, 0 ≤ y ≤1].

Therefore, the bounds for y are:

Lower bound: y_lower = kx - (k² +1)/2

Upper bound: y_upper = (k² +1)/2 -kx

But we need to ensure that these bounds are within the rectangle, i.e., between 0 and 1.

Therefore, we need to find the range of x where y_lower ≥0 and y_upper ≤1. However, since the region is between these two lines, and in the first quadrant, we need to find the intersection of these inequalities with the rectangle.

Alternatively, maybe we can find the region where both inequalities hold, which is the area between the two lines.

But to find that area, perhaps we need to find where these lines intersect the rectangle.

First, let's find where y_lower = kx - (k² +1)/2 intersects the rectangle.

Similarly for y_upper.

Let me analyze y_lower = kx - (k² +1)/2.

Since in the first quadrant, y ≥ 0. So solving for when y_lower = 0:

kx - (k² +1)/2 = 0 => x = (k² +1)/(2k)

Similarly, y_upper = (k² +1)/2 -kx. Setting y_upper = 1:

(k² +1)/2 -kx = 1 => -kx = 1 - (k² +1)/2 = (2 - k² -1)/2 = (1 - k²)/2 => x = (k² -1)/(2k)

Therefore, the line y_upper intersects the top edge y=1 at x=(k² -1)/(2k), and the line y_lower intersects the bottom edge y=0 at x=(k² +1)/(2k).

But depending on k, these x-values may be within the rectangle [0, k].

Let me analyze for different values of k.

First, consider k >1.

If k >1, then (k² -1)/(2k) is positive because k² -1 >0. Also, (k² +1)/(2k) is greater than (k²)/(2k) = k/2. Since k >1, let's see if (k² +1)/(2k) is less than k?

(k² +1)/(2k) < k ?

Multiply both sides by 2k (positive):

k² +1 < 2k² => 1 < k² => which is true since k >1.

Therefore, for k >1, the line y_upper intersects y=1 at x=(k² -1)/(2k), and lies between 0 and k. Similarly, the line y_lower intersects y=0 at x=(k² +1)/(2k), which is between k/2 and k (since (k² +1)/(2k) = k/2 + 1/(2k), which is greater than k/2).

Therefore, in the first quadrant, the region between the lines y_lower and y_upper is bounded between x values from x_low to x_high, where x_low is where y_upper meets y=1, and x_high is where y_lower meets y=0. Wait, but how exactly?

Wait, perhaps the region between the two lines is a parallelogram or some polygon. Let me sketch mentally.

For k >1:

The line y_upper = (k² +1)/2 -kx starts at the top edge at x=(k² -1)/(2k), y=1, and goes down to where?

At x=0, y_upper would be (k² +1)/2 - 0 = (k² +1)/2 >1, since for k >1, (k² +1)/2 > (1 +1)/2 =1. Therefore, the line y_upper intersects the left edge x=0 above y=1, which is outside the rectangle. Therefore, within the rectangle, the line y_upper goes from ( (k² -1)/(2k), 1 ) down to some point where it meets y_lower?

Wait, no. Wait, the two lines y_upper and y_lower intersect each other?

Let me find where y_lower = y_upper:

kx - (k² +1)/2 = (k² +1)/2 -kx

=> kx +kx = (k² +1)/2 + (k² +1)/2

=> 2kx = k² +1 => x = (k² +1)/(2k)

Which is the same x as where y_lower meets y=0. So the two lines intersect at x=(k² +1)/(2k), y=0?

Wait, plug x=(k² +1)/(2k) into y_lower:

y_lower = k*( (k² +1)/(2k) ) - (k² +1)/2 = (k² +1)/2 - (k² +1)/2 = 0. So yes, they intersect at point ((k² +1)/(2k), 0). Similarly, if we plug x=(k² -1)/(2k) into y_upper:

y_upper = (k² +1)/2 -k*( (k² -1)/(2k) ) = (k² +1)/2 - (k² -1)/2 = [k² +1 -k² +1]/2 = 2/2 = 1. So that's the point.

Therefore, in the first quadrant, the region between the two lines is a quadrilateral with vertices at:

1. Intersection of y_upper and y=1: ((k² -1)/(2k), 1)

2. Intersection of y_upper and y_lower: ((k² +1)/(2k), 0)

3. Intersection of y_lower and y=0: ((k² +1)/(2k), 0) – same as above?

Wait, no, y_lower is the lower bound, so it goes from ((k² +1)/(2k), 0) to... Wait, but if we think of the region between y_upper and y_lower, it's bounded between these two lines. But since for x between 0 and (k² -1)/(2k), y_upper is above y=1, which is outside the rectangle. Therefore, within the rectangle, the region between the two lines is from x = (k² -1)/(2k) to x = (k² +1)/(2k), between the two lines.

Wait, let me think again.

When x is less than (k² -1)/(2k), the upper bound y_upper would be greater than 1, which is outside the rectangle. Therefore, within the rectangle, the upper bound is y=1. Similarly, for x greater than (k² +1)/(2k), the lower bound y_lower would be negative, which is outside the rectangle, so the lower bound is y=0.

Therefore, within the rectangle, the region closer to O than to B or C is bounded between:

- For x between 0 and x1 = (k² -1)/(2k): between y=0 and y=1 (since the lines are outside the rectangle). Wait, no, because we have constraints from both B and C.

Wait, no. Wait, we need both distance to O < distance to B and distance to O < distance to C. So for x between 0 and x1, which is where y_upper meets y=1, we have that the upper bound is y_upper which is above y=1, so within the rectangle, the upper bound is y=1. However, the lower bound is y_lower which may be below y=0 or not?

Wait, for x between 0 and x1, the lower bound y_lower = kx - (k² +1)/2. For x=0, that is - (k² +1)/2 <0. Therefore, within the rectangle, the lower bound is y=0. Therefore, for x between 0 and x1, the region is between y=0 and y=1, but also must satisfy the inequalities from B and C. However, we already derived that the inequalities reduce to between the lines which for x in [0, x1], the region is between y=0 and y=1? Wait, no, that can't be.

Wait, no. The inequalities are:

For all points in the region, they must satisfy both inequalities:

y < (k² +1)/2 -kx (from B)

and

y > kx - (k² +1)/2 (from C)

But since in this region, for x between 0 and x1, (k² +1)/2 -kx is greater than 1, so y < (k² +1)/2 -kx is automatically satisfied since y ≤1. Similarly, y > kx - (k² +1)/2. For x between 0 and x1, kx - (k² +1)/2 is less than 0, since at x= x1, it's zero. Therefore, y > [negative number], which is automatically satisfied since y ≥0. Therefore, for x between 0 and x1, all points satisfy both inequalities. Hence, the entire region from x=0 to x=x1, y=0 to y=1, is within the desired region.

Then, for x between x1 and x2 = (k² +1)/(2k), the upper bound is y_upper = (k² +1)/2 -kx, which decreases from 1 to 0. The lower bound is y_lower = kx - (k² +1)/2, which increases from 0 to 0. Therefore, the region between these lines is a sort of "triangle" between x1 and x2.

But wait, when x is between x1 and x2, the bounds are between y_lower and y_upper, which meet at (x2, 0). Therefore, the area between x1 and x2 is bounded by the two lines and the x-axis?

Wait, actually, since y_lower is increasing from 0 to 0 (wait, at x=x2, y_lower is 0), while y_upper is decreasing from 1 to 0. Therefore, the region between x1 and x2 is bounded above by y_upper and below by y_lower, forming a region that is a triangle?

Wait, let me take specific values. Let me take k=2, which is >1.

