Explanation:
Okay, so I need to find the total area of the shaded region in this figure made up of twelve 1x1 squares. Let me try to visualize the Asymptote code they provided.

First, the Asymptote code draws a grid. The vertical lines are from j=0 to j=4, so that's 4 vertical lines, which would make 3 columns? Wait, no. Wait, if you have vertical lines at x=0,1,2,3, that's 4 vertical lines, creating 3 columns? Wait, no. Wait, if you have vertical lines from x=0 to x=3, with increments of 1, then there are 4 vertical lines (at x=0,1,2,3) which divide the rectangle into 3 columns. Similarly, the horizontal lines are from j=0 to j=5, so that's 5 horizontal lines (at y=0,1,2,3,4), creating 4 rows. So the entire figure is a rectangle that's 3 units wide (since there are 3 columns) and 4 units tall (since there are 4 rows). Therefore, the total area is 3*4=12, which matches the problem statement of twelve 1x1 squares.

Now, the shaded regions are filled with gray. The Asymptote code has two fill commands. Let me parse them:

First fill: (0,4)--(0,0)--(2,0)--cycle. So that's a triangle with vertices at (0,4), (0,0), and (2,0). Let me sketch this mentally. Starting at the top-left corner (0,4), going down to (0,0), which is the bottom-left corner, then moving right to (2,0), which is two units to the right along the bottom, and then back to (0,4). So this is a right triangle with base along the bottom from (0,0) to (2,0), which is length 2, and height from (0,0) up to (0,4), which is 4 units. Wait, but the triangle is formed by connecting (0,4) to (0,0) to (2,0). So actually, the base is from (0,0) to (2,0), which is 2 units, and the height is from (0,0) up to (0,4), which is 4 units. But since it's a triangle, the area would be (base * height)/2. Wait, but in this case, the triangle is a right triangle with legs of length 2 and 4? Wait, no. Wait, if you connect (0,4) to (0,0) to (2,0), then the triangle has vertices at (0,4), (0,0), and (2,0). So the base is from (0,0) to (2,0), which is 2 units, and the height is the vertical distance from that base to the point (0,4). Wait, but since the base is horizontal, the height would be the vertical distance from the base to the opposite vertex. However, the opposite vertex here is (0,4), which is directly above (0,0). So the height is 4 units. Therefore, the area is (base * height)/2 = (2 * 4)/2 = 4. So that triangle has area 4.

Wait, but hold on. Alternatively, since the triangle is formed by the points (0,4), (0,0), (2,0), maybe I can think of it as a right triangle with legs of length 4 and 2? Wait, if you consider the right angle at (0,0), then yes, the legs would be along the y-axis from (0,0) to (0,4) which is 4 units, and along the x-axis from (0,0) to (2,0) which is 2 units. Therefore, the area is indeed (4*2)/2 = 4. So that first shaded region is 4 square units.

Then the second fill command is (0,4)--(3,4)--(3,0)--cycle. So that's another triangle. Let me parse the coordinates. Starting at (0,4), which is the top-left corner, then going to (3,4), which is three units to the right along the top edge, then down to (3,0), which is the bottom-right corner of the entire figure, and then back to (0,4). So this is a large triangle with vertices at (0,4), (3,4), and (3,0). So this is a right triangle as well, with legs along the top edge from (0,4) to (3,4) which is 3 units, and along the right edge from (3,4) down to (3,0) which is 4 units. Therefore, the area is (3*4)/2 = 6. So that triangle has area 6.

Therefore, the total shaded area would be 4 + 6 = 10? Wait, but hold on. Wait, the problem says "the shaded region," but maybe there is overlapping? Wait, the Asymptote code fills two regions: the first triangle and the second triangle. Are these two triangles overlapping? Let me check.

First triangle: from (0,4) to (0,0) to (2,0). Second triangle: from (0,4) to (3,4) to (3,0). So the first triangle is on the left side, from the top-left corner down to the bottom-left corner and then over to (2,0). The second triangle is from the top-left corner to the top-right corner (3,4) and then down to the bottom-right corner (3,0). So these two triangles share the line from (0,4) to (3,0)? Wait, no. Wait, the first triangle is (0,4), (0,0), (2,0). The second triangle is (0,4), (3,4), (3,0). So they both have the point (0,4) and both have a line from (0,4) to (3,0)? Wait, no. The first triangle has a line from (0,4) to (0,0) to (2,0). The second triangle has a line from (0,4) to (3,4) to (3,0). So the only overlapping region would be if they share any common area. Let me think. The first triangle is in the left part of the figure, from x=0 to x=2 on the bottom, and the second triangle is from x=0 to x=3 on the top. So maybe they overlap somewhere? Let me check coordinates.

The first triangle is bounded by x from 0 to 2 on the bottom, and the second triangle is bounded by x from 0 to 3 on the top. However, the first triangle is below the line from (0,4) to (2,0), and the second triangle is above the line from (0,4) to (3,0). Wait, actually, the second triangle is the triangle formed by (0,4), (3,4), (3,0). So that triangle includes the area from the top edge down to the right edge. The first triangle is from the left edge down to the bottom edge. So maybe they overlap in some region? Let me check if there's an overlapping area.

Let me think of the two triangles. The first triangle is the area under the line from (0,4) to (2,0). The second triangle is the area under the line from (0,4) to (3,0). Wait, actually, the second triangle is the area bounded by the top edge from (0,4) to (3,4), the right edge from (3,4) to (3,0), and the diagonal from (3,0) back to (0,4). So the second triangle is a large triangle that covers the entire right side of the figure. The first triangle is a smaller triangle on the left side. So maybe they don't overlap? Wait, but the first triangle goes from (0,4) down to (0,0) and then to (2,0). The second triangle goes from (0,4) to (3,4) to (3,0). So the only overlapping point is at (0,4) and maybe along the line from (0,4) to (3,0)? Wait, but the first triangle is only up to x=2 on the bottom. The second triangle includes the entire right side. So maybe they don't overlap? Let me check if there's any overlapping area.

Suppose we have two triangles: one is the triangle with vertices (0,4), (0,0), (2,0), and the other is (0,4), (3,4), (3,0). The only common point is (0,4). The first triangle is entirely on the left side, and the second triangle is on the right side. Wait, but actually, the second triangle includes the area from (0,4) to (3,4) to (3,0). So if you imagine that triangle, it's a large triangle that covers from the top-left corner to the top-right corner to the bottom-right corner. The first triangle is from the top-left corner down to the bottom-left corner and then over to (2,0). So maybe they share the line from (0,4) to (something)? Wait, the first triangle has a line from (0,4) to (0,0), and the second triangle has a line from (0,4) to (3,0). So these two lines intersect at (0,4) and maybe somewhere else? Wait, the line from (0,4) to (0,0) is the left edge, and the line from (0,4) to (3,0) is a diagonal. These two lines only meet at (0,4). Therefore, the two triangles only share the point (0,4), so there is no overlapping area. Therefore, the total shaded area is 4 + 6 = 10. But wait, the entire figure is 12 squares, so 10 shaded? That seems possible, but let me check again.

Wait, but maybe I made a mistake in calculating the areas. Let me verify again.

First triangle: (0,4), (0,0), (2,0). So this is a right triangle with base 2 and height 4. Area is (2*4)/2 = 4. Correct.

Second triangle: (0,4), (3,4), (3,0). This is a right triangle with base 3 and height 4. Area is (3*4)/2 = 6. Correct. So total shaded area is 4 + 6 = 10. But let me check if the Asymptote code draws these two triangles. Wait, the Asymptote code also draws diagonals: draw((0, 4)--(3, 0)); draw((0, 4)--(2, 0)); So those are the diagonals mentioned. So the first triangle is formed by (0,4)--(0,0)--(2,0)--cycle, which is the triangle with vertices at those points. The second triangle is (0,4)--(3,4)--(3,0)--cycle. So that's correct.

But let me think again. Maybe there's an overlap? Wait, if you imagine the figure, the first triangle is on the left side, and the second triangle is on the right side. But maybe the line from (0,4) to (3,0) is drawn, which is the hypotenuse of the second triangle. Also, the line from (0,4) to (2,0) is drawn, which is the hypotenuse of the first triangle. So these two lines intersect somewhere? Wait, the line from (0,4) to (3,0) and the line from (0,4) to (2,0) both start at (0,4). So they only meet at (0,4). Therefore, there is no overlapping area between the two triangles. Therefore, adding their areas is correct. So total shaded area is 4 + 6 = 10. But wait, the problem says "the shaded region," but maybe there's something else?

Wait, but let me check the Asymptote code again. The first fill is (0,4)--(0,0)--(2,0)--cycle. So that's a triangle. The second fill is (0,4)--(3,4)--(3,0)--cycle. So that's another triangle. So the shaded regions are two triangles. But maybe there's another shaded region? Wait, the Asymptote code only has two fill commands. So maybe the answer is 10? But let me think again. Wait, the entire figure is 12 squares. If the shaded area is 10, that would mean 10/12 = 5/6 of the area is shaded. But maybe I need to check again.

Alternatively, maybe I need to calculate the area using coordinates. Let me try to compute the area of each triangle again.

First triangle: points (0,4), (0,0), (2,0). Using the formula for the area of a triangle with coordinates. The formula is |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|. Let me apply that.

Let me label the points:

A(0,4), B(0,0), C(2,0).

So plugging into the formula:

Area = |(0*(0 - 0) + 0*(0 - 4) + 2*(4 - 0))/2|

= |0 + 0 + 2*4 / 2| = |8/2| = 4. Correct.

Second triangle: points (0,4), (3,4), (3,0). Let me call them D(0,4), E(3,4), F(3,0).

Area = |0*(4 - 0) + 3*(0 - 4) + 3*(4 - 4)| / 2

Wait, let me use the same formula:

Area = |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| / 2

So plugging in:

x1 = 0, y1 = 4

x2 = 3, y2 = 4

x3 = 3, y3 = 0

So:

Area = |0*(4 - 0) + 3*(0 - 4) + 3*(4 - 4)| / 2

= |0 + 3*(-4) + 3*(0)| / 2

= | -12 + 0 | / 2 = 12 / 2 = 6. Correct. So that's 6. So total area is 4 + 6 = 10.

But let me think again. The problem says "the shaded region," but maybe there's an overlap? Wait, if I imagine the figure, the first triangle is on the left side, and the second triangle is on the right side. However, the line from (0,4) to (3,0) is drawn, which is the hypotenuse of the second triangle. Also, the line from (0,4) to (2,0) is drawn, which is the hypotenuse of the first triangle. So maybe these two lines intersect somewhere? Wait, but they both start at (0,4). The first line goes to (2,0), and the second goes to (3,0). So they don't intersect anywhere else. Therefore, there is no overlapping area. Therefore, the total shaded area is indeed 10.

But wait, the problem says "twelve 1 by 1 squares form a rectangle," so the total area is 12. If the shaded area is 10, that would mean 10 units. But maybe I need to check if the figure is 3x4? Wait, the grid is drawn with vertical lines from j=0 to j=4 (so x=0,1,2,3) and horizontal lines from j=0 to j=5 (so y=0,1,2,3,4). Wait, hold on. Wait, the vertical lines are from j=0 to j=4, so that's 4 vertical lines, which would make 3 columns (since each column is between two vertical lines). Similarly, horizontal lines from j=0 to j=5, which is 5 horizontal lines, making 4 rows. Therefore, the rectangle is 3 units wide (columns) and 4 units tall (rows). Therefore, the total area is 3*4=12, which matches. So each square is 1x1, so the entire figure is 3x4.

But then, if the shaded area is 10, that would be 10 square units. But maybe I need to check if the Asymptote code actually shades those regions. Wait, the Asymptote code draws two triangles. Let me think again. The first triangle is from (0,4) to (0,0) to (2,0). So that's a triangle that covers part of the left side. The second triangle is from (0,4) to (3,4) to (3,0). So that's a triangle that covers the right side. But maybe there's another shaded region? Wait, the Asymptote code only has two fill commands. So maybe the answer is 10? But let me think again.

