Explanation:
Okay, so I need to find the degree measure of the smallest angle in triangle PQS. The diagram shows four circles of radius 1 with centers P, Q, R, and S, all tangent to each other and to the sides of triangle ABC. Hmm, let me try to visualize this. There's a triangle ABC with three circles tangent to its sides and each other, and another circle? Wait, no, four circles. Maybe each circle is tangent to two sides of the triangle and to the other circles? Or maybe they are arranged in some specific way?

Looking at the Asymptote code, the coordinates are given for points R, Q, S, and P. Let me parse that. R is at (0,0), Q is at (-2,0), S is at (2,0), and P is at (1, 1.732). Then points A, B, C are given with coordinates that seem more complicated. But maybe I don't need the exact coordinates? Or maybe I can use them?

Wait, the problem says the circles are tangent to one another and to the sides of triangle ABC. So each circle is tangent to two sides of the triangle? Or maybe each circle is tangent to two other circles and the sides? Let me think. If there are four circles, maybe they are arranged in a way similar to the Soddy circles or something? Or maybe they are the incircle and three mixtilinear incircles? But the problem says four circles tangent to each other and to the sides. Maybe they are all tangent to two sides of the triangle and to each other?

Alternatively, maybe the triangle ABC is such that each side has a circle tangent to it and to the other circles. But with four circles, perhaps two circles on the base BC, one on AB, and one on AC? But the Asymptote code shows R at (0,0), Q at (-2,0), S at (2,0), so maybe R and Q and S are on the base BC? Wait, but there are three circles on the base? Wait, but there are four circles. Maybe R is at the bottom, Q and S are on the base? Wait, the Asymptote code draws the triangle ABC with coordinates A=(1.366,3.098), B=(-5.73,-1), C=(3.732,-1). So the base BC is from (-5.73,-1) to (3.732,-1), which is a horizontal line. Then R is at (0,0), which is above the base BC? Wait, but the base BC is at y=-1. So R is at (0,0), which is above the base. Similarly, Q is at (-2,0), S is at (2,0). So maybe Q and S are on the sides AB and AC? Or maybe they are the centers of circles tangent to the sides?

Wait, the problem says the circles are tangent to one another and to the sides of triangle ABC. So each circle is tangent to two sides of the triangle and to the other circles. So maybe there are three circles tangent to two sides each, and the fourth circle is tangent to all three sides? But that would be the incircle. But there are four circles. Maybe the incircle and three mixtilinear incircles? But I need to think more carefully.

Alternatively, maybe the triangle ABC is such that there are four circles, each tangent to two sides and to the other circles. For example, in a triangle, you can have circles tangent to two sides and to another circle. Maybe like the Apollonius circles?

But maybe I can use coordinates. The Asymptote code gives coordinates for points P, Q, R, S. Let me check:

R=(0,0); Q=(-2,0); S=(2,0); P=(1,1.732). So P is at (1, sqrt(3)) since 1.732 is approximately sqrt(3). So P is at (1, sqrt(3)), Q is at (-2,0), R is at (0,0), S is at (2,0). Then triangle ABC is drawn with A=(1.366,3.098), B=(-5.73,-1), C=(3.732,-1). Let me check if these coordinates make sense.

First, let me note that sqrt(3) is approximately 1.732, so P is at (1, sqrt(3)). The points Q, R, S are on the x-axis. The triangle ABC has vertices at A, B, C. Let me check if the circles are tangent to the sides.

Each circle has radius 1. So the distance between centers of tangent circles should be 2, since they are tangent. Let me check the distance between Q and R: Q is (-2,0), R is (0,0). The distance is 2 units, which is equal to 2*radius, so they are tangent. Similarly, distance between R and S is 2 units, so they are tangent. Distance between Q and S is 4 units, which would be 2*radius*2? Wait, if they are tangent, the distance between centers should be 2*radius. But radius is 1, so distance should be 2. But Q is at (-2,0), S is at (2,0), so distance is 4. That's not 2. So they are not tangent? Wait, but the problem says all four circles are tangent to one another. So maybe my assumption is wrong?

Wait, maybe the circles are tangent to each other in some other way? Maybe not all pairs? Wait, the problem says "four circles of radius 1 with centres P, Q, R, and S are tangent to one another and to the sides of triangle ABC". So all four circles are tangent to each other? That would mean each pair of circles is tangent. But if Q and S are 4 units apart, and each has radius 1, then the distance between centers should be 2 units for them to be tangent. But 4 is not 2. Therefore, there must be something wrong here.

Wait, maybe the Asymptote code is just an illustration, and the coordinates are approximate? Because if the circles are tangent to each other, then the distance between centers should be 2. But in the Asymptote code, Q is at (-2,0), R at (0,0), S at (2,0), so QR is 2 units, RS is 2 units, but QS is 4 units. Therefore, if the circles at Q and S are tangent, they should be 2 units apart, but they are 4 units apart. Therefore, maybe the Asymptote code is not to scale? Or maybe the circles are not all tangent to each other? Wait, the problem says "four circles... are tangent to one another and to the sides of triangle ABC". So maybe not all pairs are tangent? Maybe each circle is tangent to two others and the sides?

Alternatively, maybe the diagram is not to scale, and the coordinates are just for drawing purposes. Maybe I shouldn't rely on them? Maybe I need to think geometrically.

Let me think again. If there are four circles inside triangle ABC, each tangent to two sides and to the other circles. Maybe they are arranged such that two circles are on the base BC, tangent to each other and to the base, and the other two circles are on the other sides?

Wait, but there are four circles. Maybe there is an incircle and three other circles? But the problem says four circles. Maybe it's similar to the problem of circles tangent to two sides and to each other. For example, in a triangle, you can have circles tangent to two sides and to another circle. Maybe there are two circles on the base, one on the left side, and one on the right side?

Alternatively, maybe the triangle is equilateral? Because the coordinates of P are (1, sqrt(3)), which is like a height of an equilateral triangle. Wait, if the triangle ABC is equilateral, then maybe?

But the coordinates given for A, B, C are not symmetric. Let me check. A is at (1.366, 3.098). Let me approximate sqrt(3) is about 1.732, so 1.366 is roughly 1.366, which is close to 1.366... Wait, 1.366 is approximately 1 + 0.366, and 0.366 is roughly 1/2.732, which is not sure. Maybe it's related to some specific triangle?

Alternatively, maybe triangle ABC is a 30-60-90 triangle? Or some other triangle. Alternatively, maybe I need to use coordinates given in the Asymptote code to find coordinates of P, Q, R, S and then compute the angles of triangle PQS?

But earlier, there was a problem with the distance between Q and S. If the Asymptote code is accurate, then maybe the circles are not all tangent to each other? Wait, the problem says "four circles... are tangent to one another and to the sides of triangle ABC". So maybe the Asymptote code is illustrative but not exact? Maybe the coordinates are approximate?

Alternatively, maybe the circles are tangent to each other in pairs? For example, P is tangent to Q, Q is tangent to R, R is tangent to S, and S is tangent to P? But then the distance between P and Q should be 2, between Q and R is 2, R and S is 2, and S and P is 2? But in the Asymptote code, P is at (1, sqrt(3)), Q is at (-2,0), so distance between P and Q is sqrt[(1 - (-2))² + (sqrt(3) - 0)²] = sqrt[3² + (sqrt(3))²] = sqrt[9 + 3] = sqrt(12) ≈ 3.464, which is more than 2. So that's not tangent. Therefore, maybe the Asymptote code is not accurate? Maybe the coordinates are just for drawing, and the actual positions are different?

Alternatively, maybe the circles are tangent to each other in some other way. For example, maybe each circle is tangent to two other circles and to two sides. For example, circle Q is tangent to sides AB and BC, circle R is tangent to BC and AC, circle S is tangent to AC and AB, and circle P is tangent to all three sides? But then P would be the incenter. But then the inradius would be 1? But if the inradius is 1, then the distance from P to each side is 1. But then the other circles are tangent to two sides and to the inradius circle? Maybe?

Alternatively, maybe there are three circles tangent to two sides and to each other, and the fourth circle is tangent to all three circles? But this is getting complicated. Maybe I need to think about the configuration.

Alternatively, since the problem is about triangle PQS, maybe I can find coordinates of P, Q, S and compute the angles?

But if I can't rely on the Asymptote coordinates because they might not be accurate, maybe I need another approach.

Alternatively, maybe the triangle ABC is such that the centers P, Q, R, S form some known configuration. For example, if the circles are tangent to each other and to the sides, maybe they are the mixtilinear incircles?

Wait, in a triangle, the mixtilinear incircles touch two sides and the circumcircle. But maybe not. Alternatively, maybe they are the A-, B-, C- and the inradius?

Alternatively, maybe the triangle ABC is such that the four circles are the incircle and three excircles? But excircles are tangent to one side and the extensions of the other two. Not sure.

Alternatively, maybe the triangle ABC is divided into smaller triangles by the circles, and the centers form triangle PQS. Maybe I need to use some properties of tangent circles.

Alternatively, since all circles have radius 1, and they are tangent to each other, the distance between centers is 2. So if I can figure out the positions of P, Q, R, S relative to each other, maybe I can find coordinates.

But without knowing the exact configuration, it's hard. Maybe I need to assume that triangle ABC is equilateral? If ABC is equilateral, then maybe the centers of the circles can be arranged symmetrically?

Wait, if ABC is equilateral, then the inradius is (a√3)/6, where a is the side length. If the inradius is 1, then a = 2√3. But if there are four circles, maybe the incircle and three other circles? Not sure.

Alternatively, maybe the triangle ABC is such that the centers P, Q, R, S form a square? But with four circles, maybe?