For k=2, x1 = (k² -1)/(2k) = (4 -1)/4 = 3/4 = 0.75

x2 = (k² +1)/(2k) = (4 +1)/4 = 5/4 = 1.25

Therefore, between x=0.75 and x=1.25, the region is between the two lines y_upper = (4 +1)/2 - 2x = 2.5 - 2x, and y_lower = 2x - (4 +1)/2 = 2x - 2.5.

Therefore, for x between 0.75 and 1.25, the area between these two lines.

At x=0.75, y_upper = 2.5 - 2*(0.75) = 2.5 - 1.5 = 1.0, and y_lower = 2*(0.75) - 2.5 = 1.5 - 2.5 = -1.0. But since y ≥0, the lower bound is 0.

Wait, this seems conflicting with previous thought. Wait, earlier I thought that between x1 and x2, the region between the lines is bounded by those lines, but at x= x1, y_lower is negative, so within the rectangle, the lower bound is y=0. Therefore, the region between x1 and x2 is between y=0 and y_upper = 2.5 - 2x. Wait, but then why was there a y_lower?

Wait, perhaps my earlier analysis was incorrect. Let me re-express.

We had two inequalities:

1. y < (k² +1)/2 -kx

2. y > kx - (k² +1)/2

But since in the rectangle, y ∈ [0,1], the effective region is:

y ∈ [max(0, kx - (k² +1)/2 ), min(1, (k² +1)/2 -kx ) ]

Therefore, for x between 0 and x1 = (k² -1)/(2k), where (k² +1)/2 -kx ≥1 and kx - (k² +1)/2 ≤0, the region is y ∈ [0,1].

For x between x1 and x2 = (k² +1)/(2k), where (k² +1)/2 -kx decreases from 1 to 0, and kx - (k² +1)/2 increases from 0 to 0, the region is y ∈ [kx - (k² +1)/2, (k² +1)/2 -kx ].

For x between x2 and k, where (k² +1)/2 -kx < 0, and kx - (k² +1)/2 >0, the region would be y ∈ [kx - (k² +1)/2, 0], but since y ≥0, that would be empty. Hence, the region is only between x=0 and x=x2.

Wait, but x2 is (k² +1)/(2k). For k >1, since (k² +1)/(2k) is less than k (as we saw earlier), so between x= x1 and x= x2, the region is between the two lines. Beyond x2, the region would be y ≥ [kx - (k² +1)/2 ] which is positive, but the upper bound would be negative, hence no region.

Therefore, in the first quadrant, the area closer to O than to B or C is:

- For x from 0 to x1: area = width * height = (x1 - 0) * 1 = x1

- For x from x1 to x2: area is the integral from x1 to x2 of [ (k² +1)/2 -kx - (kx - (k² +1)/2 ) ] dx = integral from x1 to x2 of [ (k² +1)/2 -kx -kx + (k² +1)/2 ] dx = integral of [k² +1 - 2kx ] dx

Wait:

Wait, the upper bound minus lower bound is [(k² +1)/2 -kx] - [kx - (k² +1)/2] = (k² +1)/2 -kx -kx + (k² +1)/2 = k² +1 - 2kx.

Therefore, the integral from x1 to x2 of (k² +1 - 2kx) dx.

So total area in first quadrant is:

A1 = x1 + ∫ from x1 to x2 (k² +1 - 2kx) dx

Compute this integral:

First, compute x1 = (k² -1)/(2k)

x2 = (k² +1)/(2k)

Compute integral ∫ (k² +1 - 2kx ) dx from x1 to x2.

Antiderivative is:

(k² +1)x - kx² evaluated from x1 to x2.

So compute at x2: (k² +1)x2 - kx2²

At x1: (k² +1)x1 - kx1²

Thus, the integral is [ (k² +1)x2 - kx2² ] - [ (k² +1)x1 - kx1² ]

Therefore, total area A1 is x1 + [ (k² +1)x2 - kx2² - (k² +1)x1 + kx1² ]

Simplify:

A1 = x1 + (k² +1)(x2 - x1) -k(x2² -x1² )

Note that x2² -x1² = (x2 -x1)(x2 +x1)

Therefore, A1 = x1 + (k² +1)(x2 -x1) -k(x2 -x1)(x2 +x1 )

Factor out (x2 -x1):

A1 = x1 + (x2 -x1)[(k² +1) -k(x2 +x1 ) ]

Now compute (x2 +x1):

x1 + x2 = [ (k² -1)/(2k) ] + [ (k² +1)/(2k) ] = [ (k² -1 +k² +1 ) / (2k) ] = (2k²)/ (2k) ) = k

Therefore, x2 +x1 = k

Therefore, A1 = x1 + (x2 -x1)[(k² +1) -k*k ] = x1 + (x2 -x1)[k² +1 -k²] = x1 + (x2 -x1)(1) = x1 + x2 -x1 = x2

Therefore, A1 = x2

Wait! That's unexpected. So the total area in first quadrant is x2? Which is (k² +1)/(2k). But let me check with k=2:

For k=2, x2 = (4 +1)/4 = 5/4 =1.25. Therefore, A1=1.25? But in the first quadrant, the area closer to O than to B or C is supposed to be... Let me check with k=2.

If k=2, then rectangle is from x=-2 to 2, y=-1 to 1. In first quadrant, area closer to O is x from 0 to 1.25, and... Wait, but according to the calculation, the area is x2 = 1.25? Which is a length? Wait, no, A1 is supposed to be an area. Wait, I think there was a miscalculation here.

Wait, going back, when I said A1 = x1 + ∫ ... which resulted in A1 = x2. But x1 and x2 are lengths along x-axis, so how can area be equal to x2? There must be a mistake in the calculation.

Let me retrace:

Original A1 = area from x=0 to x=x1 (which is a rectangle with width x1 and height 1) plus the integral from x=x1 to x=x2 of [k² +1 - 2kx] dx.

Then after computing, we found that A1 = x2. But when k=2, x2 is 5/4, but the first part x1 was 3/4, and then the integral? Let me compute it numerically for k=2.

For k=2, x1 = 3/4, x2 =5/4.

Compute A1 = x1 + ∫_{x1}^{x2} [k² +1 - 2kx] dx.

k² +1 = 4 +1 =5, 2k =4.

Thus, integral becomes ∫_{3/4}^{5/4} (5 -4x) dx.

Antiderivative is 5x - 2x² evaluated from 3/4 to 5/4.

At 5/4: 5*(5/4) - 2*(25/16) = 25/4 - 50/16 = 25/4 - 25/8 = 25/8.

At 3/4: 5*(3/4) - 2*(9/16) = 15/4 - 9/8 = 30/8 - 9/8 = 21/8.

Therefore, the integral is 25/8 -21/8 = 4/8 = 1/2.

Then A1 = x1 + integral = 3/4 + 1/2 = 5/4, which is equal to x2. Therefore, for k=2, A1=5/4. But the area in first quadrant is 5/4? Wait, but the entire first quadrant has area k*1 =2*1=2. Wait, no! Wait, the rectangle in first quadrant is [0, k] x [0,1], so area is k*1 =k. For k=2, area is 2. But according to this, A1 =5/4 which is less than 2. So that seems possible.

Wait, but let me check if this answer makes sense. For k=2, the region we found is 5/4. But let me think of another way. If we consider that the region closer to O than to B or C is the set of points between the two lines. For k=2, the area is 5/4. But if we consider the entire first quadrant area is 2, so probability would be 5/4 divided by 2 = 5/8. But let me check if that's correct.

Alternatively, let's think of the area where distance to O is less than distance to B and C. For k=2, it's the area between the two lines which we found to be 5/4. But maybe I can compute it another way.

Alternatively, the region closer to O than to B and C is the intersection of two regions. Each region is a half-plane. The intersection is the area between the two lines we found. But according to the calculation, that area is x2 = (k² +1)/(2k). Wait, for k=2, that is 5/4, which seems correct. But how does that relate to the rectangle?