Alternatively, maybe I need to count the squares? Wait, but the figure is made of 1x1 squares, but the shaded regions are triangles, so maybe I need to calculate the area of the triangles. But since the triangles are not aligned with the grid, maybe there's a different way. Wait, but I already calculated the area using coordinates, and it's 4 + 6 = 10.

Alternatively, maybe the figure is different? Let me try to imagine the Asymptote code. The grid is 3 columns (x=0 to 3) and 4 rows (y=0 to 4). The first shaded triangle is from (0,4) down to (0,0) and then to (2,0). So that triangle is in the left half of the bottom part. The second shaded triangle is from (0,4) to (3,4) to (3,0). So that's a large triangle covering the top and right side.

Alternatively, maybe there's an overlap? Wait, if I imagine the two triangles, the first triangle is on the left, and the second is on the right. But maybe the line from (0,4) to (2,0) is part of the first triangle, and the line from (0,4) to (3,0) is part of the second triangle. These two lines intersect at (0,4) and maybe somewhere else? Wait, no. The line from (0,4) to (2,0) has a slope of (0 - 4)/(2 - 0) = -2. The line from (0,4) to (3,0) has a slope of (0 - 4)/(3 - 0) = -4/3. These two lines only intersect at (0,4). Therefore, there is no overlapping area. Therefore, the total shaded area is indeed 4 + 6 = 10.

But let me check again. Maybe I made a mistake in interpreting the coordinates. Let me think of the figure. The first triangle is from the top-left corner (0,4) down to the bottom-left corner (0,0), then over to (2,0). So that triangle is in the left part of the figure, covering from column 0 to column 2 on the bottom. The second triangle is from the top-left corner (0,4) to the top-right corner (3,4) to the bottom-right corner (3,0). So that triangle covers the entire right side of the figure.

But if I imagine the entire figure, the shaded regions are two triangles. However, maybe there's another shaded region? Wait, the Asymptote code only has two fill commands. So maybe the answer is 10. But let me check with another approach.

Alternatively, maybe I can calculate the area by counting the squares. But since the shaded regions are triangles, not aligned with the grid, it's more complicated. However, maybe I can divide the triangles into smaller parts.

First triangle: from (0,4) to (0,0) to (2,0). This triangle can be divided into squares? Let me think. From (0,0) to (2,0) is two units. The height is 4 units. If I imagine this triangle, it's a right triangle with base 2 and height 4. If I divide it into unit squares, how many squares are covered? Well, the triangle is not aligned with the grid, so it's not straightforward. However, since we already calculated the area as 4, which is 4 square units, and since each square is 1x1, that would mean that the area is 4. Similarly, the second triangle is 6 square units.

Alternatively, maybe I can use coordinate geometry to find the area. For the first triangle, coordinates (0,4), (0,0), (2,0). The area is 4. For the second triangle, coordinates (0,4), (3,4), (3,0). The area is 6. So total area is 10.

But let me check once more. Maybe the figure is different? Wait, the Asymptote code draws the grid with vertical lines from j=0 to j=4 (so x=0,1,2,3) and horizontal lines from j=0 to j=5 (so y=0,1,2,3,4). Therefore, the rectangle is 3 units wide and 4 units tall. The diagonals are drawn from (0,4) to (3,0) and from (0,4) to (2,0). The shaded regions are the two triangles mentioned.

Alternatively, maybe the problem is in the Asymptote code? Let me check the Asymptote code again:

// Filled portions
fill((0, 4)--(0, 0)--(2, 0)--cycle, gray(0.75));
fill((0, 4)--(3, 4)--(3, 0)--cycle, gray(0.75));

So the first fill is a triangle with vertices at (0,4), (0,0), (2,0). The second fill is a triangle with vertices at (0,4), (3,4), (3,0). So that's correct.

Alternatively, maybe the figure is not 3x4? Wait, the grid is drawn with vertical lines from x=0 to x=3 (since for j from 0 to 4, draw (j, 0)--(j, 4)), so vertical lines at x=0,1,2,3. So the width is 3 units. Horizontal lines from y=0 to y=4, with j from 0 to 5, so horizontal lines at y=0,1,2,3,4. Therefore, the height is 4 units. So the rectangle is 3x4.

Therefore, the total area is 12. The shaded area is 10. But maybe the answer is different? Wait, maybe I need to check if the triangles are overlapping? Wait, if I imagine the two triangles, the first triangle is on the left side, and the second triangle is on the right side. However, the line from (0,4) to (3,0) passes through the first triangle? Wait, the line from (0,4) to (3,0) is the hypotenuse of the second triangle. The first triangle has a hypotenuse from (0,4) to (2,0). So the line from (0,4) to (3,0) passes through the first triangle? Let me check.

The line from (0,4) to (3,0) has the equation y = (-4/3)x + 4. The line from (0,4) to (2,0) has the equation y = -2x + 4. So these two lines intersect at (0,4) and... Wait, they both start at (0,4). The line y = -2x + 4 goes from (0,4) to (2,0). The line y = -4/3 x + 4 goes from (0,4) to (3,0). So they don't intersect anywhere else. Therefore, the two triangles only share the point (0,4). Therefore, no overlapping area.

Therefore, the total shaded area is indeed 4 + 6 = 10. But let me check once more. Maybe the problem is in the Asymptote code? The Asymptote code draws the diagonals, but maybe those diagonals are part of the shaded regions? Wait, the fill commands are for the triangles, and the diagonals are drawn separately. The diagonals are just lines, not shaded.

Alternatively, maybe I need to check if the shaded regions are actually different. For example, maybe the first triangle is not entirely shaded? Wait, the first fill command is (0,4)--(0,0)--(2,0)--cycle, which is a triangle. The second fill command is (0,4)--(3,4)--(3,0)--cycle, another triangle. So those are two distinct triangles.

Alternatively, maybe the figure is different? Let me try to imagine the Asymptote code. The grid is 3 columns (width 3) and 4 rows (height 4). The first shaded triangle is from the top-left corner down to the bottom-left corner and then to (2,0). So that triangle is in the left part of the figure. The second shaded triangle is from the top-left corner to the top-right corner to the bottom-right corner. So that triangle covers the entire right side.

But if I imagine the entire figure, the shaded area would be the left triangle (area 4) and the right triangle (area 6), totaling 10. However, since the entire figure is 12, maybe the answer is 10? But I need to make sure.

Alternatively, maybe I can calculate the area by subtracting unshaded regions? Let me try that.

Total area is 12. If I can find the unshaded area, then subtract from 12.

But how? The unshaded area would be the rest of the figure. However, without seeing the figure, it's a bit challenging, but maybe I can deduce it.

Alternatively, maybe I can think of the two shaded triangles. The first triangle is 4, the second is 6. So total shaded is 10.

Alternatively, maybe the answer is different. Wait, let me think again. Maybe the first triangle is not 4? Wait, if the triangle is from (0,4) to (0,0) to (2,0), then the area is ½ * base * height. The base is 2 units (from (0,0) to (2,0)), and the height is 4 units (from (0,0) up to (0,4)). Wait, but in a triangle, the base and height must be perpendicular. If I take the base as 2 units along the x-axis, then the height is the vertical distance from that base to the opposite vertex, which is (0,4). The vertical distance from the base (which is along the x-axis from (0,0) to (2,0)) to the point (0,4) is indeed 4 units. Therefore, area is ½ * 2 * 4 = 4. Correct.

Similarly, the second triangle has base 3 units (from (0,4) to (3,4)) and height 4 units (from (3,4) down to (3,0)), so area is ½ * 3 * 4 = 6. Correct.

Therefore, total shaded area is 4 + 6 = 10.

But let me check once more with another method. For example, using coordinates to calculate the area of each triangle.

First triangle: points (0,4), (0,0), (2,0). Using the shoelace formula:

List the coordinates:

(0,4), (0,0), (2,0), (0,4)

Compute shoelace sum:

Sum1 = (0*0 + 0*0 + 2*4) = 0 + 0 + 8 = 8

Sum2 = (4*0 + 0*2 + 0*0) = 0 + 0 + 0 = 0

Area = |Sum1 - Sum2| / 2 = |8 - 0| / 2 = 4. Correct.

Second triangle: points (0,4), (3,4), (3,0), (0,4)

Shoelace formula:

Sum1 = 0*4 + 3*0 + 3*4 = 0 + 0 + 12 = 12

Sum2 = 4*3 + 4*3 + 0*0 = 12 + 12 + 0 = 24

Area = |12 - 24| / 2 = 12 / 2 = 6. Correct.

Therefore, total area is 4 + 6 = 10.

Therefore, I think the answer is 10. However, since the problem says "the shaded region," maybe there's something I'm missing? Wait, maybe the figure has overlapping shaded regions? But we already checked that there is no overlapping.

Alternatively, maybe the Asymptote code draws something else? Let me check the Asymptote code again:

// Filled portions
fill((0, 4)--(0, 0)--(2, 0)--cycle, gray(0.75));
fill((0, 4)--(3, 4)--(3, 0)--cycle, gray(0.75));

// grid
int j;
for (j = 0; j < 4; ++j) {draw((j, 0)--(j, 4));}
for (j = 0; j < 5; ++j) {draw((0, j)--(3, j));}

//diagonals
draw((0, 4)--(3, 0)); draw((0, 4)--(2, 0));

So the grid is drawn, then the two shaded triangles, then the diagonals. The diagonals are drawn, but they are not shaded. So the shaded regions are only the two triangles.

Therefore, I think my calculation is correct. The total shaded area is 10. However, since the problem is from an Olympiad or similar, maybe the answer is different? Wait, let me think again.

Wait, the entire figure is 3x4=12. If the shaded area is 10, then the unshaded area is 2. But maybe that's possible? Let me check.

Looking at the figure, the unshaded area would be the parts not covered by the two triangles. The first triangle covers area 4, the second covers 6. But maybe there's some overlapping? Wait, no.

Alternatively, maybe the figure is different. For example, maybe the first triangle is not from (0,4) to (0,0) to (2,0), but something else? Wait, the Asymptote code says fill((0, 4)--(0, 0)--(2, 0)--cycle, gray(0.75)); which is exactly that triangle.

Alternatively, maybe the figure is 4x3? Wait, no, the grid is 3 columns and 4 rows.

Alternatively, maybe I need to check if the triangles are actually in the figure. For example, the first triangle is from (0,4) down to (0,0) and then to (2,0). So that triangle is in the left part of the figure. The second triangle is from (0,4) to (3,4) to (3,0). So that triangle is in the right part.

But if I imagine the entire figure, the shaded area would be those two triangles. Therefore, 4 + 6 = 10.

Alternatively, maybe the answer is 6? Wait, no. Because the first triangle is 4.

Alternatively, maybe I made a mistake in the coordinates? For example, maybe the first triangle is not 2 units wide? Wait, from (0,0) to (2,0) is 2 units. Yes.

Alternatively, maybe the problem is in the Asymptote code's coordinate system? For example, maybe the y-axis is inverted? In Asymptote, the coordinate system usually has (0,0) at the bottom-left, but sometimes it can be different. Wait, in the Asymptote code, they draw vertical lines from (j, 0) to (j, 4), so for j from 0 to 3, which would be vertical lines at x=0,1,2,3, going from y=0 to y=4. Similarly, horizontal lines from (0, j) to (3, j), for j from 0 to 4, so horizontal lines at y=0,1,2,3,4, going from x=0 to x=3. Therefore, the coordinate system is standard, with (0,0) at the bottom-left, increasing y upwards. Therefore, the point (0,4) is the top-left corner, (3,0) is the bottom-right corner.