Alternatively, since the Asymptote code has R at (0,0), Q at (-2,0), S at (2,0), and P at (1, sqrt(3)), maybe the triangle ABC is such that BC is the base from (-5.73,-1) to (3.732,-1), which is a horizontal line. Then R is at (0,0), which is above the base. Q is at (-2,0), S is at (2,0). So maybe Q and S are centers of circles tangent to the base BC and to the sides AB and AC? Wait, but if they are tangent to the base BC, then their y-coordinate should be equal to the radius, which is 1. But in the Asymptote code, Q is at (-2,0), which is y=0, but radius is 1. So the distance from Q to BC should be 1. If BC is at y=-1, then the distance from Q to BC is |0 - (-1)| = 1, which is correct. Similarly for S at (2,0), distance to BC is 1. So Q and S are tangent to BC. Then R is at (0,0), which is also at distance 1 from BC, so R is also tangent to BC? But then there are three circles tangent to BC: Q, R, S. But they are also tangent to each other? Q and R are 2 units apart, so they are tangent. R and S are 2 units apart, so they are tangent. But Q and S are 4 units apart, so they are not tangent. So maybe Q and S are not tangent? But the problem says all four circles are tangent to one another. So maybe there is a mistake here?

Alternatively, maybe the Asymptote code is illustrative but not exact. Maybe the coordinates are approximate. For example, if Q is at (-2,0), R at (0,0), S at (2,0), and P at (1, sqrt(3)), then maybe the triangle ABC is such that the circles are tangent to the sides. Let me check if P is tangent to the other circles. The distance from P to Q is sqrt[(1 - (-2))² + (sqrt(3) - 0)^2] = sqrt[9 + 3] = sqrt(12) ≈ 3.464. Since the radius is 1, the distance between centers should be 2 if they are tangent. So they are not tangent. Therefore, maybe the Asymptote code is not accurate? Maybe the coordinates are just for drawing, and the actual positions are different?

Alternatively, maybe the circles are tangent to each other in a different way. For example, maybe P is tangent to Q and R, Q is tangent to R and S, and S is tangent to P? But then the distances would need to be 2. Let me try to imagine.

Alternatively, maybe the centers form a square? If P, Q, R, S are arranged in a square with side length 2, but then the distance between P and Q would be 2, but in the Asymptote code, P is at (1, sqrt(3)), Q is at (-2,0). The distance between them is sqrt( (1 + 2)^2 + (sqrt(3) - 0)^2 ) = sqrt(9 + 3) = sqrt(12) ≈ 3.464, which is not 2. So that doesn't fit.

Alternatively, maybe triangle PQS is a specific triangle. The problem asks for the smallest angle in triangle PQS. Maybe I can find coordinates of P, Q, S based on the Asymptote code, even if they are not perfectly accurate, and compute the angles?

Let me try that. Let me take the coordinates given in the Asymptote code:

P = (1, 1.732) ≈ (1, sqrt(3))

Q = (-2, 0)

S = (2, 0)

So triangle PQS has coordinates:

P(1, sqrt(3)), Q(-2, 0), S(2, 0)

Let me compute the lengths of the sides of triangle PQS.

First, PQ: distance between P(1, sqrt(3)) and Q(-2, 0):

dx = 1 - (-2) = 3

dy = sqrt(3) - 0 = sqrt(3)

PQ = sqrt(3² + (sqrt(3))²) = sqrt(9 + 3) = sqrt(12) = 2*sqrt(3) ≈ 3.464

QS: distance between Q(-2, 0) and S(2, 0):

dx = 2 - (-2) = 4

dy = 0 - 0 = 0

QS = 4

PS: distance between P(1, sqrt(3)) and S(2, 0):

dx = 2 - 1 = 1

dy = 0 - sqrt(3) = -sqrt(3)

PS = sqrt(1² + (sqrt(3))²) = sqrt(1 + 3) = sqrt(4) = 2

So triangle PQS has sides of length 2, 4, and 2*sqrt(3). Wait, let me check:

Wait, PQ is 2*sqrt(3), QS is 4, PS is 2. So sides are 2, 2√3, 4.

Wait, but in a triangle, the sum of any two sides must be greater than the third. Let's check:

2 + 2√3 ≈ 2 + 3.464 ≈ 5.464 > 4

2 + 4 = 6 > 2√3 ≈ 3.464

2√3 + 4 ≈ 3.464 + 4 ≈ 7.464 > 2

So it's a valid triangle.

Now, to find the smallest angle in triangle PQS. The smallest angle is opposite the shortest side. The sides are 2, 2√3, and 4. The shortest side is 2, which is PS. Therefore, the smallest angle is angle at Q? Wait, no. Wait, the side opposite angle Q is PS. Wait, let me recall:

In triangle PQS, the sides are:

- PQ = 2√3

- QS = 4

- PS = 2

So the side opposite angle P is QS = 4

The side opposite angle Q is PS = 2

The side opposite angle S is PQ = 2√3

Therefore, the smallest side is PS = 2, which is opposite angle Q. Therefore, the smallest angle is angle Q.

Wait, but let me confirm:

Wait, angle at Q is between sides PQ and QS. The side opposite angle Q is PS. Since PS is the shortest side, angle Q is the smallest angle.

But let me compute the angles to be sure.

Using the Law of Cosines.

First, let's compute angle at Q (angle Q):

cos(angle Q) = (PQ² + QS² - PS²) / (2 * PQ * QS)

Wait, no. Wait, Law of Cosines: for angle Q, which is between sides PQ and QS, the side opposite is PS.

So:

PS² = PQ² + QS² - 2 * PQ * QS * cos(angle Q)

Therefore:

cos(angle Q) = (PQ² + QS² - PS²) / (2 * PQ * QS)

Plugging in the numbers:

PQ² = (2√3)² = 12

QS² = 4² = 16

PS² = 2² = 4

So:

cos(angle Q) = (12 + 16 - 4) / (2 * 2√3 * 4) = (24) / (16√3) = 24 / (16√3) = 3 / (2√3) = √3 / 2

Therefore, angle Q is arccos(√3 / 2) = 30 degrees.

Wait, arccos(√3 / 2) is 30 degrees. So angle Q is 30 degrees.

Then, let's check angle at S:

Using Law of Cosines again.

cos(angle S) = (PS² + QS² - PQ²) / (2 * PS * QS)

Plugging in:

(4 + 16 - 12) / (2 * 2 * 4) = (8) / 16 = 0.5

Therefore, angle S is arccos(0.5) = 60 degrees.

Then angle at P:

Since the sum of angles is 180 degrees, angle P = 180 - 30 - 60 = 90 degrees.

Wait, so triangle PQS has angles 30°, 60°, 90°, with the smallest angle being 30°, which is angle Q.

But the problem says "the degree measure of the smallest angle in triangle PQS". So is it 30 degrees? But wait, let me check if my calculation is correct.

Wait, angle Q is 30°, angle S is 60°, angle P is 90°, so yes, the smallest angle is 30°. Therefore, the answer is 30 degrees?

But wait, the problem says "the smallest angle in triangle PQS". But according to this, it's 30 degrees. However, I need to make sure that the coordinates given in the Asymptote code are accurate enough. Because if the Asymptote code is illustrative but not exact, maybe the actual configuration is different?

But given that the problem is from an Olympiad or competition, maybe the answer is 30 degrees. However, let me think again.

Wait, but in the Asymptote code, the coordinates of P are (1, 1.732), which is approximately (1, sqrt(3)), and Q is (-2, 0), S is (2, 0). So triangle PQS is a triangle with coordinates as above, leading to angles 30°, 60°, 90°. Therefore, the smallest angle is 30°, so the answer is 30 degrees. But let me check if there is a mistake in assuming the coordinates.

Alternatively, maybe the Asymptote code is accurate, and the problem is designed such that triangle PQS is a 30-60-90 triangle. Therefore, the answer is 30 degrees. However, let me think again about the configuration.

But why would the problem mention that the circles are tangent to the sides of triangle ABC? Maybe there is a relation between triangle ABC and triangle PQS? For example, if ABC is a larger triangle, and PQS is inside it. But if I use the coordinates given, then triangle ABC has coordinates A=(1.366,3.098), B=(-5.73,-1), C=(3.732,-1). Let me check if the circles are tangent to the sides.

For example, circle Q is at (-2, 0) with radius 1. The distance from Q to side BC should be 1. Since BC is the base at y=-1, the distance from Q to BC is |0 - (-1)| = 1, which is correct. Similarly for circle S at (2,0), distance to BC is 1. For circle R at (0,0), distance to BC is 1. So all three circles Q, R, S are tangent to BC. What about the other sides?

For circle Q at (-2, 0), radius 1. It should be tangent to another side, say AB. Let me check if the distance from Q to AB is 1. Similarly for circle S to AC.

But without knowing the equation of AB and AC, it's hard to check. However, given that the Asymptote code draws the triangle ABC with those coordinates, maybe we can compute equations of AB and AC.

Let me try. Coordinates:

A=(1.366, 3.098), B=(-5.73, -1), C=(3.732, -1)

First, let me note that B and C are on the line y = -1, as their y-coordinates are -1. So BC is horizontal. Let me compute the equation of AB and AC.

First, equation of AB: points A(1.366, 3.098) and B(-5.73, -1).

Compute the slope of AB:

m_AB = (y_A - y_B)/(x_A - x_B) = (3.098 - (-1))/(1.366 - (-5.73)) = (4.098)/(7.096) ≈ 0.577. Which is approximately 1/√3 ≈ 0.577. So slope is 1/√3. Therefore, the equation of AB is y - y_B = m_AB(x - x_B). Let me write it:

y - (-1) = (1/√3)(x - (-5.73))

y + 1 = (1/√3)(x + 5.73)

Similarly, equation of AC: points A(1.366, 3.098) and C(3.732, -1).