Wait, maybe it's better to accept that the area in first quadrant is x2 = (k² +1)/(2k). Therefore, total area over all quadrants would be 4*(k² +1)/(2k) = 2(k² +1)/k. But wait, no. Wait, in the first quadrant, we found A1 = x2 = (k² +1)/(2k). Therefore, total area is 4*A1? Wait, no. Wait, the region is symmetric in all four quadrants? Wait, no. Wait, we did the analysis for the first quadrant, and the same applies to other quadrants. Therefore, total area is 4*A1. However, A1 = (k² +1)/(2k). Therefore, total area is 4*(k² +1)/(2k) = 2(k² +1)/k. However, the total area of the rectangle is 2k*2 = 4k. Therefore, probability is [2(k² +1)/k ] / (4k) ) = [ (k² +1)/k ] / (2k ) = (k² +1)/(2k² )

Wait, but for k=2, probability is (4 +1)/ (2*4) = 5/8, which matches our earlier calculation. But let me check if that answer makes sense. For k=1, which makes the rectangle a square. For k=1, probability is (1 +1)/2 =1, which is 1. Which doesn't make sense? Wait, wait. If k=1, the rectangle is a square with sides 2 and 2, area 4. The region closer to O than any vertex... Wait, in a square, the set of points closer to center than to any vertex is known to be a smaller square or another shape?

Wait, for a square, the region closer to center than to any vertex is the set of points where distance to center < distance to vertices. For a square, the set of points equidistant to center and a vertex is the perpendicular bisector between center and vertex, which in case of square would be a line. For example, for the vertex (1,1), the perpendicular bisector would be the line y = x, but wait:

Wait, let's take square with center at (0,0), vertices at (1,1), (-1,1), etc. For the vertex (1,1), the perpendicular bisector between (0,0) and (1,1) is the line y = -x + 1? Wait, midpoint is (0.5, 0.5). The line from (0,0) to (1,1) has slope 1, so perpendicular bisector has slope -1. Equation: y - 0.5 = -1(x -0.5) => y = -x +1. Similarly for other vertices. So the region closer to center than to any vertex is the intersection of four regions, bounded by these lines. The intersection would be a smaller square rotated by 45 degrees, whose vertices are at the intersections of these bisectors.

For example, the intersection of y = -x +1 and y = x +1 (from the other bisector) is at x=0, y=1. Similarly, other intersections would form a square. But the area of this inner square can be calculated.

For the square case (k=1), our formula gives probability = (1 +1)/(2*1² )= 2/2 =1, which can't be correct. Because the region closer to center than to any vertex can't be the entire square. For example, the corner points are exactly at distance sqrt(2)/2 from the center and distance 0 from the vertex, so those points are closer to the vertex. Therefore, the region closer to center is definitely smaller than the whole square. Hence, my calculation must be wrong.

Wait, so where is the mistake?

Earlier, I concluded that total area is 4*A1, where A1 = x2 = (k² +1)/(2k). But for k=1, this would be (1 +1)/2*1=1, so total area 4*1=4, which is equal to the area of the rectangle. Which is wrong. Hence, there's a mistake in reasoning.

Wait, when I considered the first quadrant area A1, I calculated it as x2. But when k=1, x2 = (1 +1)/(2*1)=1. Therefore, A1=1. But in the first quadrant, which is a square of area 1*1=1, so A1=1 would mean the entire quadrant is included, which is not correct. Hence, my mistake must be in the calculation of A1.

Going back, when I calculated A1 for general k, I had:

A1 = x1 + integral ... which became A1 = x2. But when k=1, this would be x2 = (1 +1)/2*1=1, but in reality, the area should be less than 1. Hence, error in the derivation. Let me check again.

Earlier steps:

I had A1 = x1 + [ (k² +1)(x2 -x1) -k(x2² -x1² ) ]

Then I factored out (x2 -x1):

A1 = x1 + (x2 -x1)[(k² +1) -k(x2 +x1 ) ]

Then I used x2 +x1 =k, so:

A1 = x1 + (x2 -x1)[k² +1 -k²] = x1 + (x2 -x1)(1) = x1 +x2 -x1 =x2.

But for k=1, x1 = (1 -1)/(2*1) =0, x2 = (1 +1)/ (2*1)=1. Hence, A1 = x2 =1, which is the entire first quadrant area. But this contradicts the square case.

But earlier, for k=1, the region closer to O than to any vertex is not the entire square. Therefore, there is a miscalculation in the steps. Where?

Wait, the error comes from the assumption that the area in the first quadrant is x2. But for k=1, this suggests that the entire first quadrant is included, which is not true. Therefore, my mistake is in the integral calculation.

Let me redo the integral calculation for general k. Let me go back.

First quadrant area A1 is:

Integral from x=0 to x=x1 of 1 dx + Integral from x=x1 to x=x2 of [ upper - lower ] dx

Which is:

x1 + ∫_{x1}^{x2} [ (k² +1)/2 -kx - (kx - (k² +1)/2 ) ] dx = x1 + ∫_{x1}^{x2} [k² +1 -2kx ] dx.

Yes, which we did. Then computed that as x2. But in reality, when k=1, this integral would be:

x1 = (1 -1)/2 =0

x2 = (1 +1)/2 =1

Integral from 0 to1 of [1 +1 - 2*1*x] dx = ∫₀¹ (2 - 2x) dx = [2x -x²]₀¹ = (2*1 -1) -0 =1. Hence, A1=0 +1=1. However, in reality, when k=1, the region closer to O than to any vertex should be the set of points where distance to O < distance to vertices. For k=1, the square has vertices at (±1, ±1). The region closer to O than to any vertex is the set of points where sqrt(x² + y²) < sqrt( (x -1)^2 + (y -1)^2 ), and similarly for other vertices. Which simplifies to the inequalities we had before.

But when k=1, the lines y_upper = (k² +1)/2 -kx = (1 +1)/2 -1*x =1 -x, and y_lower =kx - (k² +1)/2 =x -1.

In the first quadrant, between x=0 and x=1, the region between y_lower =x -1 and y_upper=1 -x. However, y_lower =x -1 is negative for x <1, and y_upper =1 -x goes from 1 to 0. Hence, the area between them in the first quadrant is the area between y=0 and y=1 -x from x=0 to x=1. Therefore, the area should be ∫₀¹ (1 -x) dx = [x - x²/2]₀¹ = 1 - 1/2 =1/2. But according to my previous calculation, A1=1, which is wrong. Hence, the error was in the previous conclusion that A1 = x2. Where is the mistake?

Ah! The mistake was in the final step where I said A1 =x1 + ... =x2. But when I substituted the numbers for k=1, I saw that A1=0 +1=1, but the actual area is 1/2. Hence, there is a miscalculation in the algebra.

Let me re-express the integral.

Earlier, I had:

The integral ∫_{x1}^{x2} [k² +1 -2kx ] dx = [ (k² +1)x -k x² ] from x1 to x2.

For k=1, that integral becomes:

[ (1 +1)x -1*x² ] from x=0 to x=1: [2x -x²]₀¹ = (2*1 -1) -0 =1, which matches. But earlier I thought that the area should be ∫₀¹ (1 -x) dx =1/2. Why the discrepancy?

Because the integrand [k² +1 -2kx ] for k=1 is [2 -2x], and I integrated it from 0 to1, which gives 1. However, the region between y_lower and y_upper is between y_lower =x -1 and y_upper=1 -x. But in reality, in the first quadrant, the lower bound is y=0, not y_lower =x -1. Therefore, my earlier assumption that the area is ∫ [upper - lower] dx from x1 to x2 is only valid when upper and lower bounds are within the rectangle. However, when k=1, for x between x1=0 and x2=1, the lower bound is not y_lower =x -1 but y=0, because y_lower is negative. Therefore, the actual region is between y=0 and y_upper=1 -x, hence area ∫₀¹ (1 -x) dx =1/2, but according to the formula [upper - lower] = [1 -x - (x -1)] = 2 -2x, which is double the correct integrand. Hence, my mistake was in the expression of the integrand.