Therefore, the triangles are correctly calculated.

Therefore, I think my answer is correct. The total shaded area is 10. However, since the problem says "the shaded region," and the answer is 10, but maybe I need to check once more.

Wait, let me think of the figure as a 3x4 grid. The first triangle covers part of the left side. Let me try to count the number of squares it covers.

The triangle from (0,4) to (0,0) to (2,0). This triangle would cover parts of the squares. For example, from column 0 (x=0 to 1) and column 1 (x=1 to 2). In terms of rows, from y=0 to y=4.

But since it's a triangle, it's not straightforward to count squares. However, since we already calculated the area as 4, which is 4 square units, and since each square is 1x1, that would mean that the area is 4. Similarly, the second triangle is 6.

Alternatively, maybe the answer is different? Wait, if I think of the entire figure, the two shaded triangles might cover some squares completely and some partially. But since the problem says "total area of the shaded region," which is a continuous area, not counting individual squares, so the area is indeed 10.

Therefore, I think the answer is 10. However, since I'm a bit unsure because sometimes problems can have tricks, but given the calculations, I think 10 is correct.

Wait, but let me check once more. If the entire figure is 3x4=12, and the shaded area is 10, then the unshaded area is 2. Let me see if that makes sense.

The unshaded area would be the parts not covered by the two triangles. The first triangle covers area 4, the second covers 6. But since they don't overlap, total shaded is 10. The unshaded area is 12 - 10 = 2. Where is this unshaded area? It would be the area not covered by the two triangles.

Looking at the figure, the first triangle covers the left side from (0,4) down to (0,0) and over to (2,0). The second triangle covers from (0,4) to (3,4) to (3,0). So the unshaded area would be the area between the two triangles. For example, the area from x=2 to x=3 on the bottom, and the area from x=0 to x=2 on the top? Wait, maybe not.

Alternatively, maybe the unshaded area is a small region. For example, the area between the two triangles. But since the two triangles don't overlap, the unshaded area is the rest.

But regardless, since the math checks out, I think the answer is 10.

But wait, hold on. Wait, the Asymptote code draws two triangles, but maybe there's another shaded region? Let me check the Asymptote code again.

The code has two fill commands:

fill((0, 4)--(0, 0)--(2, 0)--cycle, gray(0.75));

fill((0, 4)--(3, 4)--(3, 0)--cycle, gray(0.75));

So only two shaded regions. Therefore, total shaded area is 4 + 6 = 10.

Therefore, I think the answer is 10. However, since I'm a bit uncertain because sometimes problems can have unexpected overlaps or something, but given the calculations, I think it's safe to go with 10.

Wait, but let me check with another approach. Let me think of the entire figure as a 3x4 grid. The first triangle is from (0,4) to (0,0) to (2,0). This triangle can be divided into smaller triangles or shapes. For example, from (0,4) to (0,0) to (2,0) is a right triangle. If I divide this triangle into unit squares, how many squares are fully or partially covered?

But since the triangle is not aligned with the grid, it's complex. However, since we already calculated the area as 4, which is 4 square units, and since the entire figure is 12, it's plausible.

Alternatively, maybe the answer is 6? If I thought the first triangle was 2 and the second 4? But no, calculations show otherwise.

Alternatively, maybe the problem is in the Asymptote code's coordinate system? For example, if the figure is 4x3 instead of 3x4? But no, the grid is drawn with vertical lines from x=0 to x=3 (so width 3) and horizontal lines from y=0 to y=4 (height 4).

Therefore, I think my answer is correct. The total shaded area is 10.

**Final Answer**
\boxed{10}

To determine the total area of the shaded region in the given figure, we analyze the two shaded triangles described in the Asymptote code.

---

### **Step 1: Understand the Grid and Coordinates**

The figure is a rectangle formed by 12 unit squares, arranged in a 3×4 grid. This means the rectangle has:

- **Width**: 3 units (from x = 0 to x = 3)
- **Height**: 4 units (from y = 0 to y = 4)

The Asymptote code draws:

- **Vertical lines** at x = 0, 1, 2, 3
- **Horizontal lines** at y = 0, 1, 2, 3, 4

This confirms the rectangle is 3 units wide and 4 units tall, with a total area of:

$$
\text{Total Area} = 3 \times 4 = 12 \text{ square units}
$$

---

### **Step 2: Analyze the Shaded Regions**

There are two shaded triangles:

#### **Triangle 1:**
- Vertices: (0, 4), (0, 0), (2, 0)
- This is a **right triangle** with:
  - **Base** = 2 units (from (0, 0) to (2, 0))
  - **Height** = 4 units (from (0, 0) to (0, 4))

$$
\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 4 = 4 \text{ square units}
$$

#### **Triangle 2:**
- Vertices: (0, 4), (3, 4), (3, 0)
- This is also a **right triangle** with:
  - **Base** = 3 units (from (0, 4) to (3, 4))
  - **Height** = 4 units (from (3, 4) to (3, 0))

$$
\text{Area}_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6 \text{ square units}
$$

---

### **Step 3: Check for Overlap**

The two triangles share only the point (0, 4) and do not overlap in any other region. Therefore, their areas can be **added directly**.

---

### **Step 4: Final Calculation**

$$
\text{Total Shaded Area} = \text{Area}_1 + \text{Area}_2 = 4 + 6 = \boxed{10}
$$

---

### ✅ **Final Answer**

$$
\boxed{10}
$$

[Assistant]
Explanation:
Okay, so I need to find the total area of the shaded region in this figure made up of twelve 1x1 squares. Let me try to visualize the Asymptote code they provided.

First, the Asymptote code draws a grid. The vertical lines are from j=0 to j=4, so that's 4 vertical lines, which would make 3 columns? Wait, no. Wait, if you have vertical lines at x=0,1,2,3, that's 4 vertical lines, creating 3 columns? Wait, no. Wait, if you have vertical lines from x=0 to x=3, with increments of 1, then there are 4 vertical lines (at x=0,1,2,3) which divide the rectangle into 3 columns. Similarly, the horizontal lines are from j=0 to j=5, so that's 5 horizontal lines (at y=0,1,2,3,4), creating 4 rows. So the entire figure is a rectangle that's 3 units wide (since there are 3 columns) and 4 units tall (since there are 4 rows). Therefore, the total area is 3*4=12, which matches the problem statement of twelve 1x1 squares.

Now, the shaded regions are filled with gray. The Asymptote code has two fill commands. Let me parse them:

First fill: (0,4)--(0,0)--(2,0)--cycle. So that's a triangle with vertices at (0,4), (0,0), and (2,0). Let me sketch this mentally. Starting at the top-left corner (0,4), going down to (0,0), which is the bottom-left corner, then moving right to (2,0), which is two units to the right along the bottom, and then back to (0,4). So this is a right triangle with base along the bottom from (0,0) to (2,0), which is length 2, and height from (0,0) up to (0,4), which is 4 units. Wait, but the triangle is formed by connecting (0,4) to (0,0) to (2,0). So actually, the base is from (0,0) to (2,0), which is 2 units, and the height is from (0,0) up to (0,4), which is 4 units. But since it's a triangle, the area would be (base * height)/2. Wait, but in this case, the triangle is a right triangle with legs of length 2 and 4? Wait, no. Wait, if you connect (0,4) to (0,0) to (2,0), then the triangle has vertices at (0,4), (0,0), and (2,0). So the base is from (0,0) to (2,0), which is 2 units, and the height is the vertical distance from that base to the point (0,4). Wait, but since the base is horizontal, the height would be the vertical distance from the base to the opposite vertex. However, the opposite vertex here is (0,4), which is directly above (0,0). So the height is 4 units. Therefore, the area is (base * height)/2 = (2 * 4)/2 = 4. So that triangle has area 4.

Wait, but hold on. Alternatively, since the triangle is formed by the points (0,4), (0,0), (2,0), maybe I can think of it as a right triangle with legs of length 4 and 2? Wait, if you consider the right angle at (0,0), then yes, the legs would be along the y-axis from (0,0) to (0,4) which is 4 units, and along the x-axis from (0,0) to (2,0) which is 2 units. Therefore, the area is indeed (4*2)/2 = 4. So that first shaded region is 4 square units.

Then the second fill command is (0,4)--(3,4)--(3,0)--cycle. So that's another triangle. Let me parse the coordinates. Starting at (0,4), which is the top-left corner, then going to (3,4), which is three units to the right along the top edge, then down to (3,0), which is the bottom-right corner of the entire figure, and then back to (0,4). So this is a large triangle with vertices at (0,4), (3,4), and (3,0). So this is a right triangle as well, with legs along the top edge from (0,4) to (3,4) which is 3 units, and along the right edge from (3,4) down to (3,0) which is 4 units. Therefore, the area is (3*4)/2 = 6. So that triangle has area 6.

Therefore, the total shaded area would be 4 + 6 = 10? Wait, but hold on. Wait, the problem says "the shaded region," but maybe there is overlapping? Wait, the Asymptote code fills two regions: the first triangle and the second triangle. Are these two triangles overlapping? Let me check.

First triangle: from (0,4) to (0,0) to (2,0). Second triangle: from (0,4) to (3,4) to (3,0). So the first triangle is on the left side, from the top-left corner down to the bottom-left corner and then over to (2,0). The second triangle is from the top-left corner to the top-right corner (3,4) and then down to the bottom-right corner (3,0). So these two triangles share the line from (0,4) to (3,0)? Wait, no. Wait, the first triangle is (0,4), (0,0), (2,0). The second triangle is (0,4), (3,4), (3,0). So they both have the point (0,4) and both have a line from (0,4) to (3,0)? Wait, no. The first triangle has a line from (0,4) to (0,0) to (2,0). The second triangle has a line from (0,4) to (3,4) to (3,0). So the only overlapping region would be if they share any common area. Let me think. The first triangle is in the left part of the figure, from x=0 to x=2 on the bottom, and the second triangle is from x=0 to x=3 on the top. So maybe they overlap somewhere? Let me check coordinates.

The first triangle is bounded by x from 0 to 2 on the bottom, and the second triangle is bounded by x from 0 to 3 on the top. However, the first triangle is below the line from (0,4) to (2,0), and the second triangle is above the line from (0,4) to (3,0). Wait, actually, the second triangle is the triangle formed by (0,4), (3,4), (3,0). So that triangle includes the area from the top edge down to the right edge. The first triangle is from the left edge down to the bottom edge. So maybe they overlap in some region? Let me check if there's an overlapping area.

Let me think of the two triangles. The first triangle is the area under the line from (0,4) to (2,0). The second triangle is the area under the line from (0,4) to (3,0). Wait, actually, the second triangle is the area bounded by the top edge from (0,4) to (3,4), the right edge from (3,4) to (3,0), and the diagonal from (3,0) back to (0,4). So the second triangle is a large triangle that covers the entire right side of the figure. The first triangle is a smaller triangle on the left side. So maybe they don't overlap? Wait, but the first triangle goes from (0,4) down to (0,0) and then to (2,0). The second triangle goes from (0,4) to (3,4) to (3,0). So the only overlapping point is at (0,4) and maybe along the line from (0,4) to (3,0)? Wait, but the first triangle is only up to x=2 on the bottom. The second triangle includes the entire right side. So maybe they don't overlap? Let me check if there's any overlapping area.

Suppose we have two triangles: one is the triangle with vertices (0,4), (0,0), (2,0), and the other is (0,4), (3,4), (3,0). The only common point is (0,4). The first triangle is entirely on the left side, and the second triangle is on the right side. Wait, but actually, the second triangle includes the area from (0,4) to (3,4) to (3,0). So if you imagine that triangle, it's a large triangle that covers from the top-left corner to the top-right corner to the bottom-right corner. The first triangle is from the top-left corner down to the bottom-left corner and then over to (2,0). So maybe they share the line from (0,4) to (something)? Wait, the first triangle has a line from (0,4) to (0,0), and the second triangle has a line from (0,4) to (3,0). So these two lines intersect at (0,4) and maybe somewhere else? Wait, the line from (0,4) to (0,0) is the left edge, and the line from (0,4) to (3,0) is a diagonal. These two lines only meet at (0,4). Therefore, the two triangles only share the point (0,4), so there is no overlapping area. Therefore, the total shaded area is 4 + 6 = 10. But wait, the entire figure is 12 squares, so 10 shaded? That seems possible, but let me check again.