Slope of AC:

m_AC = (y_A - y_C)/(x_A - x_C) = (3.098 - (-1))/(1.366 - 3.732) = (4.098)/(-2.366) ≈ -1.732, which is approximately -√3. Therefore, the equation of AC is y - y_C = m_AC(x - x_C):

y - (-1) = -√3(x - 3.732)

y + 1 = -√3(x - 3.732)

Now, check if circle Q(-2, 0) is tangent to AB. The distance from Q to AB should be equal to radius 1.

The formula for distance from point (x0, y0) to line ax + by + c = 0 is |ax0 + by0 + c| / sqrt(a² + b²).

First, let me write equation of AB in standard form.

From earlier:

y + 1 = (1/√3)(x + 5.73)

Multiply both sides by √3:

√3(y + 1) = x + 5.73

Rearranged:

x - √3 y + (5.73 - √3) = 0

Wait, let me do it step by step.

Starting from y + 1 = (1/√3)(x + 5.73)

Multiply both sides by √3:

√3 y + √3 = x + 5.73

Bring all terms to left:

x - √3 y + (5.73 - √3) = 0

So standard form is x - √3 y + (5.73 - √3) = 0

Therefore, coefficients are a = 1, b = -√3, c = 5.73 - √3

Now, distance from Q(-2, 0) to AB is |a*(-2) + b*(0) + c| / sqrt(a² + b²) = | -2 + 0 + 5.73 - √3 | / sqrt(1 + 3) = |3.73 - √3| / 2

Compute 3.73 - √3 ≈ 3.73 - 1.732 ≈ 1.998 ≈ 2. So the distance is approximately 2 / 2 = 1. Which is equal to the radius. Therefore, the distance from Q to AB is 1, so circle Q is tangent to AB. Similarly, circle S(2, 0) should be tangent to AC.

Let me check that. Equation of AC is y + 1 = -√3(x - 3.732). Let me write it in standard form.

y + 1 = -√3 x + 3.732√3

Bring all terms to left:

√3 x + y + 1 - 3.732√3 = 0

So standard form is √3 x + y + (1 - 3.732√3) = 0

Therefore, coefficients a = √3, b = 1, c = 1 - 3.732√3

Distance from S(2, 0) to AC is |√3*2 + 1*0 + c| / sqrt( (√3)^2 + 1^2 ) = |2√3 + 1 - 3.732√3| / 2

Compute numerator:

2√3 - 3.732√3 + 1 = (-1.732√3) + 1 ≈ -1.732*1.732 + 1 ≈ -3 + 1 = -2. Absolute value is 2. Therefore, distance is 2 / 2 = 1. Which is equal to the radius. Therefore, circle S is tangent to AC. Similarly, circle R(0,0) is tangent to BC (distance 1) and maybe to another side?

Wait, circle R is at (0,0). Let me check if it's tangent to AB or AC. Let me check distance from R(0,0) to AB.

Using the equation of AB: x - √3 y + (5.73 - √3) = 0

Distance is |0 - √3*0 + 5.73 - √3| / 2 = |5.73 - √3| / 2 ≈ |5.73 - 1.732| / 2 ≈ |3.998| / 2 ≈ 1.999 ≈ 2. Which is not equal to 1. Therefore, circle R is not tangent to AB. Similarly, check distance to AC:

Equation of AC: √3 x + y + (1 - 3.732√3) = 0

Distance from R(0,0) is |√3*0 + 0 + 1 - 3.732√3| / 2 = |1 - 3.732√3| / 2 ≈ |1 - 3.732*1.732| / 2 ≈ |1 - 6.464| / 2 ≈ | -5.464 | / 2 ≈ 2.732, which is not 1. Therefore, circle R is not tangent to AB or AC. But earlier, we saw that circles Q, R, S are all tangent to BC. So maybe circle R is tangent to another side? Wait, but the problem says all four circles are tangent to the sides of triangle ABC. So maybe circle R is tangent to AB and AC? But according to the distance, it's not. Wait, maybe there is a mistake here?

Wait, maybe circle R is tangent to AB and AC? If so, then it would be the incenter. But the inradius is 1. Let me check the inradius of triangle ABC.

First, let me compute the coordinates of ABC. A=(1.366, 3.098), B=(-5.73, -1), C=(3.732, -1). Let me compute the lengths of the sides.

First, compute AB:

Coordinates of A(1.366, 3.098) and B(-5.73, -1). The distance AB is sqrt[(1.366 + 5.73)^2 + (3.098 + 1)^2] = sqrt[(7.096)^2 + (4.098)^2]. Let me approximate:

7.096² ≈ 50.35, 4.098² ≈ 16.79, total ≈ 67.14, sqrt ≈ 8.19.

Similarly, AC: distance from A(1.366, 3.098) to C(3.732, -1):

sqrt[(3.732 - 1.366)^2 + (-1 - 3.098)^2] = sqrt[(2.366)^2 + (-4.098)^2] ≈ sqrt[5.598 + 16.79] ≈ sqrt[22.388] ≈ 4.73.

BC: distance from B(-5.73, -1) to C(3.732, -1) is sqrt[(3.732 + 5.73)^2 + 0] = sqrt[(9.462)^2] = 9.462.

So sides are approximately AB ≈ 8.19, AC ≈ 4.73, BC ≈ 9.46.

Now, the inradius formula is area / semiperimeter.

First, compute semiperimeter:

s = (AB + AC + BC)/2 ≈ (8.19 + 4.73 + 9.46)/2 ≈ (22.38)/2 ≈ 11.19.

Now, compute area. Since BC is the base, length 9.46, and height is the y-coordinate of A minus the y-coordinate of BC. Since BC is at y=-1, and A is at y≈3.098, the height is 3.098 - (-1) = 4.098.

Therefore, area ≈ (base * height)/2 ≈ (9.46 * 4.098)/2 ≈ (38.7)/2 ≈ 19.35.

Therefore, inradius r = area / semiperimeter ≈ 19.35 / 11.19 ≈ 1.732, which is approximately sqrt(3). So the inradius is about 1.732, which is larger than 1. Therefore, the inradius is not 1, so circle R is not the inradius. Therefore, circle R is tangent to BC and...? Earlier, we saw that circle R is at (0,0), which is distance 1 from BC, but distance from R to AB and AC is about 2, which is twice the radius. So maybe circle R is tangent to AB and AC? Wait, if the distance from R to AB is 2, which is twice the radius, then maybe there is another circle tangent to AB and AC with radius 1? But that would be the inradius, which is larger. So this seems conflicting.

Alternatively, maybe circle R is tangent to AB and BC? But distance from R to AB is 2, which is twice the radius. If the circle is tangent to AB and BC, then the distance from center to AB and BC should be equal to radius. But distance to BC is 1, which is correct, but distance to AB is 2, which is not. Therefore, circle R is only tangent to BC? But the problem says all four circles are tangent to the sides of triangle ABC. So maybe each circle is tangent to two sides? For example, circle Q is tangent to AB and BC, circle S is tangent to AC and BC, and circle R is tangent to...? Maybe AB and AC? But distance from R to AB and AC is 2, which is twice the radius. So maybe there is a circle tangent to AB and AC with radius 1? But the inradius is larger. Maybe it's an ex-circle? Or maybe there is a mistake in the Asymptote code?

Alternatively, maybe the Asymptote code is illustrative, and the actual configuration is different. However, given that the problem is from a competition, and the Asymptote code is provided, maybe we are supposed to use the coordinates given, even if they are approximate, and compute the angles of triangle PQS as 30°, 60°, 90°, leading to the smallest angle being 30°.

Alternatively, maybe there is a different approach. Let me think again.

If all four circles have radius 1 and are tangent to each other and to the sides of triangle ABC, then maybe triangle ABC is such that it's tangent to four circles. This might be a special triangle. For example, if we have three circles tangent to two sides and to each other, and a fourth circle tangent to all three circles. But this is getting complex.

Alternatively, since the problem is about triangle PQS, and if we can find the coordinates of P, Q, S, then compute the angles. Given that in the Asymptote code, the coordinates lead to triangle PQS with angles 30°, 60°, 90°, and the problem is likely designed with that in mind, then the answer is 30 degrees.

But to be thorough, let me check if there's another way. For example, if triangle PQS is a right triangle with legs 2 and 2√3, hypotenuse 4? Wait, no. Wait, sides are 2, 2√3, and 4. Wait, 2² + (2√3)² = 4 + 12 = 16 = 4². Yes! So triangle PQS is a right triangle with right angle at P. Wait, earlier I calculated angle at P as 90 degrees. Wait, but according to the Law of Cosines, angle at P was 90 degrees. Wait, but if sides are 2, 2√3, 4, then 2² + (2√3)² = 4 + 12 = 16 = 4², so yes, it's a right triangle with right angle at P. Therefore, angles are 30°, 60°, 90°, with the smallest angle being 30°, which is at Q.

Therefore, the answer is 30 degrees. Therefore, I think the answer is 30°, so boxed answer is \boxed{30}.

But let me check once more. If triangle PQS is right-angled at P, then angle at P is 90°, angle at Q is 30°, angle at S is 60°, so yes, the smallest angle is 30°. Therefore, the answer is 30 degrees.