Wait, how did we get [upper - lower] as k² +1 -2kx? Let's recalculate.

We had upper bound y_upper = (k² +1)/2 -kx, lower bound y_lower =kx - (k² +1)/2. Therefore, upper - lower = [(k² +1)/2 -kx] - [kx - (k² +1)/2] = (k² +1)/2 -kx -kx + (k² +1)/2 = k² +1 -2kx.

But in the case of k=1, this is 2 -2x. However, in reality, the actual upper bound is y_upper = 1 -x and lower bound is y=0, so upper - lower = 1 -x -0 =1 -x. But according to the formula, it is 2 -2x. Hence, discrepancy. Why?

Because for x in [x1, x2], the lower bound is not y_lower =kx - (k² +1)/2, but y=0 when y_lower <0. Hence, my previous assumption that upper - lower =k² +1 -2kx is only valid when y_lower ≥0, i.e., when x ≥x2? Wait, no. Wait, for x between x1 and x2, since x1 is where y_upper meets y=1, and x2 is where y_lower meets y=0.

Wait, for k=1, x1=0, x2=1. For x between 0 and 1, y_lower =x -1 is negative, so effective lower bound is 0. Upper bound is y_upper =1 -x. Therefore, the integrand should be y_upper - 0 =1 -x. However, according to the formula, upper - lower =2 -2x, which is double of 1 -x. Hence, the mistake was in the previous derivation where I assumed that both upper and lower bounds are active, but actually for x between x1 and x2, only the upper bound is active (since lower bound is y=0). Hence, my earlier approach is flawed.

This suggests that my earlier analysis was incorrect. Let me correct it.

The correct region in the first quadrant is:

For x between 0 and x1, the region is between y=0 and y=1 (since both inequalities are automatically satisfied).

For x between x1 and x2, the region is between y=0 and y_upper = (k² +1)/2 -kx.

Wait, why? Because in this range, the lower bound y_lower is negative, so the effective lower bound is y=0. The upper bound is still y_upper = (k² +1)/2 -kx.

Therefore, the area in first quadrant is:

A1 = ∫₀^{x1} 1 dx + ∫_{x1}^{x2} [ (k² +1)/2 -kx - 0 ] dx

= x1 + ∫_{x1}^{x2} [ (k² +1)/2 -kx ] dx

Let me compute this correctly.

For k=1, this becomes:

x1 = 0, x2 =1

Integral from 0 to1 of [ (1 +1)/2 -1*x ] dx = ∫₀¹ [1 -x] dx = [x - x²/2]₀¹ = 1 - 1/2 =1/2, which matches the correct area.

Previously, I incorrectly took upper - lower as k² +1 -2kx, but it should be upper - 0 = (k² +1)/2 -kx. Hence, the previous error.

Therefore, let's recalculate the general case.

A1 = x1 + ∫_{x1}^{x2} [ (k² +1)/2 -kx ] dx

Compute this integral.

First, compute x1 = (k² -1)/(2k)

x2 = (k² +1)/(2k)

Integral ∫ [ (k² +1)/2 -kx ] dx from x1 to x2

Antiderivative is (k² +1)/2 * x - (k/2)x²

Evaluate at x2:

(k² +1)/2 * x2 - (k/2)x2²

At x1:

(k² +1)/2 * x1 - (k/2)x1²

Thus, integral is:

[(k² +1)/2 * x2 - (k/2)x2²] - [(k² +1)/2 * x1 - (k/2)x1²]

Factor out (k/2):

= (k/2)[ ( (k² +1)/k * x2 - x2² ) - ( (k² +1)/k * x1 - x1² ) ]

Wait, perhaps it's easier to compute numerically.

Let me compute the difference:

Let me denote C = (k² +1)/2

Then integral = [C*x2 - (k/2)x2²] - [C*x1 - (k/2)x1²] = C(x2 -x1) - (k/2)(x2² -x1² )

As before, x2 -x1 = [(k² +1)/(2k) - (k² -1)/(2k)] = [ (k² +1 -k² +1 ) / (2k) ] = 2/(2k) = 1/k

Also, x2² -x1² = (x2 -x1)(x2 +x1) = (1/k)*k =1

Therefore, integral = C*(1/k) - (k/2)(1) = [(k² +1)/2]*(1/k) - k/2 = (k² +1)/(2k) - k/2 = [k² +1 -k²]/(2k) = 1/(2k)

Therefore, A1 = x1 + 1/(2k)

But x1 = (k² -1)/(2k)

So A1 = (k² -1)/(2k) + 1/(2k) = (k² -1 +1)/ (2k) = k² / (2k) = k/2

Therefore, the area in the first quadrant is k/2. Hence, total area over all four quadrants is 4*(k/2) = 2k.

Wait, for k=1, this gives total area 2*1=2. The area of the rectangle is 4*1=4. So probability is 2/4=1/2. But earlier analysis for k=1 suggested the region is smaller. Let me check.

For k=1, the total area closer to O than any vertex is 2. The rectangle has area 4. So probability 1/2. Let me check this with the square.

In a square, the region closer to center than to any vertex is a smaller square rotated by 45 degrees, whose vertices are at the midpoints of the sides of the original square. Wait, no. Let me think.

For a square with side length 2 (from -1 to 1), the set of points closer to center than to any vertex is the set of points where distance to center < distance to any vertex. The region is the intersection of four regions, each bounded by perpendicular bisectors. These bisectors are lines like y = -x +1, etc. The intersection of these regions is a square rotated by 45 degrees, with vertices at (0,1), (1,0), (0,-1), (-1,0). Wait, no. Let me compute the intersection of the regions.

For example, the region closer to O than to vertex (1,1) is the region below y = -x +1. Similarly, closer to O than to (-1,1) is y < x +1, closer to O than to (-1,-1) is y > -x -1, closer to O than to (1,-1) is y > x -1. The intersection of these is the region where -x -1 < y < -x +1 and x -1 < y < x +1, which is the region between the lines y = |x| -1 and y = -|x| +1. Wait, this is a square with vertices at (1,0), (0,1), (-1,0), (0,-1). The area of this square is 2 (since from -1 to 1 in x and y, but it's a diamond). The area is 2*1*1 =2? Wait, the area of this diamond is 2*(base*height)/2 = 2*1*1 =2? Wait, actually, the area can be calculated as follows: it's a square with diagonals of length 2 (from -1 to 1 on both axes). The area of a square with diagonal d is d²/2. So here, d=2, area is 4/2=2. Yes, which matches. So the area is 2, and the total area of the square is 4, so probability is 2/4=1/2. Which matches our result of k=1, so A1= k/2 =1/2, total area 2k=2*1=2, probability 2/4=1/2. Okay, this makes sense.

For k=2, total area is 2k=4, while the rectangle has area 4k=8, so probability is 4/8=1/2. Wait, but earlier calculation with k=2 suggested probability 5/8? Wait, no, when I thought I found probability 5/8, that was based on a wrong calculation. Now, with corrected calculation, probability is 2k / (4k) = 1/2. Wait, this suggests the probability is always 1/2, regardless of k? But that seems counterintuitive. For example, if the rectangle is very long and skinny (k very large), intuitively, the region closer to center might be smaller. But according to this, it's always 1/2.

Wait, but let's test with k approaching infinity. Suppose k is very large, so the rectangle is very long horizontally. What is the region closer to O than to any vertex? The vertices are at (k,1), (-k,1), etc. For a point in the middle of the rectangle, say (0,0), it's closer to O. For a point near the center, it's closer to O. For a point near the edge, say (x, y) near (k, 1), it's closer to vertex B(k,1) than to O. But the region closer to O would be the set of points where distance to O < distance to any vertex. For large k, what does this region look like?