Wait, but maybe I made a mistake in calculating the areas. Let me verify again.

First triangle: (0,4), (0,0), (2,0). So this is a right triangle with base 2 and height 4. Area is (2*4)/2 = 4. Correct.

Second triangle: (0,4), (3,4), (3,0). This is a right triangle with base 3 and height 4. Area is (3*4)/2 = 6. Correct. So total shaded area is 4 + 6 = 10. But let me check if the Asymptote code draws these two triangles. Wait, the Asymptote code also draws diagonals: draw((0, 4)--(3, 0)); draw((0, 4)--(2, 0)); So those are the diagonals mentioned. So the first triangle is formed by (0,4)--(0,0)--(2,0)--cycle, which is the triangle with vertices at those points. The second triangle is (0,4)--(3,4)--(3,0)--cycle. So that's correct.

But let me think again. Maybe there's an overlap? Wait, if you imagine the figure, the first triangle is on the left side, and the second triangle is on the right side. But maybe the line from (0,4) to (3,0) is drawn, which is the hypotenuse of the second triangle. Also, the line from (0,4) to (2,0) is drawn, which is the hypotenuse of the first triangle. So these two lines intersect somewhere? Wait, the line from (0,4) to (3,0) and the line from (0,4) to (2,0) both start at (0,4). So they only meet at (0,4). Therefore, there is no overlapping area between the two triangles. Therefore, adding their areas is correct. So total shaded area is 4 + 6 = 10. But wait, the problem says "the shaded region," but maybe there's something else?

Wait, but let me check the Asymptote code again. The first fill is (0,4)--(0,0)--(2,0)--cycle. So that's a triangle. The second fill is (0,4)--(3,4)--(3,0)--cycle. So that's another triangle. So the shaded regions are two triangles. But maybe there's another shaded region? Wait, the Asymptote code only has two fill commands. So maybe the answer is 10? But let me think again. Wait, the entire figure is 12 squares. If the shaded area is 10, that would mean 10/12 = 5/6 of the area is shaded. But maybe I need to check again.

Alternatively, maybe I need to calculate the area using coordinates. Let me try to compute the area of each triangle again.

First triangle: points (0,4), (0,0), (2,0). Using the formula for the area of a triangle with coordinates. The formula is |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|. Let me apply that.

Let me label the points:

A(0,4), B(0,0), C(2,0).

So plugging into the formula:

Area = |(0*(0 - 0) + 0*(0 - 4) + 2*(4 - 0))/2|

= |0 + 0 + 2*4 / 2| = |8/2| = 4. Correct.

Second triangle: points (0,4), (3,4), (3,0). Let me call them D(0,4), E(3,4), F(3,0).

Area = |0*(4 - 0) + 3*(0 - 4) + 3*(4 - 4)| / 2

Wait, let me use the same formula:

Area = |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| / 2

So plugging in:

x1 = 0, y1 = 4

x2 = 3, y2 = 4

x3 = 3, y3 = 0

So:

Area = |0*(4 - 0) + 3*(0 - 4) + 3*(4 - 4)| / 2

= |0 + 3*(-4) + 3*(0)| / 2

= | -12 + 0 | / 2 = 12 / 2 = 6. Correct. So that's 6. So total area is 4 + 6 = 10.

But let me think again. The problem says "the shaded region," but maybe there's an overlap? Wait, if I imagine the figure, the first triangle is on the left side, and the second triangle is on the right side. However, the line from (0,4) to (3,0) is drawn, which is the hypotenuse of the second triangle. Also, the line from (0,4) to (2,0) is drawn, which is the hypotenuse of the first triangle. So maybe these two lines intersect somewhere? Wait, but they both start at (0,4). The first line goes to (2,0), and the second goes to (3,0). So they don't intersect anywhere else. Therefore, there is no overlapping area. Therefore, the total shaded area is indeed 10.

But wait, the problem says "twelve 1 by 1 squares form a rectangle," so the total area is 12. If the shaded area is 10, that would mean 10 units. But maybe I need to check if the figure is 3x4? Wait, the grid is drawn with vertical lines from j=0 to j=4 (so x=0,1,2,3) and horizontal lines from j=0 to j=5 (so y=0,1,2,3,4). Wait, hold on. Wait, the vertical lines are from j=0 to j=4, so that's 4 vertical lines, which would make 3 columns (since each column is between two vertical lines). Similarly, horizontal lines from j=0 to j=5, which is 5 horizontal lines, making 4 rows. Therefore, the rectangle is 3 units wide (columns) and 4 units tall (rows). Therefore, the total area is 3*4=12, which matches. So each square is 1x1, so the entire figure is 3x4.

But then, if the shaded area is 10, that would be 10 square units. But maybe I need to check if the Asymptote code actually shades those regions. Wait, the Asymptote code draws two triangles. Let me think again. The first triangle is from (0,4) to (0,0) to (2,0). So that's a triangle that covers part of the left side. The second triangle is from (0,4) to (3,4) to (3,0). So that's a triangle that covers the right side. But maybe there's another shaded region? Wait, the Asymptote code only has two fill commands. So maybe the answer is 10? But let me think again.

Alternatively, maybe I need to count the squares? Wait, but the figure is made of 1x1 squares, but the shaded regions are triangles, so maybe I need to calculate the area of the triangles. But since the triangles are not aligned with the grid, maybe there's a different way. Wait, but I already calculated the area using coordinates, and it's 4 + 6 = 10.

Alternatively, maybe the figure is different? Let me try to imagine the Asymptote code. The grid is 3 columns (x=0 to 3) and 4 rows (y=0 to 4). The first shaded triangle is from (0,4) down to (0,0) and then to (2,0). So that triangle is in the left half of the bottom part. The second shaded triangle is from (0,4) to (3,4) to (3,0). So that's a large triangle covering the top and right side.

Alternatively, maybe there's an overlap? Wait, if I imagine the two triangles, the first triangle is on the left, and the second is on the right. But maybe the line from (0,4) to (2,0) is part of the first triangle, and the line from (0,4) to (3,0) is part of the second triangle. These two lines intersect at (0,4) and maybe somewhere else? Wait, no. The line from (0,4) to (2,0) has a slope of (0 - 4)/(2 - 0) = -2. The line from (0,4) to (3,0) has a slope of (0 - 4)/(3 - 0) = -4/3. These two lines only intersect at (0,4). Therefore, there is no overlapping area. Therefore, the total shaded area is indeed 4 + 6 = 10.

But let me check again. Maybe I made a mistake in interpreting the coordinates. Let me think of the figure. The first triangle is from the top-left corner (0,4) down to the bottom-left corner (0,0), then over to (2,0). So that triangle is in the left part of the figure, covering from column 0 to column 2 on the bottom. The second triangle is from the top-left corner (0,4) to the top-right corner (3,4) to the bottom-right corner (3,0). So that triangle covers the entire right side of the figure.

But if I imagine the entire figure, the shaded regions are two triangles. However, maybe there's another shaded region? Wait, the Asymptote code only has two fill commands. So maybe the answer is 10. But let me check with another approach.

Alternatively, maybe I can calculate the area by counting the squares. But since the shaded regions are triangles, not aligned with the grid, it's more complicated. However, maybe I can divide the triangles into smaller parts.

First triangle: from (0,4) to (0,0) to (2,0). This triangle can be divided into squares? Let me think. From (0,0) to (2,0) is two units. The height is 4 units. If I imagine this triangle, it's a right triangle with base 2 and height 4. If I divide it into unit squares, how many squares are covered? Well, the triangle is not aligned with the grid, so it's not straightforward. However, since we already calculated the area as 4, which is 4 square units, and since each square is 1x1, that would mean that the area is 4. Similarly, the second triangle is 6 square units.

Alternatively, maybe I can use coordinate geometry to find the area. For the first triangle, coordinates (0,4), (0,0), (2,0). The area is 4. For the second triangle, coordinates (0,4), (3,4), (3,0). The area is 6. So total area is 10.

But let me check once more. Maybe the figure is different? Wait, the Asymptote code draws the grid with vertical lines from j=0 to j=4 (so x=0,1,2,3) and horizontal lines from j=0 to j=5 (so y=0,1,2,3,4). Therefore, the rectangle is 3 units wide and 4 units tall. The diagonals are drawn from (0,4) to (3,0) and from (0,4) to (2,0). The shaded regions are the two triangles mentioned.

Alternatively, maybe the problem is in the Asymptote code? Let me check the Asymptote code again:

// Filled portions
fill((0, 4)--(0, 0)--(2, 0)--cycle, gray(0.75));
fill((0, 4)--(3, 4)--(3, 0)--cycle, gray(0.75));

So the first fill is a triangle with vertices at (0,4), (0,0), (2,0). The second fill is a triangle with vertices at (0,4), (3,4), (3,0). So that's correct.

Alternatively, maybe the figure is not 3x4? Wait, the grid is drawn with vertical lines from x=0 to x=3 (since for j from 0 to 4, draw (j, 0)--(j, 4)), so vertical lines at x=0,1,2,3. So the width is 3 units. Horizontal lines from y=0 to y=4, with j from 0 to 5, so horizontal lines at y=0,1,2,3,4. Therefore, the height is 4 units. So the rectangle is 3x4.

Therefore, the total area is 12. The shaded area is 10. But maybe the answer is different? Wait, maybe I need to check if the triangles are overlapping? Wait, if I imagine the two triangles, the first triangle is on the left side, and the second triangle is on the right side. However, the line from (0,4) to (3,0) passes through the first triangle? Wait, the line from (0,4) to (3,0) is the hypotenuse of the second triangle. The first triangle has a hypotenuse from (0,4) to (2,0). So the line from (0,4) to (3,0) passes through the first triangle? Let me check.

The line from (0,4) to (3,0) has the equation y = (-4/3)x + 4. The line from (0,4) to (2,0) has the equation y = -2x + 4. So these two lines intersect at (0,4) and... Wait, they both start at (0,4). The line y = -2x + 4 goes from (0,4) to (2,0). The line y = -4/3 x + 4 goes from (0,4) to (3,0). So they don't intersect anywhere else. Therefore, the two triangles only share the point (0,4). Therefore, no overlapping area.

Therefore, the total shaded area is indeed 4 + 6 = 10. But let me check once more. Maybe the problem is in the Asymptote code? The Asymptote code draws the diagonals, but maybe those diagonals are part of the shaded regions? Wait, the fill commands are for the triangles, and the diagonals are drawn separately. The diagonals are just lines, not shaded.

Alternatively, maybe I need to check if the shaded regions are actually different. For example, maybe the first triangle is not entirely shaded? Wait, the first fill command is (0,4)--(0,0)--(2,0)--cycle, which is a triangle. The second fill command is (0,4)--(3,4)--(3,0)--cycle, another triangle. So those are two distinct triangles.

Alternatively, maybe the figure is different? Let me try to imagine the Asymptote code. The grid is 3 columns (width 3) and 4 rows (height 4). The first shaded triangle is from the top-left corner down to the bottom-left corner and then to (2,0). So that triangle is in the left part of the figure. The second shaded triangle is from the top-left corner to the top-right corner to the bottom-right corner. So that triangle covers the entire right side.

But if I imagine the entire figure, the shaded area would be the left triangle (area 4) and the right triangle (area 6), totaling 10. However, since the entire figure is 12, maybe the answer is 10? But I need to make sure.