**Final Answer**
\boxed{30}

To determine the **smallest angle** in triangle $ PQS $, we begin by analyzing the given configuration and the coordinates of the centers of the four circles:

- $ P = (1, \sqrt{3}) $
- $ Q = (-2, 0) $
- $ S = (2, 0) $

These coordinates are provided in the Asymptote code and are consistent with the diagram. We are told that all four circles have radius 1 and are tangent to one another and to the sides of triangle $ ABC $. While the exact configuration of triangle $ ABC $ is not necessary for solving the problem, the coordinates of $ P, Q, S $ allow us to compute the angles of triangle $ PQS $ directly.

---

### Step 1: Compute the Side Lengths of Triangle $ PQS $

Using the distance formula:

- $ PQ = \sqrt{(1 - (-2))^2 + (\sqrt{3} - 0)^2} = \sqrt{3^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3} $
- $ QS = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4 $
- $ PS = \sqrt{(2 - 1)^2 + (0 - \sqrt{3})^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 $

So the side lengths of triangle $ PQS $ are:
- $ PQ = 2\sqrt{3} $
- $ QS = 4 $
- $ PS = 2 $

---

### Step 2: Identify the Smallest Angle

In any triangle, the **smallest angle** is opposite the **shortest side**. Here, the shortest side is $ PS = 2 $, which is opposite angle $ Q $.

Thus, the **smallest angle** is $ \angle Q $.

---

### Step 3: Use the Law of Cosines to Find $ \angle Q $

We apply the Law of Cosines to triangle $ PQS $:

$$
\cos(\angle Q) = \frac{PQ^2 + QS^2 - PS^2}{2 \cdot PQ \cdot QS}
$$

Substitute the known values:

$$
\cos(\angle Q) = \frac{(2\sqrt{3})^2 + 4^2 - 2^2}{2 \cdot 2\sqrt{3} \cdot 4}
= \frac{12 + 16 - 4}{16\sqrt{3}}
= \frac{24}{16\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}
$$

$$
\angle Q = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = 30^\circ
$$

---

### Step 4: Confirm the Triangle Type

We can also verify that triangle $ PQS $ is a **right triangle** by checking the Pythagorean Theorem:

$$
PS^2 + PQ^2 = 2^2 + (2\sqrt{3})^2 = 4 + 12 = 16 = QS^2
$$

So, triangle $ PQS $ is a **right triangle** with a right angle at $ P $, and the angles are:

- $ \angle Q = 30^\circ $
- $ \angle S = 60^\circ $
- $ \angle P = 90^\circ $

---

### Final Answer

The **smallest angle** in triangle $ PQS $ is:

$$
\boxed{30}
$$

[Assistant]
Explanation:
Okay, so I need to find the degree measure of the smallest angle in triangle PQS. The diagram shows four circles of radius 1 with centers P, Q, R, and S, all tangent to each other and to the sides of triangle ABC. Hmm, let me try to visualize this. There's a triangle ABC with three circles tangent to its sides and each other, and another circle? Wait, no, four circles. Maybe each circle is tangent to two sides of the triangle and to the other circles? Or maybe they are arranged in some specific way?

Looking at the Asymptote code, the coordinates are given for points R, Q, S, and P. Let me parse that. R is at (0,0), Q is at (-2,0), S is at (2,0), and P is at (1, 1.732). Then points A, B, C are given with coordinates that seem more complicated. But maybe I don't need the exact coordinates? Or maybe I can use them?

Wait, the problem says the circles are tangent to one another and to the sides of triangle ABC. So each circle is tangent to two sides of the triangle? Or maybe each circle is tangent to two other circles and the sides? Let me think. If there are four circles, maybe they are arranged in a way similar to the Soddy circles or something? Or maybe they are the incircle and three mixtilinear incircles? But the problem says four circles tangent to each other and to the sides. Maybe they are all tangent to two sides of the triangle and to each other?

Alternatively, maybe the triangle ABC is such that each side has a circle tangent to it and to the other circles. But with four circles, perhaps two circles on the base BC, one on AB, and one on AC? But the Asymptote code shows R at (0,0), Q at (-2,0), S at (2,0), so maybe R and Q and S are on the base BC? Wait, but there are three circles on the base? Wait, but there are four circles. Maybe R is at the bottom, Q and S are on the base? Wait, the Asymptote code draws the triangle ABC with coordinates A=(1.366,3.098), B=(-5.73,-1), C=(3.732,-1). So the base BC is from (-5.73,-1) to (3.732,-1), which is a horizontal line. Then R is at (0,0), which is above the base BC? Wait, but the base BC is at y=-1. So R is at (0,0), which is above the base. Similarly, Q is at (-2,0), S is at (2,0). So maybe Q and S are on the sides AB and AC? Or maybe they are the centers of circles tangent to the sides?

Wait, the problem says the circles are tangent to one another and to the sides of triangle ABC. So each circle is tangent to two sides of the triangle and to the other circles. So maybe there are three circles tangent to two sides each, and the fourth circle is tangent to all three sides? But that would be the incircle. But there are four circles. Maybe the incircle and three mixtilinear incircles? But I need to think more carefully.

Alternatively, maybe the triangle ABC is such that there are four circles, each tangent to two sides and to the other circles. For example, in a triangle, you can have circles tangent to two sides and to another circle. Maybe like the Apollonius circles?

But maybe I can use coordinates. The Asymptote code gives coordinates for points P, Q, R, S. Let me check:

R=(0,0); Q=(-2,0); S=(2,0); P=(1,1.732). So P is at (1, sqrt(3)) since 1.732 is approximately sqrt(3). So P is at (1, sqrt(3)), Q is at (-2,0), R is at (0,0), S is at (2,0). Then triangle ABC is drawn with A=(1.366,3.098), B=(-5.73,-1), C=(3.732,-1). Let me check if these coordinates make sense.

First, let me note that sqrt(3) is approximately 1.732, so P is at (1, sqrt(3)). The points Q, R, S are on the x-axis. The triangle ABC has vertices at A, B, C. Let me check if the circles are tangent to the sides.

Each circle has radius 1. So the distance between centers of tangent circles should be 2, since they are tangent. Let me check the distance between Q and R: Q is (-2,0), R is (0,0). The distance is 2 units, which is equal to 2*radius, so they are tangent. Similarly, distance between R and S is 2 units, so they are tangent. Distance between Q and S is 4 units, which would be 2*radius*2? Wait, if they are tangent, the distance between centers should be 2*radius. But radius is 1, so distance should be 2. But Q is at (-2,0), S is at (2,0), so distance is 4. That's not 2. So they are not tangent? Wait, but the problem says all four circles are tangent to one another. So maybe my assumption is wrong?

Wait, maybe the circles are tangent to each other in some other way? Maybe not all pairs? Wait, the problem says "four circles of radius 1 with centres P, Q, R, and S are tangent to one another and to the sides of triangle ABC". So all four circles are tangent to each other? That would mean each pair of circles is tangent. But if Q and S are 4 units apart, and each has radius 1, then the distance between centers should be 2 units for them to be tangent. But 4 is not 2. Therefore, there must be something wrong here.

Wait, maybe the Asymptote code is just an illustration, and the coordinates are approximate? Because if the circles are tangent to each other, then the distance between centers should be 2. But in the Asymptote code, Q is at (-2,0), R at (0,0), S at (2,0), so QR is 2 units, RS is 2 units, but QS is 4 units. Therefore, if the circles at Q and S are tangent, they should be 2 units apart, but they are 4 units apart. Therefore, maybe the Asymptote code is not to scale? Or maybe the circles are not all tangent to each other? Wait, the problem says "four circles... are tangent to one another and to the sides of triangle ABC". So maybe not all pairs are tangent? Maybe each circle is tangent to two others and the sides?

Alternatively, maybe the diagram is not to scale, and the coordinates are just for drawing purposes. Maybe I shouldn't rely on them? Maybe I need to think geometrically.

Let me think again. If there are four circles inside triangle ABC, each tangent to two sides and to the other circles. Maybe they are arranged such that two circles are on the base BC, tangent to each other and to the base, and the other two circles are on the other sides?

Wait, but there are four circles. Maybe there is an incircle and three other circles? But the problem says four circles. Maybe it's similar to the problem of circles tangent to two sides and to each other. For example, in a triangle, you can have circles tangent to two sides and to another circle. Maybe there are two circles on the base, one on the left side, and one on the right side?

Alternatively, maybe the triangle is equilateral? Because the coordinates of P are (1, sqrt(3)), which is like a height of an equilateral triangle. Wait, if the triangle ABC is equilateral, then maybe?

But the coordinates given for A, B, C are not symmetric. Let me check. A is at (1.366, 3.098). Let me approximate sqrt(3) is about 1.732, so 1.366 is roughly 1.366, which is close to 1.366... Wait, 1.366 is approximately 1 + 0.366, and 0.366 is roughly 1/2.732, which is not sure. Maybe it's related to some specific triangle?

Alternatively, maybe triangle ABC is a 30-60-90 triangle? Or some other triangle. Alternatively, maybe I need to use coordinates given in the Asymptote code to find coordinates of P, Q, R, S and then compute the angles of triangle PQS?

But earlier, there was a problem with the distance between Q and S. If the Asymptote code is accurate, then maybe the circles are not all tangent to each other? Wait, the problem says "four circles... are tangent to one another and to the sides of triangle ABC". So maybe the Asymptote code is illustrative but not exact? Maybe the coordinates are approximate?

Alternatively, maybe the circles are tangent to each other in pairs? For example, P is tangent to Q, Q is tangent to R, R is tangent to S, and S is tangent to P? But then the distance between P and Q should be 2, between Q and R is 2, R and S is 2, and S and P is 2? But in the Asymptote code, P is at (1, sqrt(3)), Q is at (-2,0), so distance between P and Q is sqrt[(1 - (-2))² + (sqrt(3) - 0)²] = sqrt[3² + (sqrt(3))²] = sqrt[9 + 3] = sqrt(12) ≈ 3.464, which is more than 2. So that's not tangent. Therefore, maybe the Asymptote code is not accurate? Maybe the coordinates are just for drawing, and the actual positions are different?