If k is large, then for a point (x,y), distance to O is sqrt(x² + y²). Distance to nearest vertex is min(distance to (k,1), distance to (-k,1), distance to (k,-1), distance to (-k,-1)).

For a point in the middle, it's closer to O. For a point near the side, like near (k,0), distance to O is sqrt(k² + 0) =k. Distance to vertex (k,1) is sqrt(0 +1) =1, which is less. So the region closer to O would be bounded away from the sides.

But according to our result, the probability is always 1/2. Let's see: if the area of the region closer to O is 2k, and the total area is 4k, then probability is 1/2. But is that true?

Wait, in our calculation, we derived that the area closer to O is 2k for any k, but let's check with k approaching infinity. If k approaches infinity, then the region closer to O should be a smaller and smaller fraction of the rectangle. But according to our result, it's always 1/2. Contradiction! Hence, there must be a mistake.

Wait, let's re-examine the calculation. For general k, we found that the area in first quadrant is k/2. So total area is 4*(k/2) =2k. Total area of rectangle is 4k. Therefore, probability is 2k / 4k =1/2. But for k approaching infinity, this suggests the probability is 1/2, which doesn't make sense. What's wrong?

Wait, but if k approaches infinity, the rectangle becomes very long and thin. The region closer to O than to any vertex should be a horizontal strip around the center. Let's calculate its area.

For a very large k, the region closer to O than to any vertex is the set of points where distance to O < distance to any vertex. Let's take a point (x,y). Distance to O: sqrt(x² + y²). Distance to nearest vertex: the closest vertex would be either (k,1), (-k,1), (k,-1), or (-k,-1). Let's assume the point is in the right half, x ≥0. The closest vertex would be (k,1) if the point is near the right side, or perhaps another.

For a point near the middle, say x=0, y=0, it's closer to O. For a point near (k,1), it's closer to that vertex. The boundary between being closer to O or to (k,1) is the perpendicular bisector between O and (k,1). Which we already derived as y = kx + (k² +1)/2. Wait, no, earlier we had for the bisector between O and A(-k,1) was y = kx + (k² +1)/2. But for the bisector between O and B(k,1), it's y = -kx + (k² +1)/2.

Wait, in the first quadrant, the bisector between O and B(k,1) is y = -kx + (k² +1)/2. For large k, this line is approximately y = -kx + (k²)/2, which simplifies to y ≈ -kx + k²/2. For x =0, y≈ k²/2, which is way outside the rectangle (which has y up to 1). For y=1, solving for x: 1 = -kx +k²/2 => kx =k²/2 -1 => x ≈k/2 - 1/k. For large k, x≈k/2. So the bisector intersects the top edge y=1 at x≈k/2. Hence, in the first quadrant, the region closer to O than to B is to the left of this line, i.e., x < (k²/2 -1)/k ≈k/2.

Similarly, the region closer to O than to C(k,-1) in first quadrant is above the line y = kx - (k² +1)/2. For large k, this is approximately y ≈kx -k²/2. Setting y=0: 0 ≈kx -k²/2 => x≈k/2. Hence, the region closer to O than to C is above y ≈kx -k²/2, which for x near k/2 is y≈0.

Thus, the region closer to O than to any vertex in first quadrant is approximately the region where x <k/2 and y > something, but for large k, this region is a rectangle from x=0 to x=k/2, y= -1 to y=1? No, because it's bounded by the bisectors.

Wait, actually, for large k, the region closer to O than to any vertex is approximately the set of points where |x| <k/2 and |y| < something, but I need to think carefully.

Actually, according to our previous calculation, the area closer to O is 2k, which for large k would be a strip from x=-k/2 to x=k/2, y=... but the height?

Wait, the area is 2k, while the total area is 4k. Hence, the region is 1/2 of the rectangle. For large k, the region closer to O is a central stripe of width k (from x=-k/2 to x=k/2), but height? If the width is k, then the height would have to be 2 to maintain area 2k. But the rectangle has height 2, so it would be the central stripe from x=-k/2 to x=k/2, full height. But this is impossible because for x=k/2, the distance to O is sqrt( (k/2)^2 + y^2 ), while distance to vertex B(k,1) is sqrt( (k/2)^2 + (1 - y)^2 ). Setting them equal: (k/2)^2 + y^2 = (k/2)^2 + (1 - y)^2 => y² = (1 - y)^2 => y² =1 -2y + y² => 0 =1 -2y => y=1/2. Hence, the boundary is at y=1/2 for x=k/2. Similarly on other sides. Hence, the region closer to O is a central square from x=-k/2 to x=k/2 and y=-1/2 to y=1/2? Wait, if that's the case, the area would be k * 1 =k, which is half of the total area 2k? No, total area is 4k. Wait, if it's a central rectangle of width k and height 1, area is k*1=k, which is k/4k =1/4. Doesn't match.

Wait, this suggests that my previous conclusion that the area is 2k is incorrect for large k. Hence, there is a consistent error in my calculation.

Let me re-examine the calculation for general k. We had:

A1 = k/2, total area = 2k, total rectangle area =4k, probability=1/2. But for k approaching infinity, this suggests the probability is always 1/2, which seems incorrect.

However, for k=1, it's correct (probability 1/2). For k approaching infinity, let me think of another approach.

Consider the rectangle with AB/AD =k, very large k. Let's assume k is very large, so the rectangle is long horizontally. The set of points closer to O than to any vertex is the set of points where distance to O < distance to any vertex. For a point (x,y), this means:

sqrt(x² + y²) < sqrt( (x -k)^2 + (y -1)^2 )

sqrt(x² + y²) < sqrt( (x +k)^2 + (y -1)^2 )

sqrt(x² + y²) < sqrt( (x -k)^2 + (y +1)^2 )

sqrt(x² + y²) < sqrt( (x +k)^2 + (y +1)^2 )

These are the four inequalities. Let's focus on the first one, which is for vertex B(k,1). Squaring:

x² + y² < (x -k)^2 + (y -1)^2 => 0 < -2kx + k² -2y +1 => 2kx + 2y < k² +1.

For large k, this becomes 2kx < k² => x < k/2.

Similarly, for the other vertices, we get similar bounds. For vertex C(k,-1), similar to B, we get 2kx + 2y < k² +1? No, let's check.

For vertex C(k,-1):

sqrt(x² + y²) < sqrt((x -k)^2 + (y +1)^2 )

Squaring: x² + y² < (x -k)^2 + (y +1)^2 => 0 < -2kx + k² + 2y +1 => 2kx - 2y < -k² -1 => 2kx - 2y < -k² -1.

For large k, this is 2kx ≈ -k² => x ≈ -k/2.

Similarly, for vertices A and D, we get x > -k/2 and x <k/2. Therefore, the region closer to O than to any vertex is the set of points where |x| <k/2 and |y| < something?

Wait, but what about y? Let's take the first inequality 2kx + 2y < k² +1. For x <k/2, this is satisfied for any y? No. For example, if x=0, it becomes 2y <k² +1, which is always true since y ≤1 and k is large. Similarly, for x=k/2, it becomes k² + 2y <k² +1 => 2y <1 => y <1/2. So the region closer to O than to B is x <k/2 and y < (k² +1 -2kx)/2. For large k, this is approximately y < (k²)/2 -kx. For x near k/2, y < (k²)/2 -k*(k/2) = (k²)/2 -k²/2 = 0. So the boundary is y=0 at x=k/2. Similarly on the other side, for x=-k/2, the boundary is y=1.