Alternatively, maybe I can calculate the area by subtracting unshaded regions? Let me try that.

Total area is 12. If I can find the unshaded area, then subtract from 12.

But how? The unshaded area would be the rest of the figure. However, without seeing the figure, it's a bit challenging, but maybe I can deduce it.

Alternatively, maybe I can think of the two shaded triangles. The first triangle is 4, the second is 6. So total shaded is 10.

Alternatively, maybe the answer is different. Wait, let me think again. Maybe the first triangle is not 4? Wait, if the triangle is from (0,4) to (0,0) to (2,0), then the area is ½ * base * height. The base is 2 units (from (0,0) to (2,0)), and the height is 4 units (from (0,0) up to (0,4)). Wait, but in a triangle, the base and height must be perpendicular. If I take the base as 2 units along the x-axis, then the height is the vertical distance from that base to the opposite vertex, which is (0,4). The vertical distance from the base (which is along the x-axis from (0,0) to (2,0)) to the point (0,4) is indeed 4 units. Therefore, area is ½ * 2 * 4 = 4. Correct.

Similarly, the second triangle has base 3 units (from (0,4) to (3,4)) and height 4 units (from (3,4) down to (3,0)), so area is ½ * 3 * 4 = 6. Correct.

Therefore, total shaded area is 4 + 6 = 10.

But let me check once more with another method. For example, using coordinates to calculate the area of each triangle.

First triangle: points (0,4), (0,0), (2,0). Using the shoelace formula:

List the coordinates:

(0,4), (0,0), (2,0), (0,4)

Compute shoelace sum:

Sum1 = (0*0 + 0*0 + 2*4) = 0 + 0 + 8 = 8

Sum2 = (4*0 + 0*2 + 0*0) = 0 + 0 + 0 = 0

Area = |Sum1 - Sum2| / 2 = |8 - 0| / 2 = 4. Correct.

Second triangle: points (0,4), (3,4), (3,0), (0,4)

Shoelace formula:

Sum1 = 0*4 + 3*0 + 3*4 = 0 + 0 + 12 = 12

Sum2 = 4*3 + 4*3 + 0*0 = 12 + 12 + 0 = 24

Area = |12 - 24| / 2 = 12 / 2 = 6. Correct.

Therefore, total area is 4 + 6 = 10.

Therefore, I think the answer is 10. However, since the problem says "the shaded region," maybe there's something I'm missing? Wait, maybe the figure has overlapping shaded regions? But we already checked that there is no overlapping.

Alternatively, maybe the Asymptote code draws something else? Let me check the Asymptote code again:

// Filled portions
fill((0, 4)--(0, 0)--(2, 0)--cycle, gray(0.75));
fill((0, 4)--(3, 4)--(3, 0)--cycle, gray(0.75));

// grid
int j;
for (j = 0; j < 4; ++j) {draw((j, 0)--(j, 4));}
for (j = 0; j < 5; ++j) {draw((0, j)--(3, j));}

//diagonals
draw((0, 4)--(3, 0)); draw((0, 4)--(2, 0));

So the grid is drawn, then the two shaded triangles, then the diagonals. The diagonals are drawn, but they are not shaded. So the shaded regions are only the two triangles.

Therefore, I think my calculation is correct. The total shaded area is 10. However, since the problem is from an Olympiad or similar, maybe the answer is different? Wait, let me think again.

Wait, the entire figure is 3x4=12. If the shaded area is 10, then the unshaded area is 2. But maybe that's possible? Let me check.

Looking at the figure, the unshaded area would be the parts not covered by the two triangles. The first triangle covers area 4, the second covers 6. But maybe there's some overlapping? Wait, no.

Alternatively, maybe the figure is different. For example, maybe the first triangle is not from (0,4) to (0,0) to (2,0), but something else? Wait, the Asymptote code says fill((0, 4)--(0, 0)--(2, 0)--cycle, gray(0.75)); which is exactly that triangle.

Alternatively, maybe the figure is 4x3? Wait, no, the grid is 3 columns and 4 rows.

Alternatively, maybe I need to check if the triangles are actually in the figure. For example, the first triangle is from (0,4) down to (0,0) and then to (2,0). So that triangle is in the left part of the figure. The second triangle is from (0,4) to (3,4) to (3,0). So that triangle is in the right part.

But if I imagine the entire figure, the shaded area would be those two triangles. Therefore, 4 + 6 = 10.

Alternatively, maybe the answer is 6? Wait, no. Because the first triangle is 4.

Alternatively, maybe I made a mistake in the coordinates? For example, maybe the first triangle is not 2 units wide? Wait, from (0,0) to (2,0) is 2 units. Yes.

Alternatively, maybe the problem is in the Asymptote code's coordinate system? For example, maybe the y-axis is inverted? In Asymptote, the coordinate system usually has (0,0) at the bottom-left, but sometimes it can be different. Wait, in the Asymptote code, they draw vertical lines from (j, 0) to (j, 4), so for j from 0 to 3, which would be vertical lines at x=0,1,2,3, going from y=0 to y=4. Similarly, horizontal lines from (0, j) to (3, j), for j from 0 to 4, so horizontal lines at y=0,1,2,3,4, going from x=0 to x=3. Therefore, the coordinate system is standard, with (0,0) at the bottom-left, increasing y upwards. Therefore, the point (0,4) is the top-left corner, (3,0) is the bottom-right corner.

Therefore, the triangles are correctly calculated.

Therefore, I think my answer is correct. The total shaded area is 10. However, since the problem says "the shaded region," and the answer is 10, but maybe I need to check once more.

Wait, let me think of the figure as a 3x4 grid. The first triangle covers part of the left side. Let me try to count the number of squares it covers.

The triangle from (0,4) to (0,0) to (2,0). This triangle would cover parts of the squares. For example, from column 0 (x=0 to 1) and column 1 (x=1 to 2). In terms of rows, from y=0 to y=4.

But since it's a triangle, it's not straightforward to count squares. However, since we already calculated the area as 4, which is 4 square units, and since each square is 1x1, that would mean that the area is 4. Similarly, the second triangle is 6.

Alternatively, maybe the answer is different? Wait, if I think of the entire figure, the two shaded triangles might cover some squares completely and some partially. But since the problem says "total area of the shaded region," which is a continuous area, not counting individual squares, so the area is indeed 10.

Therefore, I think the answer is 10. However, since I'm a bit unsure because sometimes problems can have tricks, but given the calculations, I think 10 is correct.

Wait, but let me check once more. If the entire figure is 3x4=12, and the shaded area is 10, then the unshaded area is 2. Let me see if that makes sense.

The unshaded area would be the parts not covered by the two triangles. The first triangle covers area 4, the second covers 6. But since they don't overlap, total shaded is 10. The unshaded area is 12 - 10 = 2. Where is this unshaded area? It would be the area not covered by the two triangles.

Looking at the figure, the first triangle covers the left side from (0,4) down to (0,0) and over to (2,0). The second triangle covers from (0,4) to (3,4) to (3,0). So the unshaded area would be the area between the two triangles. For example, the area from x=2 to x=3 on the bottom, and the area from x=0 to x=2 on the top? Wait, maybe not.

Alternatively, maybe the unshaded area is a small region. For example, the area between the two triangles. But since the two triangles don't overlap, the unshaded area is the rest.

But regardless, since the math checks out, I think the answer is 10.

But wait, hold on. Wait, the Asymptote code draws two triangles, but maybe there's another shaded region? Let me check the Asymptote code again.

The code has two fill commands:

fill((0, 4)--(0, 0)--(2, 0)--cycle, gray(0.75));

fill((0, 4)--(3, 4)--(3, 0)--cycle, gray(0.75));

So only two shaded regions. Therefore, total shaded area is 4 + 6 = 10.

Therefore, I think the answer is 10. However, since I'm a bit uncertain because sometimes problems can have unexpected overlaps or something, but given the calculations, I think it's safe to go with 10.

Wait, but let me check with another approach. Let me think of the entire figure as a 3x4 grid. The first triangle is from (0,4) to (0,0) to (2,0). This triangle can be divided into smaller triangles or shapes. For example, from (0,4) to (0,0) to (2,0) is a right triangle. If I divide this triangle into unit squares, how many squares are fully or partially covered?

But since the triangle is not aligned with the grid, it's complex. However, since we already calculated the area as 4, which is 4 square units, and since the entire figure is 12, it's plausible.

Alternatively, maybe the answer is 6? If I thought the first triangle was 2 and the second 4? But no, calculations show otherwise.

Alternatively, maybe the problem is in the Asymptote code's coordinate system? For example, if the figure is 4x3 instead of 3x4? But no, the grid is drawn with vertical lines from x=0 to x=3 (so width 3) and horizontal lines from y=0 to y=4 (height 4).

Therefore, I think my answer is correct. The total shaded area is 10.

**Final Answer**
\boxed{10}

To determine the total area of the shaded region in the given figure, we analyze the two shaded triangles described in the Asymptote code.

---

### **Step 1: Understand the Grid and Coordinates**

The figure is a rectangle formed by 12 unit squares, arranged in a 3×4 grid. This means the rectangle has:

- **Width**: 3 units (from x = 0 to x = 3)
- **Height**: 4 units (from y = 0 to y = 4)

The Asymptote code draws:

- **Vertical lines** at x = 0, 1, 2, 3
- **Horizontal lines** at y = 0, 1, 2, 3, 4

This confirms the rectangle is 3 units wide and 4 units tall, with a total area of:

$$
\text{Total Area} = 3 \times 4 = 12 \text{ square units}
$$

---

### **Step 2: Analyze the Shaded Regions**

There are two shaded triangles:

#### **Triangle 1:**
- Vertices: (0, 4), (0, 0), (2, 0)
- This is a **right triangle** with:
  - **Base** = 2 units (from (0, 0) to (2, 0))
  - **Height** = 4 units (from (0, 0) to (0, 4))

$$
\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 4 = 4 \text{ square units}
$$

#### **Triangle 2:**
- Vertices: (0, 4), (3, 4), (3, 0)
- This is also a **right triangle** with:
  - **Base** = 3 units (from (0, 4) to (3, 4))
  - **Height** = 4 units (from (3, 4) to (3, 0))

$$
\text{Area}_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6 \text{ square units}
$$

---

### **Step 3: Check for Overlap**

The two triangles share only the point (0, 4) and do not overlap in any other region. Therefore, their areas can be **added directly**.

---

### **Step 4: Final Calculation**

$$
\text{Total Shaded Area} = \text{Area}_1 + \text{Area}_2 = 4 + 6 = \boxed{10}
$$

---

### ✅ **Final Answer**

$$
\boxed{10}
$$

Solution:
The total area of the rectangle is $3 \times 4 =12$.

The total area of the shaded regions equals the total area of the rectangle (12) minus the area of the unshaded region.

The unshaded region is a triangle with base of length 1 and height 4; the area of this region is $\frac{1}{2}(1)(4)=2$.

Therefore, the total area of the shaded regions is $12 - 2 = \boxed{10}$.


Explanation:
Okay, so I need to compute the infinite sum $\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}}$, where $F_n$ is the Fibonacci sequence with $F_0 = 0$ and $F_1 = 1$. Let me start by recalling some properties of Fibonacci numbers and see if I can find a pattern or a telescoping series here.

First, let me write down the first few Fibonacci numbers to get a sense of the sequence:

- $F_0 = 0$
- $F_1 = 1$
- $F_2 = 1$
- $F_3 = 2$
- $F_4 = 3$
- $F_5 = 5$
- $F_6 = 8$
- $F_7 = 13$
- $F_8 = 21$
- $F_9 = 34$
- $F_{10} = 55$
- and so on.

So, the Fibonacci sequence is defined by $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$, with the initial terms given.