Alternatively, maybe the circles are tangent to each other in some other way. For example, maybe each circle is tangent to two other circles and to two sides. For example, circle Q is tangent to sides AB and BC, circle R is tangent to BC and AC, circle S is tangent to AC and AB, and circle P is tangent to all three sides? But then P would be the incenter. But then the inradius would be 1? But if the inradius is 1, then the distance from P to each side is 1. But then the other circles are tangent to two sides and to the inradius circle? Maybe?

Alternatively, maybe there are three circles tangent to two sides and to each other, and the fourth circle is tangent to all three circles? But this is getting complicated. Maybe I need to think about the configuration.

Alternatively, since the problem is about triangle PQS, maybe I can find coordinates of P, Q, S and compute the angles?

But if I can't rely on the Asymptote coordinates because they might not be accurate, maybe I need another approach.

Alternatively, maybe the triangle ABC is such that the centers P, Q, R, S form some known configuration. For example, if the circles are tangent to each other and to the sides, maybe they are the mixtilinear incircles?

Wait, in a triangle, the mixtilinear incircles touch two sides and the circumcircle. But maybe not. Alternatively, maybe they are the A-, B-, C- and the inradius?

Alternatively, maybe the triangle ABC is such that the four circles are the incircle and three excircles? But excircles are tangent to one side and the extensions of the other two. Not sure.

Alternatively, maybe the triangle ABC is divided into smaller triangles by the circles, and the centers form triangle PQS. Maybe I need to use some properties of tangent circles.

Alternatively, since all circles have radius 1, and they are tangent to each other, the distance between centers is 2. So if I can figure out the positions of P, Q, R, S relative to each other, maybe I can find coordinates.

But without knowing the exact configuration, it's hard. Maybe I need to assume that triangle ABC is equilateral? If ABC is equilateral, then maybe the centers of the circles can be arranged symmetrically?

Wait, if ABC is equilateral, then the inradius is (a√3)/6, where a is the side length. If the inradius is 1, then a = 2√3. But if there are four circles, maybe the incircle and three other circles? Not sure.

Alternatively, maybe the triangle ABC is such that the centers P, Q, R, S form a square? But with four circles, maybe?

Alternatively, since the Asymptote code has R at (0,0), Q at (-2,0), S at (2,0), and P at (1, sqrt(3)), maybe the triangle ABC is such that BC is the base from (-5.73,-1) to (3.732,-1), which is a horizontal line. Then R is at (0,0), which is above the base. Q is at (-2,0), S is at (2,0). So maybe Q and S are centers of circles tangent to the base BC and to the sides AB and AC? Wait, but if they are tangent to the base BC, then their y-coordinate should be equal to the radius, which is 1. But in the Asymptote code, Q is at (-2,0), which is y=0, but radius is 1. So the distance from Q to BC should be 1. If BC is at y=-1, then the distance from Q to BC is |0 - (-1)| = 1, which is correct. Similarly for S at (2,0), distance to BC is 1. So Q and S are tangent to BC. Then R is at (0,0), which is also at distance 1 from BC, so R is also tangent to BC? But then there are three circles tangent to BC: Q, R, S. But they are also tangent to each other? Q and R are 2 units apart, so they are tangent. R and S are 2 units apart, so they are tangent. But Q and S are 4 units apart, so they are not tangent. So maybe Q and S are not tangent? But the problem says all four circles are tangent to one another. So maybe there is a mistake here?

Alternatively, maybe the Asymptote code is illustrative but not exact. Maybe the coordinates are approximate. For example, if Q is at (-2,0), R at (0,0), S at (2,0), and P at (1, sqrt(3)), then maybe the triangle ABC is such that the circles are tangent to the sides. Let me check if P is tangent to the other circles. The distance from P to Q is sqrt[(1 - (-2))² + (sqrt(3) - 0)^2] = sqrt[9 + 3] = sqrt(12) ≈ 3.464. Since the radius is 1, the distance between centers should be 2 if they are tangent. So they are not tangent. Therefore, maybe the Asymptote code is not accurate? Maybe the coordinates are just for drawing, and the actual positions are different?

Alternatively, maybe the circles are tangent to each other in a different way. For example, maybe P is tangent to Q and R, Q is tangent to R and S, and S is tangent to P? But then the distances would need to be 2. Let me try to imagine.

Alternatively, maybe the centers form a square? If P, Q, R, S are arranged in a square with side length 2, but then the distance between P and Q would be 2, but in the Asymptote code, P is at (1, sqrt(3)), Q is at (-2,0). The distance between them is sqrt( (1 + 2)^2 + (sqrt(3) - 0)^2 ) = sqrt(9 + 3) = sqrt(12) ≈ 3.464, which is not 2. So that doesn't fit.

Alternatively, maybe triangle PQS is a specific triangle. The problem asks for the smallest angle in triangle PQS. Maybe I can find coordinates of P, Q, S based on the Asymptote code, even if they are not perfectly accurate, and compute the angles?

Let me try that. Let me take the coordinates given in the Asymptote code:

P = (1, 1.732) ≈ (1, sqrt(3))

Q = (-2, 0)

S = (2, 0)

So triangle PQS has coordinates:

P(1, sqrt(3)), Q(-2, 0), S(2, 0)

Let me compute the lengths of the sides of triangle PQS.

First, PQ: distance between P(1, sqrt(3)) and Q(-2, 0):

dx = 1 - (-2) = 3

dy = sqrt(3) - 0 = sqrt(3)

PQ = sqrt(3² + (sqrt(3))²) = sqrt(9 + 3) = sqrt(12) = 2*sqrt(3) ≈ 3.464

QS: distance between Q(-2, 0) and S(2, 0):

dx = 2 - (-2) = 4

dy = 0 - 0 = 0

QS = 4

PS: distance between P(1, sqrt(3)) and S(2, 0):

dx = 2 - 1 = 1

dy = 0 - sqrt(3) = -sqrt(3)

PS = sqrt(1² + (sqrt(3))²) = sqrt(1 + 3) = sqrt(4) = 2

So triangle PQS has sides of length 2, 4, and 2*sqrt(3). Wait, let me check:

Wait, PQ is 2*sqrt(3), QS is 4, PS is 2. So sides are 2, 2√3, 4.

Wait, but in a triangle, the sum of any two sides must be greater than the third. Let's check:

2 + 2√3 ≈ 2 + 3.464 ≈ 5.464 > 4

2 + 4 = 6 > 2√3 ≈ 3.464

2√3 + 4 ≈ 3.464 + 4 ≈ 7.464 > 2

So it's a valid triangle.

Now, to find the smallest angle in triangle PQS. The smallest angle is opposite the shortest side. The sides are 2, 2√3, and 4. The shortest side is 2, which is PS. Therefore, the smallest angle is angle at Q? Wait, no. Wait, the side opposite angle Q is PS. Wait, let me recall:

In triangle PQS, the sides are:

- PQ = 2√3

- QS = 4

- PS = 2

So the side opposite angle P is QS = 4

The side opposite angle Q is PS = 2

The side opposite angle S is PQ = 2√3

Therefore, the smallest side is PS = 2, which is opposite angle Q. Therefore, the smallest angle is angle Q.

Wait, but let me confirm:

Wait, angle at Q is between sides PQ and QS. The side opposite angle Q is PS. Since PS is the shortest side, angle Q is the smallest angle.

But let me compute the angles to be sure.

Using the Law of Cosines.

First, let's compute angle at Q (angle Q):

cos(angle Q) = (PQ² + QS² - PS²) / (2 * PQ * QS)

Wait, no. Wait, Law of Cosines: for angle Q, which is between sides PQ and QS, the side opposite is PS.

So:

PS² = PQ² + QS² - 2 * PQ * QS * cos(angle Q)

Therefore:

cos(angle Q) = (PQ² + QS² - PS²) / (2 * PQ * QS)

Plugging in the numbers:

PQ² = (2√3)² = 12

QS² = 4² = 16

PS² = 2² = 4

So:

cos(angle Q) = (12 + 16 - 4) / (2 * 2√3 * 4) = (24) / (16√3) = 24 / (16√3) = 3 / (2√3) = √3 / 2

Therefore, angle Q is arccos(√3 / 2) = 30 degrees.

Wait, arccos(√3 / 2) is 30 degrees. So angle Q is 30 degrees.

Then, let's check angle at S:

Using Law of Cosines again.

cos(angle S) = (PS² + QS² - PQ²) / (2 * PS * QS)

Plugging in:

(4 + 16 - 12) / (2 * 2 * 4) = (8) / 16 = 0.5

Therefore, angle S is arccos(0.5) = 60 degrees.

Then angle at P:

Since the sum of angles is 180 degrees, angle P = 180 - 30 - 60 = 90 degrees.

Wait, so triangle PQS has angles 30°, 60°, 90°, with the smallest angle being 30°, which is angle Q.

But the problem says "the degree measure of the smallest angle in triangle PQS". So is it 30 degrees? But wait, let me check if my calculation is correct.

Wait, angle Q is 30°, angle S is 60°, angle P is 90°, so yes, the smallest angle is 30°. Therefore, the answer is 30 degrees?

But wait, the problem says "the smallest angle in triangle PQS". But according to this, it's 30 degrees. However, I need to make sure that the coordinates given in the Asymptote code are accurate enough. Because if the Asymptote code is illustrative but not exact, maybe the actual configuration is different?