Wait, this suggests that the region closer to O than to any vertex is the central rectangle from x=-k/2 to x=k/2 and y from -1/2 to 1/2? No, because for x=0, the constraint is y < (k² +1)/2, which is always true since y ≤1. Hence, the region is x between -k/2 and k/2, and y between -1 and 1? But this would make the area k * 2 = 2k, which matches our previous calculation. But this would mean the probability is 2k / 4k =1/2, which seems to suggest that for any k, the region is the central rectangle of width k and height 2, which is half the area of the rectangle. But this seems to contradict the intuition for large k, but maybe it's correct?

Wait, if the region is the central rectangle from x=-k/2 to x=k/2 and y from -1 to 1, then for any k, this is indeed half the area of the rectangle (since total area is 2k*2 =4k, and central area is k*2=2k). But why is this the case?

Because the boundaries of the region closer to O are at x=±k/2 and y=±1/2? No, for x=0, the constraint is y < (k² +1)/2, which is always true. But for x=k/2, the constraint is y < 1/2. Similarly, for x=-k/2, y < -1/2. Hence, the region closer to O than to any vertex is the set of points where |x| <k/2 and |y| < 1/2? Wait, no, because for x=0, y can go up to 1. Wait, no, for x=0, the constraint from vertex B is 2*0 + 2y < k² +1 => y < (k² +1)/2, which is always true for y ≤1. Similarly for other vertices. Hence, the only constraints are from the vertices at the ends. For example, for x=k/2, the constraint from vertex B is y < 1/2, and from vertex C is y > -1/2. Hence, the region is x between -k/2 and k/2, and y between -1/2 and 1/2? No, because for x=0, y can be from -1 to 1. Wait, no. Let me think.

For a point (x, y) to be closer to O than to any vertex, it must satisfy all four inequalities. For example, the inequality from vertex B: 2kx + 2y < k² +1, from vertex C: 2kx - 2y < -k² -1, from vertex A: -2kx + 2y < -k² -1, and from vertex D: -2kx - 2y < -k² -1. Let me rewrite these inequalities:

1. 2kx + 2y < k² +1

2. 2kx - 2y < -k² -1

3. -2kx + 2y < -k² -1

4. -2kx - 2y < -k² -1

Let me analyze these inequalities. For large k, let's divide by 2k:

1. x + y/k < k/2 + 1/(2k)

Since y/k is small, this approximates to x < k/2

2. x - y/k < -k/2 - 1/(2k) ≈ -k/2

Which implies x < -k/2 - y/k - 1/(2k) ≈ -k/2 (since y/k is small)

Similarly, the other inequalities:

3. -x + y/k < -k/2 - 1/(2k) ≈ -k/2 => -x < -k/2 => x > k/2

4. -x - y/k < -k/2 - 1/(2k) ≈ -k/2 => -x < -k/2 => x > k/2

Wait, this seems conflicting. Wait, let me do it properly.

Let me take inequality 2: 2kx - 2y < -k² -1 => 2kx < 2y -k² -1. For large k, this implies 2kx is negative, so x is negative. Let me divide by 2k: x - (y)/k < -k/2 - 1/(2k). For large k, x < -k/2 - (small term), so x < -k/2.

Similarly, inequality 3: -2kx + 2y < -k² -1 => -2kx < -k² -1 -2y => 2kx > k² +1 +2y. For large k, this implies x > k/2 + (small term), so x > k/2.

Therefore, combining inequalities 2 and 3, we have x < -k/2 and x > k/2, which is impossible. This suggests a contradiction, which means my earlier conclusion was wrong.

Wait, this indicates that for large k, there is no region satisfying all inequalities, which is impossible because the center (0,0) should satisfy them. Hence, there must be a mistake in interpreting the inequalities.

Let me re-derive the four inequalities for distance to O < distance to each vertex.

For vertex B(k,1): sqrt(x² + y²) < sqrt((x -k)^2 + (y -1)^2)

Squared: x² + y² < x² - 2kx + k² + y² - 2y + 1

Simplifies to 0 < -2kx + k² - 2y + 1 => 2kx + 2y < k² +1 (1)

Similarly for vertex C(k,-1): sqrt(x² + y²) < sqrt((x -k)^2 + (y +1)^2)

Squared: x² + y² < x² - 2kx + k² + y² + 2y + 1

Simplifies to 0 < -2kx + k² + 2y +1 => 2kx - 2y < -k² -1 (2)

For vertex A(-k,1): sqrt(x² + y²) < sqrt((x +k)^2 + (y -1)^2)

Squared: x² + y² < x² + 2kx + k² + y² - 2y + 1

Simplifies to 0 < 2kx + k² - 2y +1 => -2kx - 2y < -k² -1 => 2kx + 2y > k² +1 (3)

For vertex D(-k,-1): sqrt(x² + y²) < sqrt((x +k)^2 + (y +1)^2)

Squared: x² + y² < x² + 2kx + k² + y² + 2y + 1

Simplifies to 0 < 2kx + k² + 2y +1 => -2kx - 2y < -k² -1 => 2kx + 2y > -k² -1 (4)

Wait, this is different from before. Earlier I had inequality (3) as 2kx + 2y > k² +1 and (4) as 2kx + 2y > -k² -1.

So for large k, the inequalities are:

1. 2kx + 2y < k² +1

2. 2kx - 2y < -k² -1

3. 2kx + 2y > k² +1

4. 2kx + 2y > -k² -1

Wait, but this is a contradiction: (1) says 2kx + 2y < k² +1 and (3) says 2kx + 2y > k² +1, which can only be true if 2kx + 2y = k² +1. Similarly for others. This suggests that the only points satisfying all inequalities are on the boundaries, which is impossible. Clearly, there is a mistake here.

Wait, let me re-derivate the inequalities for vertex A.

For vertex A(-k,1):

Distance to O: sqrt(x² + y²)

Distance to A: sqrt((x + k)^2 + (y - 1)^2)

Setting distance to O < distance to A:

x² + y² < (x + k)^2 + (y - 1)^2

Expanding right side: x² + 2kx + k² + y² - 2y + 1

Subtract left side: 0 < 2kx + k² - 2y + 1

Which gives 2kx - 2y + k² + 1 > 0 => 2kx - 2y > -k² -1

Wait, earlier I had 2kx + 2y > k² +1, which was incorrect. Let me correct this.

Original inequality: 0 < 2kx + k² - 2y + 1 => 2kx - 2y > -k² -1. So inequality (3) is 2kx - 2y > -k² -1.

Similarly, for vertex D(-k,-1):

Distance to O < distance to D:

x² + y² < (x + k)^2 + (y + 1)^2

Expanding: x² + y² < x² + 2kx + k² + y² + 2y + 1

Subtract: 0 < 2kx + k² + 2y + 1 => 2kx + 2y > -k² -1. So inequality (4) is 2kx + 2y > -k² -1.

Therefore, the four inequalities are:

1. 2kx + 2y < k² +1

2. 2kx - 2y < -k² -1

3. 2kx - 2y > -k² -1

4. 2kx + 2y > -k² -1

Wait, now this makes more sense. Let me re-express:

Inequality (1): 2kx + 2y < k² +1

Inequality (2): 2kx - 2y < -k² -1

Inequality (3): 2kx - 2y > -k² -1

Inequality (4): 2kx + 2y > -k² -1

Wait, but (2) and (3) are inequalities in the same expression: 2kx - 2y is between -k² -1 and -k² -1? No, (2) is 2kx - 2y < -k² -1 and (3) is 2kx - 2y > -k² -1? That would mean 2kx - 2y = -k² -1, which is a line, not a region. This suggests that there is a mistake in the derivation.

Wait, let me re-derive inequality (3). For vertex A(-k,1):

0 < 2kx + k² - 2y + 1 => 2kx + k² - 2y + 1 > 0 => 2kx - 2y > -k² -1. So this is inequality (3): 2kx - 2y > -k² -1

Similarly, for vertex D(-k,-1), we had 2kx + 2y > -k² -1.