Now, the sum in question is $\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}}$. Let me try to compute the first few terms to see if there's a pattern or if it telescopes.

Let me compute the first term when $n=1$:

Term 1: $\frac{F_2}{F_1 F_3} = \frac{1}{1 \times 2} = \frac{1}{2}$

Term 2: $\frac{F_3}{F_2 F_4} = \frac{2}{1 \times 3} = \frac{2}{3}$

Term 3: $\frac{F_4}{F_3 F_5} = \frac{3}{2 \times 5} = \frac{3}{10}$

Term 4: $\frac{F_5}{F_4 F_6} = \frac{5}{3 \times 8} = \frac{5}{24}$

Term 5: $\frac{F_6}{F_5 F_7} = \frac{8}{5 \times 13} = \frac{8}{65}$

Term 6: $\frac{F_7}{F_6 F_8} = \frac{13}{8 \times 21} = \frac{13}{168}$

Hmm, adding these up:

First term: 1/2 = 0.5

After term 2: 0.5 + 2/3 ≈ 0.5 + 0.666... ≈ 1.166...

After term 3: ≈ 1.166... + 0.3 = 1.466...

After term 4: ≈ 1.466... + 0.208... ≈ 1.674...

After term 5: ≈ 1.674... + 0.123... ≈ 1.797...

After term 6: ≈ 1.797... + 0.077... ≈ 1.874...

So it seems like the sum is converging to something around 1.8 or 1.9? Maybe approaching 2? But I need to find the exact value.

Alternatively, maybe there's a telescoping nature here. Let me think about how to manipulate the general term $\frac{F_{n + 1}}{F_n F_{n + 2}}$.

First, recall that Fibonacci numbers satisfy the identity $F_{n + 2} = F_{n + 1} + F_n$. Maybe I can use this identity to rewrite the denominator.

Let me write the denominator as $F_n F_{n + 2} = F_n (F_{n + 1} + F_n)$. Therefore, the term becomes:

$$
\frac{F_{n + 1}}{F_n (F_{n + 1} + F_n)} = \frac{F_{n + 1}}{F_n F_{n + 1} + F_n^2}
$$

Hmm, not sure if that helps directly. Maybe I can split the fraction into partial fractions? Let me try that.

Alternatively, perhaps express the term as a difference of reciprocals or something similar. Let me think.

Let me recall that Fibonacci numbers have some reciprocal identities. For example, there's an identity involving reciprocals of Fibonacci numbers, but I don't remember exactly. Alternatively, maybe using the property of Fibonacci numbers in terms of continued fractions or something else.

Alternatively, let me consider the general term:

$$
\frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{F_{n + 1}}{F_n (F_{n + 1} + F_n)} = \frac{F_{n + 1}}{F_n F_{n + 1} + F_n^2}
$$

Alternatively, divide numerator and denominator by $F_n F_{n + 1}$:

$$
\frac{1}{F_n + \frac{F_n^2}{F_{n + 1}}}
$$

Wait, that seems messy. Maybe another approach.

Alternatively, let me think about the identity involving reciprocals. Let me recall that for Fibonacci numbers, there's an identity:

$$
\frac{1}{F_n F_{n + 1}} = \frac{1}{F_n F_{n + 1}} = \frac{1}{F_n F_{n + 1}}.
$$

But not sure. Wait, maybe using the identity $F_{n + 1} F_{n - 1} - F_n^2 = (-1)^n$. This is Cassini's identity. Maybe that can be helpful here?

But how?

Alternatively, let me try to manipulate the term $\frac{F_{n + 1}}{F_n F_{n + 2}}$.

Since $F_{n + 2} = F_{n + 1} + F_n$, so:

$$
\frac{F_{n + 1}}{F_n (F_{n + 1} + F_n)} = \frac{F_{n + 1}}{F_n F_{n + 1} + F_n^2} = \frac{1}{F_n + \frac{F_n^2}{F_{n + 1}}}
$$

Hmm, not helpful. Maybe another approach.

Alternatively, let me consider writing the term as a difference between consecutive terms. For telescoping series, we often look for expressions like $a_n - a_{n+1}$, so that when we sum over n, most terms cancel.

Let me try to express $\frac{F_{n + 1}}{F_n F_{n + 2}}$ as a difference. Let me think about the reciprocal of Fibonacci numbers.

Alternatively, let me recall that Fibonacci numbers satisfy the relation $F_{n + 2} = F_{n + 1} + F_n$, so maybe I can write:

Let me consider the term $\frac{F_{n + 1}}{F_n F_{n + 2}}$. Let me try to express this as $\frac{A}{F_n} + \frac{B}{F_{n + 2}}$ or something similar. Wait, but partial fractions with Fibonacci numbers?

Alternatively, maybe consider the difference between $\frac{1}{F_n F_{n + 1}}$ and $\frac{1}{F_{n + 1} F_{n + 2}}$?

Let me compute that difference:

$$
\frac{1}{F_n F_{n + 1}} - \frac{1}{F_{n + 1} F_{n + 2}} = \frac{F_{n + 2} - F_n}{F_n F_{n + 1} F_{n + 2}}}
$$

But $F_{n + 2} - F_n = F_{n + 1}$, so this becomes:

$$
\frac{F_{n + 1}}{F_n F_{n + 1} F_{n + 2}}} = \frac{1}{F_n F_{n + 2}}}
$$

Wait, so that gives:

$$
\frac{1}{F_n F_{n + 1}} - \frac{1}{F_{n + 1} F_{n + 2}} = \frac{1}{F_n F_{n + 2}}}
$$

But our term is $\frac{F_{n + 1}}{F_n F_{n + 2}}$, which is $F_{n + 1} \times \frac{1}{F_n F_{n + 2}}$. So if I have the above identity, then multiplying both sides by $F_{n + 1}$:

$$
F_{n + 1} \left( \frac{1}{F_n F_{n + 1}} - \frac{1}{F_{n + 1} F_{n + 2}} \right) = \frac{F_{n + 1}}{F_n F_{n + 2}}}
$$

Which simplifies to:

$$
\frac{1}{F_n} - \frac{1}{F_{n + 2}} = \frac{F_{n + 1}}{F_n F_{n + 2}}}
$$

Wait, that's exactly our term! So:

$$
\frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{1}{F_n} - \frac{1}{F_{n + 2}}
$$

Oh! That's a telescoping term! So the general term of the series can be written as the difference between reciprocals of Fibonacci numbers. Therefore, the sum becomes:

$$
\sum_{n = 1}^\infty \left( \frac{1}{F_n} - \frac{1}{F_{n + 2}} \right)
$$

This is a telescoping series. Let me write out the first few terms to see the cancellation:

For n=1: $\frac{1}{F_1} - \frac{1}{F_3} = \frac{1}{1} - \frac{1}{2}$

For n=2: $\frac{1}{F_2} - \frac{1}{F_4} = \frac{1}{1} - \frac{1}{3}$

For n=3: $\frac{1}{F_3} - \frac{1}{F_5} = \frac{1}{2} - \frac{1}{5}$

For n=4: $\frac{1}{F_4} - \frac{1}{F_6} = \frac{1}{3} - \frac{1}{8}$

For n=5: $\frac{1}{F_5} - \frac{1}{F_7} = \frac{1}{5} - \frac{1}{13}$

And so on.

So when we add them up, the series becomes:

$$
\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{1}{5} \right) + \left( \frac{1}{3} - \frac{1}{8} \right) + \left( \frac{1}{5} - \frac{1}{13} \right) + \cdots
$$

Now, let's see the cancellation:

- The first term has $-1/2$, then the third term has $+1/2$, so they cancel.
- The second term has $-1/3$, then the fourth term has $+1/3$, so they cancel.
- The third term has $-1/5$, then the fifth term has $+1/5$, so they cancel.
- Similarly, the fourth term has $-1/8$, which will be canceled by the sixth term, etc.

So, all the negative terms cancel with the positive terms two steps ahead. Therefore, the sum telescopes to the sum of the first few positive terms that don't get canceled.

Let me write the partial sum up to N terms:

$$
\sum_{n=1}^N \left( \frac{1}{F_n} - \frac{1}{F_{n + 2}} \right) = \left( \frac{1}{F_1} + \frac{1}{F_2} + \frac{1}{F_3} + \cdots + \frac{1}{F_N} \right) - \left( \frac{1}{F_3} + \frac{1}{F_4} + \cdots + \frac{1}{F_{N + 2}} \right)
$$

Simplifying this, the terms from $\frac{1}{F_3}$ to $\frac{1}{F_N}$ cancel out, leaving:

$$
\frac{1}{F_1} + \frac{1}{F_2} - \frac{1}{F_{N + 1}} - \frac{1}{F_{N + 2}}
$$

Therefore, the partial sum $S_N$ is:

$$
S_N = \frac{1}{F_1} + \frac{1}{F_2} - \frac{1}{F_{N + 1}} - \frac{1}{F_{N + 2}}
$$

Given that $F_1 = 1$ and $F_2 = 1$, this simplifies to:

$$
S_N = 1 + 1 - \frac{1}{F_{N + 1}} - \frac{1}{F_{N + 2}} = 2 - \left( \frac{1}{F_{N + 1}} + \frac{1}{F_{N + 2}} \right)
$$

Now, taking the limit as $N$ approaches infinity:

$$
\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}} = \lim_{N \to \infty} \left[ 2 - \left( \frac{1}{F_{N + 1}} + \frac{1}{F_{N + 2}} \right) \right]
$$

We need to evaluate the limit of $\frac{1}{F_{N + 1}} + \frac{1}{F_{N + 2}}$ as $N$ approaches infinity. Since Fibonacci numbers grow exponentially (they grow like $\phi^n / \sqrt{5}$, where $\phi = (1 + \sqrt{5})/2$), the reciprocals $\frac{1}{F_{N + 1}}$ and $\frac{1}{F_{N + 2}}$ approach zero as $N$ becomes large. Therefore, the limit becomes:

$$
2 - 0 - 0 = 2
$$

Therefore, the sum converges to 2.

Wait, but let me check with the partial sums I calculated earlier. When I added up the first six terms, I got approximately 1.874, which is less than 2. But as N increases, the partial sums approach 2. So that seems plausible.

But let me verify with another approach. Let me check for N approaching infinity, the terms $\frac{1}{F_{N + 1}} + \frac{1}{F_{N + 2}}$ go to zero. Since Fibonacci numbers grow exponentially, their reciprocals decay exponentially, so yes, the limit is zero. Therefore, the total sum is 2.

But let me check with another perspective. Let me recall that the sum telescopes to 2 minus the limit of those reciprocals. Therefore, the answer is 2.

But let me check with the first few partial sums. For example, when N=1:

S_1 = 2 - (1/F_2 + 1/F_3) = 2 - (1/1 + 1/2) = 2 - 1.5 = 0.5, which matches the first term.

Wait, but earlier when I calculated the first term (n=1) as 1/2, so S_1 = 1/2. But according to the formula, S_1 = 2 - (1/F_2 + 1/F_3) = 2 - (1 + 1/2) = 2 - 3/2 = 1/2. Correct.

Similarly, for N=2:

S_2 = 2 - (1/F_3 + 1/F_4) = 2 - (1/2 + 1/3) = 2 - 5/6 = 7/6 ≈ 1.166..., which matches the sum of first two terms (1/2 + 2/3 = 7/6). Correct.

For N=3:

S_3 = 2 - (1/F_4 + 1/F_5) = 2 - (1/3 + 1/5) = 2 - 8/15 = 22/15 ≈ 1.466..., which matches the sum of first three terms (1/2 + 2/3 + 3/10 = 1/2 + 2/3 + 3/10 = let's compute:

Convert to common denominator, say 30:

1/2 = 15/30, 2/3 = 20/30, 3/10 = 9/30. Total: 15 + 20 + 9 = 44/30 = 22/15 ≈ 1.466... Correct.