But given that the problem is from an Olympiad or competition, maybe the answer is 30 degrees. However, let me think again.

Wait, but in the Asymptote code, the coordinates of P are (1, 1.732), which is approximately (1, sqrt(3)), and Q is (-2, 0), S is (2, 0). So triangle PQS is a triangle with coordinates as above, leading to angles 30°, 60°, 90°. Therefore, the smallest angle is 30°, so the answer is 30 degrees. But let me check if there is a mistake in assuming the coordinates.

Alternatively, maybe the Asymptote code is accurate, and the problem is designed such that triangle PQS is a 30-60-90 triangle. Therefore, the answer is 30 degrees. However, let me think again about the configuration.

But why would the problem mention that the circles are tangent to the sides of triangle ABC? Maybe there is a relation between triangle ABC and triangle PQS? For example, if ABC is a larger triangle, and PQS is inside it. But if I use the coordinates given, then triangle ABC has coordinates A=(1.366,3.098), B=(-5.73,-1), C=(3.732,-1). Let me check if the circles are tangent to the sides.

For example, circle Q is at (-2, 0) with radius 1. The distance from Q to side BC should be 1. Since BC is the base at y=-1, the distance from Q to BC is |0 - (-1)| = 1, which is correct. Similarly for circle S at (2,0), distance to BC is 1. For circle R at (0,0), distance to BC is 1. So all three circles Q, R, S are tangent to BC. What about the other sides?

For circle Q at (-2, 0), radius 1. It should be tangent to another side, say AB. Let me check if the distance from Q to AB is 1. Similarly for circle S to AC.

But without knowing the equation of AB and AC, it's hard to check. However, given that the Asymptote code draws the triangle ABC with those coordinates, maybe we can compute equations of AB and AC.

Let me try. Coordinates:

A=(1.366, 3.098), B=(-5.73, -1), C=(3.732, -1)

First, let me note that B and C are on the line y = -1, as their y-coordinates are -1. So BC is horizontal. Let me compute the equation of AB and AC.

First, equation of AB: points A(1.366, 3.098) and B(-5.73, -1).

Compute the slope of AB:

m_AB = (y_A - y_B)/(x_A - x_B) = (3.098 - (-1))/(1.366 - (-5.73)) = (4.098)/(7.096) ≈ 0.577. Which is approximately 1/√3 ≈ 0.577. So slope is 1/√3. Therefore, the equation of AB is y - y_B = m_AB(x - x_B). Let me write it:

y - (-1) = (1/√3)(x - (-5.73))

y + 1 = (1/√3)(x + 5.73)

Similarly, equation of AC: points A(1.366, 3.098) and C(3.732, -1).

Slope of AC:

m_AC = (y_A - y_C)/(x_A - x_C) = (3.098 - (-1))/(1.366 - 3.732) = (4.098)/(-2.366) ≈ -1.732, which is approximately -√3. Therefore, the equation of AC is y - y_C = m_AC(x - x_C):

y - (-1) = -√3(x - 3.732)

y + 1 = -√3(x - 3.732)

Now, check if circle Q(-2, 0) is tangent to AB. The distance from Q to AB should be equal to radius 1.

The formula for distance from point (x0, y0) to line ax + by + c = 0 is |ax0 + by0 + c| / sqrt(a² + b²).

First, let me write equation of AB in standard form.

From earlier:

y + 1 = (1/√3)(x + 5.73)

Multiply both sides by √3:

√3(y + 1) = x + 5.73

Rearranged:

x - √3 y + (5.73 - √3) = 0

Wait, let me do it step by step.

Starting from y + 1 = (1/√3)(x + 5.73)

Multiply both sides by √3:

√3 y + √3 = x + 5.73

Bring all terms to left:

x - √3 y + (5.73 - √3) = 0

So standard form is x - √3 y + (5.73 - √3) = 0

Therefore, coefficients are a = 1, b = -√3, c = 5.73 - √3

Now, distance from Q(-2, 0) to AB is |a*(-2) + b*(0) + c| / sqrt(a² + b²) = | -2 + 0 + 5.73 - √3 | / sqrt(1 + 3) = |3.73 - √3| / 2

Compute 3.73 - √3 ≈ 3.73 - 1.732 ≈ 1.998 ≈ 2. So the distance is approximately 2 / 2 = 1. Which is equal to the radius. Therefore, the distance from Q to AB is 1, so circle Q is tangent to AB. Similarly, circle S(2, 0) should be tangent to AC.

Let me check that. Equation of AC is y + 1 = -√3(x - 3.732). Let me write it in standard form.

y + 1 = -√3 x + 3.732√3

Bring all terms to left:

√3 x + y + 1 - 3.732√3 = 0

So standard form is √3 x + y + (1 - 3.732√3) = 0

Therefore, coefficients a = √3, b = 1, c = 1 - 3.732√3

Distance from S(2, 0) to AC is |√3*2 + 1*0 + c| / sqrt( (√3)^2 + 1^2 ) = |2√3 + 1 - 3.732√3| / 2

Compute numerator:

2√3 - 3.732√3 + 1 = (-1.732√3) + 1 ≈ -1.732*1.732 + 1 ≈ -3 + 1 = -2. Absolute value is 2. Therefore, distance is 2 / 2 = 1. Which is equal to the radius. Therefore, circle S is tangent to AC. Similarly, circle R(0,0) is tangent to BC (distance 1) and maybe to another side?

Wait, circle R is at (0,0). Let me check if it's tangent to AB or AC. Let me check distance from R(0,0) to AB.

Using the equation of AB: x - √3 y + (5.73 - √3) = 0

Distance is |0 - √3*0 + 5.73 - √3| / 2 = |5.73 - √3| / 2 ≈ |5.73 - 1.732| / 2 ≈ |3.998| / 2 ≈ 1.999 ≈ 2. Which is not equal to 1. Therefore, circle R is not tangent to AB. Similarly, check distance to AC:

Equation of AC: √3 x + y + (1 - 3.732√3) = 0

Distance from R(0,0) is |√3*0 + 0 + 1 - 3.732√3| / 2 = |1 - 3.732√3| / 2 ≈ |1 - 3.732*1.732| / 2 ≈ |1 - 6.464| / 2 ≈ | -5.464 | / 2 ≈ 2.732, which is not 1. Therefore, circle R is not tangent to AB or AC. But earlier, we saw that circles Q, R, S are all tangent to BC. So maybe circle R is tangent to another side? Wait, but the problem says all four circles are tangent to the sides of triangle ABC. So maybe circle R is tangent to AB and AC? But according to the distance, it's not. Wait, maybe there is a mistake here?

Wait, maybe circle R is tangent to AB and AC? If so, then it would be the incenter. But the inradius is 1. Let me check the inradius of triangle ABC.

First, let me compute the coordinates of ABC. A=(1.366, 3.098), B=(-5.73, -1), C=(3.732, -1). Let me compute the lengths of the sides.

First, compute AB:

Coordinates of A(1.366, 3.098) and B(-5.73, -1). The distance AB is sqrt[(1.366 + 5.73)^2 + (3.098 + 1)^2] = sqrt[(7.096)^2 + (4.098)^2]. Let me approximate:

7.096² ≈ 50.35, 4.098² ≈ 16.79, total ≈ 67.14, sqrt ≈ 8.19.

Similarly, AC: distance from A(1.366, 3.098) to C(3.732, -1):

sqrt[(3.732 - 1.366)^2 + (-1 - 3.098)^2] = sqrt[(2.366)^2 + (-4.098)^2] ≈ sqrt[5.598 + 16.79] ≈ sqrt[22.388] ≈ 4.73.

BC: distance from B(-5.73, -1) to C(3.732, -1) is sqrt[(3.732 + 5.73)^2 + 0] = sqrt[(9.462)^2] = 9.462.

So sides are approximately AB ≈ 8.19, AC ≈ 4.73, BC ≈ 9.46.

Now, the inradius formula is area / semiperimeter.

First, compute semiperimeter:

s = (AB + AC + BC)/2 ≈ (8.19 + 4.73 + 9.46)/2 ≈ (22.38)/2 ≈ 11.19.

Now, compute area. Since BC is the base, length 9.46, and height is the y-coordinate of A minus the y-coordinate of BC. Since BC is at y=-1, and A is at y≈3.098, the height is 3.098 - (-1) = 4.098.

Therefore, area ≈ (base * height)/2 ≈ (9.46 * 4.098)/2 ≈ (38.7)/2 ≈ 19.35.

Therefore, inradius r = area / semiperimeter ≈ 19.35 / 11.19 ≈ 1.732, which is approximately sqrt(3). So the inradius is about 1.732, which is larger than 1. Therefore, the inradius is not 1, so circle R is not the inradius. Therefore, circle R is tangent to BC and...? Earlier, we saw that circle R is at (0,0), which is distance 1 from BC, but distance from R to AB and AC is about 2, which is twice the radius. So maybe circle R is tangent to AB and AC? Wait, if the distance from R to AB is 2, which is twice the radius, then maybe there is another circle tangent to AB and AC with radius 1? But that would be the inradius, which is larger. So this seems conflicting.

Alternatively, maybe circle R is tangent to AB and BC? But distance from R to AB is 2, which is twice the radius. If the circle is tangent to AB and BC, then the distance from center to AB and BC should be equal to radius. But distance to BC is 1, which is correct, but distance to AB is 2, which is not. Therefore, circle R is only tangent to BC? But the problem says all four circles are tangent to the sides of triangle ABC. So maybe each circle is tangent to two sides? For example, circle Q is tangent to AB and BC, circle S is tangent to AC and BC, and circle R is tangent to...? Maybe AB and AC? But distance from R to AB and AC is 2, which is twice the radius. So maybe there is a circle tangent to AB and AC with radius 1? But the inradius is larger. Maybe it's an ex-circle? Or maybe there is a mistake in the Asymptote code?