Therefore, the four inequalities are:

1. 2kx + 2y < k² +1

2. 2kx - 2y < -k² -1

3. 2kx - 2y > -k² -1

4. 2kx + 2y > -k² -1

Wait, this seems contradictory. inequalities (2) and (3) are 2kx - 2y < -k² -1 and 2kx - 2y > -k² -1, which can only be true if 2kx - 2y = -k² -1, which is a line. Similarly for (1) and (4): 2kx + 2y < k² +1 and 2kx + 2y > -k² -1. This suggests that the region is between two lines for each pair, but how?

Actually, it seems that the correct inequalities are:

For each vertex, the inequality is a half-plane. For example, for vertex B: 2kx + 2y < k² +1, for vertex A: 2kx - 2y > -k² -1, for vertex C: 2kx - 2y < -k² -1, and for vertex D: 2kx + 2y > -k² -1. Wait, but this would mean:

- From vertex B: 2kx + 2y < k² +1

- From vertex A: 2kx - 2y > -k² -1

- From vertex C: 2kx - 2y < -k² -1

- From vertex D: 2kx + 2y > -k² -1

This is a contradiction unless the inequalities for A and C are swapped. Wait, let me recheck.

For vertex C(k, -1), the inequality was 2kx - 2y < -k² -1. For vertex A(-k,1), the inequality was 2kx - 2y > -k² -1. So these are two inequalities: 2kx - 2y > -k² -1 and 2kx - 2y < -k² -1, which is impossible unless 2kx - 2y = -k² -1. This suggests that there is an error in the derivation.

Wait, let me re-derive the inequality for vertex A(-k,1):

Original inequality: 0 < 2kx + k² - 2y +1 => 2kx - 2y + k² +1 > 0 => 2kx - 2y > -k² -1. Correct.

For vertex C(k, -1): original inequality was 0 < -2kx + k² + 2y +1 => -2kx + 2y > -k² -1 => 2kx - 2y < k² +1. Wait, wait, let me redo this.

For vertex C(k, -1):

0 < -2kx + k² + 2y +1 => -2kx + 2y > -k² -1 => 2kx - 2y < k² +1. Wait, this is different from what I had before. Previously, I had for vertex C: 2kx - 2y < -k² -1, which was incorrect. Let me re-derive:

For vertex C(k, -1):

Distance to O < distance to C:

x² + y² < (x -k)^2 + (y + 1)^2

Expand: x² + y² < x² - 2kx + k² + y² + 2y + 1

Simplify: 0 < -2kx + k² + 2y + 1 => -2kx + 2y > -k² -1 => 2kx - 2y < k² +1.

Yes, that's correct. So for vertex C, the inequality is 2kx - 2y < k² +1. Similarly, for vertex A, it's 2kx - 2y > -k² -1. Hence, the inequalities for the two vertices A and C are:

-2kx - 2y > -k² -1 (from A) and 2kx - 2y < k² +1 (from C). Wait, no, for vertex A, it's 2kx - 2y > -k² -1, and for vertex C, it's 2kx - 2y < k² +1.

Similarly, for vertex B, it's 2kx + 2y < k² +1, and for vertex D, it's 2kx + 2y > -k² -1.

Therefore, the four inequalities are:

1. 2kx + 2y < k² +1 (from B)

2. 2kx + 2y > -k² -1 (from D)

3. 2kx - 2y > -k² -1 (from A)

4. 2kx - 2y < k² +1 (from C)

Therefore, combining these, the region is defined by:

- For 2kx + 2y: between -k² -1 and k² +1

- For 2kx - 2y: between -k² -1 and k² +1

Which, for large k, translates to:

- 2kx + 2y is between -k² -1 and k² +1

- 2kx - 2y is between -k² -1 and k² +1

Dividing by 2k:

For the first pair: x + y/k is between -k/2 - 1/(2k) and k/2 + 1/(2k)

For the second pair: x - y/k is between -k/2 - 1/(2k) and k/2 + 1/(2k)

For large k, this approximates to:

x is between -k/2 and k/2, and y is between -k/2 and k/2. But since the original rectangle has y between -1 and 1, this suggests that for large k, the region is x between -k/2 and k/2, y between -1 and 1, but this contradicts the earlier conclusion.

Wait, no. Let me think of it this way: the inequalities 2kx + 2y < k² +1 and 2kx + 2y > -k² -1 can be rewritten as:

- (k² +1)/2 < kx + y < (k² +1)/2

Similarly for the other pair:

- (k² +1)/2 < kx - y < (k² +1)/2

For large k, this becomes:

- (k²)/2 < kx + y < k²/2

- (k²)/2 < kx - y < k²/2

Let me solve these inequalities for y.

From the first pair:

- k²/2 < kx + y < k²/2 => -k²/2 - kx < y < k²/2 - kx

From the second pair:

- k²/2 < kx - y < k²/2 => -k²/2 - kx < -y < k²/2 - kx => -k²/2 + kx < y < k²/2 + kx

Combining both:

y must satisfy both:

- k²/2 - kx < y < k²/2 - kx

and

- k²/2 + kx < y < k²/2 + kx

For large k, the intersection of these regions is y between max(-k²/2 - kx, -k²/2 + kx) and min(k²/2 - kx, k²/2 + kx).

Since k is positive, -k²/2 - kx < -k²/2 + kx if x > 0, and vice versa.

For x > 0, the lower bound is -k²/2 + kx and the upper bound is k²/2 - kx. For y to satisfy both, we need:

- k²/2 + kx < y < k²/2 - kx

Which requires that -k²/2 + kx < k²/2 - kx => -k²/2 - k²/2 < -kx -kx => -k² < -2kx => k² > 2kx => x < k/2

Similarly for x < 0, the lower bound is -k²/2 - kx and the upper bound is k²/2 + kx, and this requires x > -k/2.

Therefore, the region is x between -k/2 and k/2, and for each x, y is between -k²/2 + kx and k²/2 - kx. For large k, this is approximately y between -k²/2 + kx and k²/2 - kx. But since the rectangle has y between -1 and 1, the actual bounds are y between -1 and 1, but only where -k²/2 + kx < 1 and k²/2 - kx > -1.

For x between -k/2 and k/2, let's find the range of y:

For the upper bound: k²/2 - kx > -1 => k²/2 > kx -1. For x < k/2, kx < k²/2, so kx -1 < k²/2 -1. Hence, k²/2 - kx > k²/2 - (k²/2 -1) =1. So the upper bound is always >1, but since y ≤1, it is effectively 1.

Similarly, the lower bound: -k²/2 + kx > -1 => kx > -1 + k²/2. For x > -k/2, kx > -k²/2. So -k²/2 + kx > -k²/2 + (-k²/2) = -k², but this is always greater than -1 for large k. Hence, the lower bound is effectively -1.

Therefore, for x between -k/2 and k/2, the region is y between -1 and 1, which is the entire central strip. Hence, the area is 2k (width from -k/2 to k/2) * 2 (height from -1 to 1) = 4k._WAIT, but this is the entire area of the rectangle! This is impossible. Clearly, there is a mistake.

Wait, no. The region defined by the inequalities is the entire rectangle? That can't be. For example, take a point near the corner (k,1). It should not be in the region. But according to this, it is. Hence, there is a fundamental error in the derivation.

This suggests that my approach is incorrect, and I need to find another way.

Let me return to the original correct calculation for k=1, where the area is 2, and the probability is 1/2. For k=1, it's correct. For k=2, if I calculate the area using the correct method, what do I get?

For k=2, the area in first quadrant is k/2 =1. Total area is 2k =4. Total rectangle area is 4k=8. Probability is 4/8=1/2. But let's check by integrating.

For k=2, the region in first quadrant is the set of points where distance to O < distance to B and C.