Similarly, for N=4:

S_4 = 2 - (1/F_5 + 1/F_6) = 2 - (1/5 + 1/8) = 2 - 13/40 = 67/40 = 1.675, which matches the sum of first four terms (previous 22/15 ≈ 1.466... + 5/24 ≈ 0.208... ≈ 1.674...). Correct.

So the formula works. Therefore, as N approaches infinity, the partial sums approach 2. Therefore, the sum is 2.

But let me check if there's any possible mistake here. For example, maybe I made a mistake in the telescoping step?

Earlier, I had:

$$
\frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{1}{F_n} - \frac{1}{F_{n + 2}}
$$

Let me verify this identity with n=1:

Left-hand side: F_2 / (F_1 F_3) = 1 / (1 * 2) = 1/2

Right-hand side: 1/F_1 - 1/F_3 = 1/1 - 1/2 = 1/2. Correct.

For n=2:

Left-hand side: F_3 / (F_2 F_4) = 2 / (1 * 3) = 2/3

Right-hand side: 1/F_2 - 1/F_4 = 1/1 - 1/3 = 2/3. Correct.

For n=3:

Left-hand side: F_4 / (F_3 F_5) = 3 / (2 * 5) = 3/10

Right-hand side: 1/F_3 - 1/F_5 = 1/2 - 1/5 = 3/10. Correct.

So the identity holds for these cases. Therefore, the telescoping is valid. Therefore, the conclusion that the sum is 2 is correct.

Therefore, the answer is 2. So the sum equals 2.

But let me check if there's another way to think about this. For example, using generating functions or other properties. But since the telescoping approach worked and verified with examples, I think this is solid.

Alternatively, maybe using the closed-form expression for Fibonacci numbers (Binet's formula). Let me recall that $F_n = \frac{\phi^n - \psi^n}{\sqrt{5}}$, where $\phi = \frac{1 + \sqrt{5}}{2}$ and $\psi = \frac{1 - \sqrt{5}}{2}$. However, since $\psi$ has absolute value less than 1, as n increases, $\psi^n$ tends to zero, so $F_n$ is approximately $\frac{\phi^n}{\sqrt{5}}$.

But maybe this is overcomplicating. Since we already have a telescoping series that gives the answer 2, and verified with partial sums, I think that's sufficient.

Therefore, the value of the infinite sum is 2.

**Final Answer**
\boxed{2}

To evaluate the infinite sum

$$
\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}},
$$

where $ F_n $ denotes the $ n $-th Fibonacci number with $ F_0 = 0 $ and $ F_1 = 1 $, we begin by analyzing the structure of the general term.

---

### Step 1: Use the Fibonacci Recurrence

Recall the Fibonacci recurrence:

$$
F_{n + 2} = F_{n + 1} + F_n.
$$

This allows us to rewrite the denominator:

$$
F_n F_{n + 2} = F_n (F_{n + 1} + F_n) = F_n F_{n + 1} + F_n^2.
$$

So the general term becomes:

$$
\frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{F_{n + 1}}{F_n F_{n + 1} + F_n^2}.
$$

This expression is not immediately helpful, so we try a different approach.

---

### Step 2: Telescoping Identity

We aim to express the term as a difference of reciprocals. Consider the identity:

$$
\frac{1}{F_n F_{n + 1}} - \frac{1}{F_{n + 1} F_{n + 2}} = \frac{F_{n + 2} - F_n}{F_n F_{n + 1} F_{n + 2}}.
$$

Using the Fibonacci recurrence $ F_{n + 2} - F_n = F_{n + 1} $, this simplifies to:

$$
\frac{F_{n + 1}}{F_n F_{n + 1} F_{n + 2}} = \frac{1}{F_n F_{n + 2}}.
$$

Multiplying both sides by $ F_{n + 1} $, we get:

$$
\frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{1}{F_n} - \frac{1}{F_{n + 2}}.
$$

---

### Step 3: Rewrite the Sum

Now, the original sum becomes:

$$
\sum_{n = 1}^\infty \left( \frac{1}{F_n} - \frac{1}{F_{n + 2}} \right).
$$

This is a **telescoping series**. Writing out the first few terms:

$$
\left( \frac{1}{F_1} - \frac{1}{F_3} \right) + \left( \frac{1}{F_2} - \frac{1}{F_4} \right) + \left( \frac{1}{F_3} - \frac{1}{F_5} \right) + \left( \frac{1}{F_4} - \frac{1}{F_6} \right) + \cdots
$$

We observe that all intermediate terms cancel, leaving:

$$
\frac{1}{F_1} + \frac{1}{F_2} - \lim_{N \to \infty} \left( \frac{1}{F_{N + 1}} + \frac{1}{F_{N + 2}} \right).
$$

Since $ F_n $ grows exponentially (as $ F_n \sim \phi^n / \sqrt{5} $, where $ \phi = \frac{1 + \sqrt{5}}{2} $), the terms $ \frac{1}{F_{N + 1}} $ and $ \frac{1}{F_{N + 2}} $ approach zero as $ N \to \infty $.

---

### Step 4: Final Computation

We now compute the sum:

$$
\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{1}{F_1} + \frac{1}{F_2} = 1 + 1 = 2.
$$

---

### Final Answer

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to compute the infinite sum $\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}}$, where $F_n$ is the Fibonacci sequence with $F_0 = 0$ and $F_1 = 1$. Let me start by recalling some properties of Fibonacci numbers and see if I can find a pattern or a telescoping series here.

First, let me write down the first few Fibonacci numbers to get a sense of the sequence:

- $F_0 = 0$
- $F_1 = 1$
- $F_2 = 1$
- $F_3 = 2$
- $F_4 = 3$
- $F_5 = 5$
- $F_6 = 8$
- $F_7 = 13$
- $F_8 = 21$
- $F_9 = 34$
- $F_{10} = 55$
- and so on.

So, the Fibonacci sequence is defined by $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$, with the initial terms given.

Now, the sum in question is $\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}}$. Let me try to compute the first few terms to see if there's a pattern or if it telescopes.

Let me compute the first term when $n=1$:

Term 1: $\frac{F_2}{F_1 F_3} = \frac{1}{1 \times 2} = \frac{1}{2}$

Term 2: $\frac{F_3}{F_2 F_4} = \frac{2}{1 \times 3} = \frac{2}{3}$

Term 3: $\frac{F_4}{F_3 F_5} = \frac{3}{2 \times 5} = \frac{3}{10}$

Term 4: $\frac{F_5}{F_4 F_6} = \frac{5}{3 \times 8} = \frac{5}{24}$

Term 5: $\frac{F_6}{F_5 F_7} = \frac{8}{5 \times 13} = \frac{8}{65}$

Term 6: $\frac{F_7}{F_6 F_8} = \frac{13}{8 \times 21} = \frac{13}{168}$

Hmm, adding these up:

First term: 1/2 = 0.5

After term 2: 0.5 + 2/3 ≈ 0.5 + 0.666... ≈ 1.166...

After term 3: ≈ 1.166... + 0.3 = 1.466...

After term 4: ≈ 1.466... + 0.208... ≈ 1.674...

After term 5: ≈ 1.674... + 0.123... ≈ 1.797...

After term 6: ≈ 1.797... + 0.077... ≈ 1.874...

So it seems like the sum is converging to something around 1.8 or 1.9? Maybe approaching 2? But I need to find the exact value.

Alternatively, maybe there's a telescoping nature here. Let me think about how to manipulate the general term $\frac{F_{n + 1}}{F_n F_{n + 2}}$.

First, recall that Fibonacci numbers satisfy the identity $F_{n + 2} = F_{n + 1} + F_n$. Maybe I can use this identity to rewrite the denominator.

Let me write the denominator as $F_n F_{n + 2} = F_n (F_{n + 1} + F_n)$. Therefore, the term becomes:

$$
\frac{F_{n + 1}}{F_n (F_{n + 1} + F_n)} = \frac{F_{n + 1}}{F_n F_{n + 1} + F_n^2}
$$

Hmm, not sure if that helps directly. Maybe I can split the fraction into partial fractions? Let me try that.

Alternatively, perhaps express the term as a difference of reciprocals or something similar. Let me think.

Let me recall that Fibonacci numbers have some reciprocal identities. For example, there's an identity involving reciprocals of Fibonacci numbers, but I don't remember exactly. Alternatively, maybe using the property of Fibonacci numbers in terms of continued fractions or something else.

Alternatively, let me consider the general term:

$$
\frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{F_{n + 1}}{F_n (F_{n + 1} + F_n)} = \frac{F_{n + 1}}{F_n F_{n + 1} + F_n^2}
$$

Alternatively, divide numerator and denominator by $F_n F_{n + 1}$:

$$
\frac{1}{F_n + \frac{F_n^2}{F_{n + 1}}}
$$

Wait, that seems messy. Maybe another approach.

Alternatively, let me think about the identity involving reciprocals. Let me recall that for Fibonacci numbers, there's an identity:

$$
\frac{1}{F_n F_{n + 1}} = \frac{1}{F_n F_{n + 1}} = \frac{1}{F_n F_{n + 1}}.
$$

But not sure. Wait, maybe using the identity $F_{n + 1} F_{n - 1} - F_n^2 = (-1)^n$. This is Cassini's identity. Maybe that can be helpful here?

But how?

Alternatively, let me try to manipulate the term $\frac{F_{n + 1}}{F_n F_{n + 2}}$.

Since $F_{n + 2} = F_{n + 1} + F_n$, so:

$$
\frac{F_{n + 1}}{F_n (F_{n + 1} + F_n)} = \frac{F_{n + 1}}{F_n F_{n + 1} + F_n^2} = \frac{1}{F_n + \frac{F_n^2}{F_{n + 1}}}
$$

Hmm, not helpful. Maybe another approach.

Alternatively, let me consider writing the term as a difference between consecutive terms. For telescoping series, we often look for expressions like $a_n - a_{n+1}$, so that when we sum over n, most terms cancel.

Let me try to express $\frac{F_{n + 1}}{F_n F_{n + 2}}$ as a difference. Let me think about the reciprocal of Fibonacci numbers.

Alternatively, let me recall that Fibonacci numbers satisfy the relation $F_{n + 2} = F_{n + 1} + F_n$, so maybe I can write:

Let me consider the term $\frac{F_{n + 1}}{F_n F_{n + 2}}$. Let me try to express this as $\frac{A}{F_n} + \frac{B}{F_{n + 2}}$ or something similar. Wait, but partial fractions with Fibonacci numbers?

Alternatively, maybe consider the difference between $\frac{1}{F_n F_{n + 1}}$ and $\frac{1}{F_{n + 1} F_{n + 2}}$?