Alternatively, maybe the Asymptote code is illustrative, and the actual configuration is different. However, given that the problem is from a competition, and the Asymptote code is provided, maybe we are supposed to use the coordinates given, even if they are approximate, and compute the angles of triangle PQS as 30°, 60°, 90°, leading to the smallest angle being 30°.

Alternatively, maybe there is a different approach. Let me think again.

If all four circles have radius 1 and are tangent to each other and to the sides of triangle ABC, then maybe triangle ABC is such that it's tangent to four circles. This might be a special triangle. For example, if we have three circles tangent to two sides and to each other, and a fourth circle tangent to all three circles. But this is getting complex.

Alternatively, since the problem is about triangle PQS, and if we can find the coordinates of P, Q, S, then compute the angles. Given that in the Asymptote code, the coordinates lead to triangle PQS with angles 30°, 60°, 90°, and the problem is likely designed with that in mind, then the answer is 30 degrees.

But to be thorough, let me check if there's another way. For example, if triangle PQS is a right triangle with legs 2 and 2√3, hypotenuse 4? Wait, no. Wait, sides are 2, 2√3, and 4. Wait, 2² + (2√3)² = 4 + 12 = 16 = 4². Yes! So triangle PQS is a right triangle with right angle at P. Wait, earlier I calculated angle at P as 90 degrees. Wait, but according to the Law of Cosines, angle at P was 90 degrees. Wait, but if sides are 2, 2√3, 4, then 2² + (2√3)² = 4 + 12 = 16 = 4², so yes, it's a right triangle with right angle at P. Therefore, angles are 30°, 60°, 90°, with the smallest angle being 30°, which is at Q.

Therefore, the answer is 30 degrees. Therefore, I think the answer is 30°, so boxed answer is \boxed{30}.

But let me check once more. If triangle PQS is right-angled at P, then angle at P is 90°, angle at Q is 30°, angle at S is 60°, so yes, the smallest angle is 30°. Therefore, the answer is 30 degrees.

**Final Answer**
\boxed{30}

To determine the **smallest angle** in triangle $ PQS $, we begin by analyzing the given configuration and the coordinates of the centers of the four circles:

- $ P = (1, \sqrt{3}) $
- $ Q = (-2, 0) $
- $ S = (2, 0) $

These coordinates are provided in the Asymptote code and are consistent with the diagram. We are told that all four circles have radius 1 and are tangent to one another and to the sides of triangle $ ABC $. While the exact configuration of triangle $ ABC $ is not necessary for solving the problem, the coordinates of $ P, Q, S $ allow us to compute the angles of triangle $ PQS $ directly.

---

### Step 1: Compute the Side Lengths of Triangle $ PQS $

Using the distance formula:

- $ PQ = \sqrt{(1 - (-2))^2 + (\sqrt{3} - 0)^2} = \sqrt{3^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3} $
- $ QS = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4 $
- $ PS = \sqrt{(2 - 1)^2 + (0 - \sqrt{3})^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 $

So the side lengths of triangle $ PQS $ are:
- $ PQ = 2\sqrt{3} $
- $ QS = 4 $
- $ PS = 2 $

---

### Step 2: Identify the Smallest Angle

In any triangle, the **smallest angle** is opposite the **shortest side**. Here, the shortest side is $ PS = 2 $, which is opposite angle $ Q $.

Thus, the **smallest angle** is $ \angle Q $.

---

### Step 3: Use the Law of Cosines to Find $ \angle Q $

We apply the Law of Cosines to triangle $ PQS $:

$$
\cos(\angle Q) = \frac{PQ^2 + QS^2 - PS^2}{2 \cdot PQ \cdot QS}
$$

Substitute the known values:

$$
\cos(\angle Q) = \frac{(2\sqrt{3})^2 + 4^2 - 2^2}{2 \cdot 2\sqrt{3} \cdot 4}
= \frac{12 + 16 - 4}{16\sqrt{3}}
= \frac{24}{16\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}
$$

$$
\angle Q = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = 30^\circ
$$

---

### Step 4: Confirm the Triangle Type

We can also verify that triangle $ PQS $ is a **right triangle** by checking the Pythagorean Theorem:

$$
PS^2 + PQ^2 = 2^2 + (2\sqrt{3})^2 = 4 + 12 = 16 = QS^2
$$

So, triangle $ PQS $ is a **right triangle** with a right angle at $ P $, and the angles are:

- $ \angle Q = 30^\circ $
- $ \angle S = 60^\circ $
- $ \angle P = 90^\circ $

---

### Final Answer

The **smallest angle** in triangle $ PQS $ is:

$$
\boxed{30}
$$

Solution:
Join $PQ$, $PR$, $PS$, $RQ$, and $RS$. Since the circles with center $Q$, $R$ and $S$ are all tangent to $BC$, then $QR$ and $RS$ are each parallel to $BC$ (as the centres $Q$, $R$ and $S$ are each 1 unit above $BC$). This tells us that $QS$ passes through $R$. When the centers of tangent circles are joined, the line segments formed pass through the associated point of tangency, and so have lengths equal to the sum of the radii of those circles. Therefore, $QR=RS=PR=PS=1+1=2$.

[asy]
size(200);
pair P, Q, R, S;
Q=(0,0);
R=(2,0);
S=(4,0);
P=(3,1.732);
label("Q", Q, SW);
label("R", R, dir(270));
label("S", S, SE);
label("P", P, N);
draw(circle(Q,1), dashed);
draw(circle(P,1), dashed);
draw(circle(R,1), dashed);
draw(circle(S,1), dashed);
draw(P--Q--S--P--R);
[/asy]

Since $PR=PS=RS$, we know $\triangle PRS$ is equilateral, so $\angle PSR=\angle PRS=60^\circ$.  Since $\angle PRS=60^\circ$ and $QRS$ is a straight line, we have $\angle QRP=180^\circ-60^\circ=120^\circ$. Since $QR=RP$, we know $\triangle QRP$ is isosceles, so $$\angle PQR = \frac{1}{2}(180^\circ-120^\circ)= 30^\circ.$$Since $\angle PQS=30^\circ$ and $\angle PSQ=60^\circ$, we have $\angle QPS = 180^\circ - 30^\circ - 60^\circ = 90^\circ$, so $\triangle PQS$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle. Thus, the answer is $\boxed{30^\circ}$.


Explanation:
Okay, so I need to figure out how many positive two-digit integers are factors of both 100 and 150. Hmm, let me start by recalling what it means for a number to be a factor of another number. A factor of a number is an integer that divides that number without leaving a remainder. So, if a number is a factor of both 100 and 150, it must divide both of them exactly.

First, maybe I should find the common factors of 100 and 150. I remember that the common factors of two numbers are the factors of their greatest common divisor (GCD). So, if I can find the GCD of 100 and 150, then all the factors of that GCD will be the common factors of 100 and 150. Then, from those factors, I need to pick the ones that are two-digit numbers.

Alright, let me calculate the GCD of 100 and 150. To do that, I can use the Euclidean algorithm. Let me recall how that works. The Euclidean algorithm involves dividing the larger number by the smaller one, then replacing the larger number with the smaller number and the smaller number with the remainder from the division, and repeating this process until the remainder is zero. The last non-zero remainder is the GCD.

So, starting with 150 and 100.

First step: divide 150 by 100. The quotient is 1, and the remainder is 50.

Now, replace the numbers with 100 and 50.

Next step: divide 100 by 50. The quotient is 2, and the remainder is 0.

Since the remainder is 0, the GCD is the last non-zero remainder, which is 50.

So, the GCD of 100 and 150 is 50. Therefore, all common factors of 100 and 150 are the factors of 50.

Now, I need to list all the factors of 50 and then identify which of those are two-digit numbers.

Let me list the factors of 50. To do that, I can start by finding pairs of numbers that multiply to 50.

Starting with 1: 1 × 50 = 50. So, 1 and 50 are factors.

Next, 2: 2 × 25 = 50. So, 2 and 25 are factors.

Then, 5: 5 × 10 = 50. So, 5 and 10 are factors.

Is there another? Let me check. The next number would be 5 again, but since 5 × 10 is already covered, and 25 is already there. So, the factors of 50 are 1, 2, 5, 10, 25, 50.

Wait, let me confirm that. Let me list them in order: 1, 2, 5, 10, 25, 50. Yes, that seems right.

Now, from these factors, I need to pick the two-digit numbers. Let me check each one:

1: single-digit, so no.

2: single-digit, no.

5: single-digit, no.

10: two-digit, yes.

25: two-digit, yes.

50: two-digit, yes.

So, the two-digit factors are 10, 25, and 50.

Wait, but hold on. Let me make sure that these numbers are indeed factors of both 100 and 150.

Let me check 10: 100 divided by 10 is 10, which is an integer. 150 divided by 10 is 15, which is also an integer. So, yes, 10 is a common factor.

25: 100 divided by 25 is 4, which is an integer. 150 divided by 25 is 6, which is also an integer. So, 25 is a common factor.

50: 100 divided by 50 is 2, which is an integer. 150 divided by 50 is 3, which is also an integer. So, 50 is a common factor.

Therefore, there are three two-digit numbers: 10, 25, and 50.

But wait, let me think again. Is there a possibility that I missed some factors? Let me check the factors of 50 again.