We had earlier, for k=2, the area in first quadrant is 1. Let's verify by integrating.

The region in first quadrant is:

For x from 0 to x1 = (k² -1)/(2k) = (4 -1)/4 = 3/4, y from 0 to 1.

For x from 3/4 to x2 = (k² +1)/(2k) = 5/4, y from 0 to y_upper = (k² +1)/2 -kx = (5/2) - 2x.

So area is:

A1 = ∫₀³/4 1 dx + ∫³/4^5/4 (5/2 - 2x) dx

First integral: 3/4

Second integral: [ (5/2)x - x² ] from 3/4 to 5/4

At 5/4: (5/2)*(5/4) - (5/4)^2 = 25/8 - 25/16 = 25/16

At 3/4: (5/2)*(3/4) - (3/4)^2 = 15/8 - 9/16 = 30/16 - 9/16 = 21/16

Difference: 25/16 - 21/16 = 4/16 = 1/4

Total A1 = 3/4 + 1/4 = 1. Which matches k/2 = 2/2 =1. Hence, for k=2, the area in first quadrant is 1, total area 4, probability 4/8=1/2.

But for k=2, the region is the central rectangle from x=-1 to x=1 and y=-1 to 1? No, because for k=2, the rectangle is from x=-2 to 2, y=-1 to 1. The area closer to O is 4, which is half of 8. This suggests that for any k, the region is half the area of the rectangle. This is surprising but mathematically consistent with the calculations for k=1 and k=2. Hence, perhaps the probability is always 1/2, regardless of k.

But why? Intuitively, for a rectangle, the set of points closer to the center than to any vertex is always half the area? This seems counterintuitive for very long rectangles, but mathematically, it seems to hold.

Let me think about symmetry. The set of points closer to O than to any vertex is symmetric with respect to the center. If we consider the rectangle and its complement, perhaps there is some symmetry.

Alternatively, consider that the region closer to O than to any vertex is the intersection of four regions, each defined by being closer to O than to a vertex. For each vertex, the region closer to O is a half-plane, and the intersection is a convex region. But why its area is always half the rectangle?

Perhaps there is a general theorem here, but I'm not sure. However, given that it works for k=1 and k=2, and the calculations are consistent, I might have to accept that the probability is 1/2 for any k.

But the problem states AB/AD =k, and asks for the probability. If the answer is always 1/2, then the probability is 1/2, and the answer is \boxed{\dfrac{1}{2}}.

But I need to confirm with another value. Let's take k=0.5, which is a rectangle width 1 and height 2, so it's taller than it is wide.

For k=0.5, the rectangle is from x=-0.5 to 0.5, y=-1 to 1. The area is 1*2=2. According to our formula, the region closer to O is 2k=1, hence probability 1/2. Let's verify.

For k=0.5, the region closer to O than to any vertex is the set of points where distance to O < distance to any vertex.

For example, take a point (0,0). It's closer to O. Take a point (0,1). Distance to O is 1, distance to A(-0.5,1) is sqrt((0 +0.5)^2 + (1 -1)^2) = 0.5 < 1. So (0,1) is closer to A, hence not in the region. Hence, the region is not the entire rectangle. But according to our calculation, the area is 1, which is half of 2. Let's calculate it.

For k=0.5, the area in first quadrant is k/2 =0.5/2=0.25. Total area 4*0.25=1. Probability 1/2.

To verify, let's calculate the area in first quadrant.

For k=0.5, x1 = (k² -1)/(2k) = (0.25 -1)/1 = -0.75. But since we are in first quadrant x ≥0, x1 is negative, so the first part is from x=0 to x=x1 (which is 0 to -0.75, but x ≥0, so no area. Hence, A1 = ∫_{0}^{x2} [ (k² +1)/2 -kx ] dx, where x2 = (k² +1)/(2k) = (0.25 +1)/1 = 1.25. But the rectangle in first quadrant is x from 0 to 0.5, y from 0 to 1. Hence, the upper limit is min(x2, 0.5) = 0.5.

Therefore, A1 = ∫₀^0.5 [ (0.25 +1)/2 - 0.5x ] dx = ∫₀^0.5 [ 1.25/2 - 0.5x ] dx = ∫₀^0.5 (0.625 - 0.5x) dx = [0.625x - 0.25x²] from 0 to 0.5 = 0.625*0.5 - 0.25*(0.25) = 0.3125 - 0.0625 = 0.25, which matches k/2 =0.5/2=0.25. Hence, total area is 4*0.25=1, probability 1/2. This seems to hold.

Thus, despite the initial intuition, it appears the probability is always 1/2, regardless of k. This is surprising, but the calculations for specific cases support it. Hence, the answer is 1/2.

**Final Answer**
\boxed{\dfrac{1}{2}}

To determine the probability that a randomly chosen point inside rectangle $ABCD$ is closer to the center $O$ than to any of the four vertices, we analyze the geometric condition that defines this region.

---

### Setup and Geometry

- **Rectangle Setup**: Place rectangle $ABCD$ with center at the origin $O(0,0)$. Let the rectangle have sides of length $2k$ (horizontal) and $2$ (vertical), resulting in:
  - Vertices: $A = (-k, 1)$, $B = (k, 1)$, $C = (k, -1)$, $D = (-k, -1)$
  - Total area of the rectangle: $A = 2k \cdot 2 = 4k$

- We are to find the area of the region within the rectangle where the distance from a point $(x, y)$ to the center $O$ is *less than* the distance from that point to any of the four vertices $A$, $B$, $C$, and $D$.

This condition defines a set that is the **intersection** of four regions, each region being the set of points closer to $O$ than to a specific vertex. So the desired region is the **intersection** of four half-planes defined by perpendicular bisectors between $O$ and each vertex.

---

### Key Insight: Symmetry and Area

Due to the symmetry of the rectangle about the center and the equal contribution of each quadrant, we can compute the area of the region in one quadrant (say, the first quadrant where $x \geq 0$, $y \geq 0$) and multiply by 4.

Using symmetry and analytical geometry, we determine that in each quadrant, the region of points closer to the center than to any vertex is a shape bounded by two perpendicular bisectors. Integrating this region in each quadrant leads to the calculation of the total area of the region closer to the center than any vertex.

After detailed calculation (and verified for several cases including $k = 1, 2, 0.5$), it turns out that the **total area of this region is**:

$$
\text{Desired Region Area} = 2k
$$

Since the **total area of the rectangle** is $4k$, the probability that a randomly chosen point is closer to the center than to any vertex is:

$$
P = \frac{\text{Desired Region Area}}{\text{Total Area}} = \frac{2k}{4k} = \frac{1}{2}
$$

---

### Final Conclusion

Regardless of the value of $k$, the probability that a point randomly chosen from the interior of rectangle $ABCD$ is closer to its center than to any of its vertices is:

$$
\boxed{\dfrac{1}{2}}
$$

Solution:
The original rectangle may be subdivided into four smaller congruent rectangles, all sharing $O$ as a vertex. Each of these rectangles is analogous, so we can consider our random point $P$ to be without loss of generality in the smaller rectangle with $A$ as a vertex. All points in this smaller rectangle are closer to $A$ than they are to $B$, $C$, or $D$, so we just need to determine the probability that $OP<AP$. [asy]
size(100);
draw((0,0)--(0,100)--(-250,100)--(-250,0)--cycle);
label("$A$",(-250,100),NW); label("$O$",(0,0),SE);
draw((-105,100)--(-145,0));
fill((-105,100)--(-145,0)--(0,0)--(0,100)--cycle, gray(.7));
[/asy] Since a $180^\circ$ rotation about the center of the smaller rectangle takes $O$ to $A$, it takes the shaded region to the unshaded region. Therefore, exactly half the area is shaded, and the overall probability is $\boxed{\frac{1}{2}}$, independent of $k$.