Let me compute that difference:

$$
\frac{1}{F_n F_{n + 1}} - \frac{1}{F_{n + 1} F_{n + 2}} = \frac{F_{n + 2} - F_n}{F_n F_{n + 1} F_{n + 2}}}
$$

But $F_{n + 2} - F_n = F_{n + 1}$, so this becomes:

$$
\frac{F_{n + 1}}{F_n F_{n + 1} F_{n + 2}}} = \frac{1}{F_n F_{n + 2}}}
$$

Wait, so that gives:

$$
\frac{1}{F_n F_{n + 1}} - \frac{1}{F_{n + 1} F_{n + 2}} = \frac{1}{F_n F_{n + 2}}}
$$

But our term is $\frac{F_{n + 1}}{F_n F_{n + 2}}$, which is $F_{n + 1} \times \frac{1}{F_n F_{n + 2}}$. So if I have the above identity, then multiplying both sides by $F_{n + 1}$:

$$
F_{n + 1} \left( \frac{1}{F_n F_{n + 1}} - \frac{1}{F_{n + 1} F_{n + 2}} \right) = \frac{F_{n + 1}}{F_n F_{n + 2}}}
$$

Which simplifies to:

$$
\frac{1}{F_n} - \frac{1}{F_{n + 2}} = \frac{F_{n + 1}}{F_n F_{n + 2}}}
$$

Wait, that's exactly our term! So:

$$
\frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{1}{F_n} - \frac{1}{F_{n + 2}}
$$

Oh! That's a telescoping term! So the general term of the series can be written as the difference between reciprocals of Fibonacci numbers. Therefore, the sum becomes:

$$
\sum_{n = 1}^\infty \left( \frac{1}{F_n} - \frac{1}{F_{n + 2}} \right)
$$

This is a telescoping series. Let me write out the first few terms to see the cancellation:

For n=1: $\frac{1}{F_1} - \frac{1}{F_3} = \frac{1}{1} - \frac{1}{2}$

For n=2: $\frac{1}{F_2} - \frac{1}{F_4} = \frac{1}{1} - \frac{1}{3}$

For n=3: $\frac{1}{F_3} - \frac{1}{F_5} = \frac{1}{2} - \frac{1}{5}$

For n=4: $\frac{1}{F_4} - \frac{1}{F_6} = \frac{1}{3} - \frac{1}{8}$

For n=5: $\frac{1}{F_5} - \frac{1}{F_7} = \frac{1}{5} - \frac{1}{13}$

And so on.

So when we add them up, the series becomes:

$$
\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{1}{5} \right) + \left( \frac{1}{3} - \frac{1}{8} \right) + \left( \frac{1}{5} - \frac{1}{13} \right) + \cdots
$$

Now, let's see the cancellation:

- The first term has $-1/2$, then the third term has $+1/2$, so they cancel.
- The second term has $-1/3$, then the fourth term has $+1/3$, so they cancel.
- The third term has $-1/5$, then the fifth term has $+1/5$, so they cancel.
- Similarly, the fourth term has $-1/8$, which will be canceled by the sixth term, etc.

So, all the negative terms cancel with the positive terms two steps ahead. Therefore, the sum telescopes to the sum of the first few positive terms that don't get canceled.

Let me write the partial sum up to N terms:

$$
\sum_{n=1}^N \left( \frac{1}{F_n} - \frac{1}{F_{n + 2}} \right) = \left( \frac{1}{F_1} + \frac{1}{F_2} + \frac{1}{F_3} + \cdots + \frac{1}{F_N} \right) - \left( \frac{1}{F_3} + \frac{1}{F_4} + \cdots + \frac{1}{F_{N + 2}} \right)
$$

Simplifying this, the terms from $\frac{1}{F_3}$ to $\frac{1}{F_N}$ cancel out, leaving:

$$
\frac{1}{F_1} + \frac{1}{F_2} - \frac{1}{F_{N + 1}} - \frac{1}{F_{N + 2}}
$$

Therefore, the partial sum $S_N$ is:

$$
S_N = \frac{1}{F_1} + \frac{1}{F_2} - \frac{1}{F_{N + 1}} - \frac{1}{F_{N + 2}}
$$

Given that $F_1 = 1$ and $F_2 = 1$, this simplifies to:

$$
S_N = 1 + 1 - \frac{1}{F_{N + 1}} - \frac{1}{F_{N + 2}} = 2 - \left( \frac{1}{F_{N + 1}} + \frac{1}{F_{N + 2}} \right)
$$

Now, taking the limit as $N$ approaches infinity:

$$
\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}} = \lim_{N \to \infty} \left[ 2 - \left( \frac{1}{F_{N + 1}} + \frac{1}{F_{N + 2}} \right) \right]
$$

We need to evaluate the limit of $\frac{1}{F_{N + 1}} + \frac{1}{F_{N + 2}}$ as $N$ approaches infinity. Since Fibonacci numbers grow exponentially (they grow like $\phi^n / \sqrt{5}$, where $\phi = (1 + \sqrt{5})/2$), the reciprocals $\frac{1}{F_{N + 1}}$ and $\frac{1}{F_{N + 2}}$ approach zero as $N$ becomes large. Therefore, the limit becomes:

$$
2 - 0 - 0 = 2
$$

Therefore, the sum converges to 2.

Wait, but let me check with the partial sums I calculated earlier. When I added up the first six terms, I got approximately 1.874, which is less than 2. But as N increases, the partial sums approach 2. So that seems plausible.

But let me verify with another approach. Let me check for N approaching infinity, the terms $\frac{1}{F_{N + 1}} + \frac{1}{F_{N + 2}}$ go to zero. Since Fibonacci numbers grow exponentially, their reciprocals decay exponentially, so yes, the limit is zero. Therefore, the total sum is 2.

But let me check with another perspective. Let me recall that the sum telescopes to 2 minus the limit of those reciprocals. Therefore, the answer is 2.

But let me check with the first few partial sums. For example, when N=1:

S_1 = 2 - (1/F_2 + 1/F_3) = 2 - (1/1 + 1/2) = 2 - 1.5 = 0.5, which matches the first term.

Wait, but earlier when I calculated the first term (n=1) as 1/2, so S_1 = 1/2. But according to the formula, S_1 = 2 - (1/F_2 + 1/F_3) = 2 - (1 + 1/2) = 2 - 3/2 = 1/2. Correct.

Similarly, for N=2:

S_2 = 2 - (1/F_3 + 1/F_4) = 2 - (1/2 + 1/3) = 2 - 5/6 = 7/6 ≈ 1.166..., which matches the sum of first two terms (1/2 + 2/3 = 7/6). Correct.

For N=3:

S_3 = 2 - (1/F_4 + 1/F_5) = 2 - (1/3 + 1/5) = 2 - 8/15 = 22/15 ≈ 1.466..., which matches the sum of first three terms (1/2 + 2/3 + 3/10 = 1/2 + 2/3 + 3/10 = let's compute:

Convert to common denominator, say 30:

1/2 = 15/30, 2/3 = 20/30, 3/10 = 9/30. Total: 15 + 20 + 9 = 44/30 = 22/15 ≈ 1.466... Correct.

Similarly, for N=4:

S_4 = 2 - (1/F_5 + 1/F_6) = 2 - (1/5 + 1/8) = 2 - 13/40 = 67/40 = 1.675, which matches the sum of first four terms (previous 22/15 ≈ 1.466... + 5/24 ≈ 0.208... ≈ 1.674...). Correct.

So the formula works. Therefore, as N approaches infinity, the partial sums approach 2. Therefore, the sum is 2.

But let me check if there's any possible mistake here. For example, maybe I made a mistake in the telescoping step?

Earlier, I had:

$$
\frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{1}{F_n} - \frac{1}{F_{n + 2}}
$$

Let me verify this identity with n=1:

Left-hand side: F_2 / (F_1 F_3) = 1 / (1 * 2) = 1/2

Right-hand side: 1/F_1 - 1/F_3 = 1/1 - 1/2 = 1/2. Correct.

For n=2:

Left-hand side: F_3 / (F_2 F_4) = 2 / (1 * 3) = 2/3

Right-hand side: 1/F_2 - 1/F_4 = 1/1 - 1/3 = 2/3. Correct.

For n=3:

Left-hand side: F_4 / (F_3 F_5) = 3 / (2 * 5) = 3/10

Right-hand side: 1/F_3 - 1/F_5 = 1/2 - 1/5 = 3/10. Correct.

So the identity holds for these cases. Therefore, the telescoping is valid. Therefore, the conclusion that the sum is 2 is correct.

Therefore, the answer is 2. So the sum equals 2.

But let me check if there's another way to think about this. For example, using generating functions or other properties. But since the telescoping approach worked and verified with examples, I think this is solid.

Alternatively, maybe using the closed-form expression for Fibonacci numbers (Binet's formula). Let me recall that $F_n = \frac{\phi^n - \psi^n}{\sqrt{5}}$, where $\phi = \frac{1 + \sqrt{5}}{2}$ and $\psi = \frac{1 - \sqrt{5}}{2}$. However, since $\psi$ has absolute value less than 1, as n increases, $\psi^n$ tends to zero, so $F_n$ is approximately $\frac{\phi^n}{\sqrt{5}}$.

But maybe this is overcomplicating. Since we already have a telescoping series that gives the answer 2, and verified with partial sums, I think that's sufficient.

Therefore, the value of the infinite sum is 2.

**Final Answer**
\boxed{2}

To evaluate the infinite sum

$$
\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}},
$$

where $ F_n $ denotes the $ n $-th Fibonacci number with $ F_0 = 0 $ and $ F_1 = 1 $, we begin by analyzing the structure of the general term.

---

### Step 1: Use the Fibonacci Recurrence

Recall the Fibonacci recurrence:

$$
F_{n + 2} = F_{n + 1} + F_n.
$$

This allows us to rewrite the denominator:

$$
F_n F_{n + 2} = F_n (F_{n + 1} + F_n) = F_n F_{n + 1} + F_n^2.
$$

So the general term becomes:

$$
\frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{F_{n + 1}}{F_n F_{n + 1} + F_n^2}.
$$

This expression is not immediately helpful, so we try a different approach.

---

### Step 2: Telescoping Identity

We aim to express the term as a difference of reciprocals. Consider the identity:

$$
\frac{1}{F_n F_{n + 1}} - \frac{1}{F_{n + 1} F_{n + 2}} = \frac{F_{n + 2} - F_n}{F_n F_{n + 1} F_{n + 2}}.
$$

Using the Fibonacci recurrence $ F_{n + 2} - F_n = F_{n + 1} $, this simplifies to:

$$
\frac{F_{n + 1}}{F_n F_{n + 1} F_{n + 2}} = \frac{1}{F_n F_{n + 2}}.
$$

Multiplying both sides by $ F_{n + 1} $, we get:

$$
\frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{1}{F_n} - \frac{1}{F_{n + 2}}.
$$

---

### Step 3: Rewrite the Sum

Now, the original sum becomes:

$$
\sum_{n = 1}^\infty \left( \frac{1}{F_n} - \frac{1}{F_{n + 2}} \right).
$$

This is a **telescoping series**. Writing out the first few terms:

$$
\left( \frac{1}{F_1} - \frac{1}{F_3} \right) + \left( \frac{1}{F_2} - \frac{1}{F_4} \right) + \left( \frac{1}{F_3} - \frac{1}{F_5} \right) + \left( \frac{1}{F_4} - \frac{1}{F_6} \right) + \cdots
$$

We observe that all intermediate terms cancel, leaving:

$$
\frac{1}{F_1} + \frac{1}{F_2} - \lim_{N \to \infty} \left( \frac{1}{F_{N + 1}} + \frac{1}{F_{N + 2}} \right).
$$

Since $ F_n $ grows exponentially (as $ F_n \sim \phi^n / \sqrt{5} $, where $ \phi = \frac{1 + \sqrt{5}}{2} $), the terms $ \frac{1}{F_{N + 1}} $ and $ \frac{1}{F_{N + 2}} $ approach zero as $ N \to \infty $.

---

### Step 4: Final Computation

We now compute the sum:

$$
\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{1}{F_1} + \frac{1}{F_2} = 1 + 1 = 2.
$$

---

### Final Answer

$$
\boxed{2}
$$

Solution:
Since $F_{n + 1} = F_{n + 2} - F_n,$
\[\frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{F_{n + 2} - F_n}{F_n F_{n + 2}} = \frac{1}{F_n} - \frac{1}{F_{n + 2}}.\]Then
\begin{align*}
\sum_{n = 1}^\infty \frac{1}{F_n F_{n + 2}} &= \left( \frac{1}{F_1} - \frac{1}{F_3} \right) + \left( \frac{1}{F_2} - \frac{1}{F_4} \right) + \left( \frac{1}{F_3} - \frac{1}{F_5} \right) + \dotsb \\
&= \frac{1}{F_1} + \frac{1}{F_2} \\
&= \boxed{2}.
\end{align*}