Factors of 50: 1, 2, 5, 10, 25, 50. That's all. Because 50 is 2 × 5². So, the exponents for prime factors are 2^1 and 5^2. Therefore, the number of factors is (1+1)(2+1) = 6, which matches the list above. So, there are six factors, and three of them are two-digit numbers.

But let me cross-verify this by another method. Maybe by listing all factors of 100 and 150 and then finding their intersection.

First, factors of 100. Let me list them.

Starting from 1: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Wait, let me check that.

1 × 100 = 100

2 × 50 = 100

4 × 25 = 100

5 × 20 = 100

10 × 10 = 100

So, factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, 100.

Now, factors of 150. Let me list them.

1 × 150 = 150

2 × 75 = 150

3 × 50 = 150

5 × 30 = 150

6 × 25 = 150

10 × 15 = 150

So, factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150.

Now, the common factors of 100 and 150 would be the numbers that are in both lists. Let me compare the two lists:

Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Factors of 150: 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150.

Common factors are numbers that appear in both. Let me check:

1 is common.

2 is common.

4 is not in the second list.

5 is common.

10 is common.

20 is not in the second list.

25 is common.

50 is common.

100 is not in the second list.

So, common factors are 1, 2, 5, 10, 25, 50. Which matches the factors of 50. So, same as before.

Therefore, the common factors are 1, 2, 5, 10, 25, 50.

So, the two-digit ones are 10, 25, 50. Therefore, there are three such numbers.

But let me check once more if there are any other two-digit numbers that I might have missed.

For example, is 15 a factor of both? Let me check. 15 divides 150, since 150 ÷ 15 = 10. But does 15 divide 100? 100 ÷ 15 is approximately 6.666..., which is not an integer. So, 15 is not a common factor.

Similarly, 20: 20 divides 100, but does it divide 150? 150 ÷ 20 = 7.5, which is not an integer. So, no.

What about 30? 30 divides 150, but 100 ÷ 30 is not an integer.

Similarly, 75 divides 150, but not 100.

So, no other two-digit numbers are common factors.

Therefore, the answer should be 3.

But let me think again. Wait, is 50 considered a two-digit number? Yes, because it's between 10 and 99 inclusive. So, 50 is two-digit.

So, 10, 25, 50. Three numbers.

Alternatively, maybe I can approach this problem by finding all two-digit numbers and checking which ones divide both 100 and 150. But that might take longer, but let me try for verification.

Two-digit numbers range from 10 to 99.

But instead of checking all of them, maybe I can find the common factors first.

But since we already did that, and cross-verified, maybe it's redundant.

Alternatively, think about the factors of 50, which are 1, 2, 5, 10, 25, 50. So, among these, the two-digit numbers are 10, 25, 50.

Therefore, the answer is 3.

But just to be thorough, let me check if there's a possibility that another number outside of the factors of 50 could divide both 100 and 150.

But since the GCD is 50, any common factor must divide 50. Therefore, there can't be any other common factors outside of the factors of 50.

Therefore, the answer is 3.

Wait, but let me check if 50 is considered a two-digit number. Yes, because it's 5 and 0, so two digits.

So, 10, 25, 50.

Therefore, the number of positive two-digit integers that are factors of both 100 and 150 is 3.

I think that's solid.

**Final Answer**
The number of positive two-digit integers that are factors of both 100 and 150 is \boxed{3}.

To determine how many **positive two-digit integers** are factors of **both 100 and 150**, we follow a structured approach based on the concept of **common factors** and the **greatest common divisor (GCD)**.

---

### Step 1: Understand the Problem

We are looking for **two-digit integers** that divide **both 100 and 150** without leaving a remainder. That is, we want the **common factors** of 100 and 150 that are **between 10 and 99 inclusive**.

---

### Step 2: Find the GCD of 100 and 150

The **common factors** of two numbers are exactly the **factors of their GCD**. So, we first compute the GCD of 100 and 150.

Using the **Euclidean algorithm**:

- $ 150 \div 100 = 1 $ remainder $ 50 $
- $ 100 \div 50 = 2 $ remainder $ 0 $

So, the GCD is **50**.

---

### Step 3: Find All Factors of the GCD (50)

Now, we find all the **factors of 50**:

$$
\text{Factors of } 50 = \{1, 2, 5, 10, 25, 50\}
$$

---

### Step 4: Identify the Two-Digit Factors

From the list of factors of 50, we select those that are **two-digit numbers**:

- 10
- 25
- 50

These are the only two-digit numbers that divide both 100 and 150.

---

### Step 5: Final Verification

Let’s verify that each of these numbers divides both 100 and 150:

- $ 100 \div 10 = 10 $, $ 150 \div 10 = 15 $ → ✅
- $ 100 \div 25 = 4 $, $ 150 \div 25 = 6 $ → ✅
- $ 100 \div 50 = 2 $, $ 150 \div 50 = 3 $ → ✅

All three numbers are valid.

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many positive two-digit integers are factors of both 100 and 150. Hmm, let me start by recalling what it means for a number to be a factor of another number. A factor of a number is an integer that divides that number without leaving a remainder. So, if a number is a factor of both 100 and 150, it must divide both of them exactly.

First, maybe I should find the common factors of 100 and 150. I remember that the common factors of two numbers are the factors of their greatest common divisor (GCD). So, if I can find the GCD of 100 and 150, then all the factors of that GCD will be the common factors of 100 and 150. Then, from those factors, I need to pick the ones that are two-digit numbers.

Alright, let me calculate the GCD of 100 and 150. To do that, I can use the Euclidean algorithm. Let me recall how that works. The Euclidean algorithm involves dividing the larger number by the smaller one, then replacing the larger number with the smaller number and the smaller number with the remainder from the division, and repeating this process until the remainder is zero. The last non-zero remainder is the GCD.

So, starting with 150 and 100.

First step: divide 150 by 100. The quotient is 1, and the remainder is 50.

Now, replace the numbers with 100 and 50.

Next step: divide 100 by 50. The quotient is 2, and the remainder is 0.

Since the remainder is 0, the GCD is the last non-zero remainder, which is 50.

So, the GCD of 100 and 150 is 50. Therefore, all common factors of 100 and 150 are the factors of 50.

Now, I need to list all the factors of 50 and then identify which of those are two-digit numbers.

Let me list the factors of 50. To do that, I can start by finding pairs of numbers that multiply to 50.

Starting with 1: 1 × 50 = 50. So, 1 and 50 are factors.

Next, 2: 2 × 25 = 50. So, 2 and 25 are factors.

Then, 5: 5 × 10 = 50. So, 5 and 10 are factors.

Is there another? Let me check. The next number would be 5 again, but since 5 × 10 is already covered, and 25 is already there. So, the factors of 50 are 1, 2, 5, 10, 25, 50.

Wait, let me confirm that. Let me list them in order: 1, 2, 5, 10, 25, 50. Yes, that seems right.

Now, from these factors, I need to pick the two-digit numbers. Let me check each one:

1: single-digit, so no.

2: single-digit, no.

5: single-digit, no.

10: two-digit, yes.

25: two-digit, yes.

50: two-digit, yes.

So, the two-digit factors are 10, 25, and 50.

Wait, but hold on. Let me make sure that these numbers are indeed factors of both 100 and 150.

Let me check 10: 100 divided by 10 is 10, which is an integer. 150 divided by 10 is 15, which is also an integer. So, yes, 10 is a common factor.

25: 100 divided by 25 is 4, which is an integer. 150 divided by 25 is 6, which is also an integer. So, 25 is a common factor.

50: 100 divided by 50 is 2, which is an integer. 150 divided by 50 is 3, which is also an integer. So, 50 is a common factor.

Therefore, there are three two-digit numbers: 10, 25, and 50.

But wait, let me think again. Is there a possibility that I missed some factors? Let me check the factors of 50 again.

Factors of 50: 1, 2, 5, 10, 25, 50. That's all. Because 50 is 2 × 5². So, the exponents for prime factors are 2^1 and 5^2. Therefore, the number of factors is (1+1)(2+1) = 6, which matches the list above. So, there are six factors, and three of them are two-digit numbers.

But let me cross-verify this by another method. Maybe by listing all factors of 100 and 150 and then finding their intersection.

First, factors of 100. Let me list them.

Starting from 1: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Wait, let me check that.

1 × 100 = 100

2 × 50 = 100

4 × 25 = 100

5 × 20 = 100

10 × 10 = 100

So, factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, 100.

Now, factors of 150. Let me list them.

1 × 150 = 150

2 × 75 = 150

3 × 50 = 150

5 × 30 = 150

6 × 25 = 150

10 × 15 = 150

So, factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150.

Now, the common factors of 100 and 150 would be the numbers that are in both lists. Let me compare the two lists:

Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Factors of 150: 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150.

Common factors are numbers that appear in both. Let me check:

1 is common.

2 is common.

4 is not in the second list.

5 is common.

10 is common.

20 is not in the second list.

25 is common.

50 is common.

100 is not in the second list.

So, common factors are 1, 2, 5, 10, 25, 50. Which matches the factors of 50. So, same as before.

Therefore, the common factors are 1, 2, 5, 10, 25, 50.

So, the two-digit ones are 10, 25, 50. Therefore, there are three such numbers.

But let me check once more if there are any other two-digit numbers that I might have missed.

For example, is 15 a factor of both? Let me check. 15 divides 150, since 150 ÷ 15 = 10. But does 15 divide 100? 100 ÷ 15 is approximately 6.666..., which is not an integer. So, 15 is not a common factor.

Similarly, 20: 20 divides 100, but does it divide 150? 150 ÷ 20 = 7.5, which is not an integer. So, no.

What about 30? 30 divides 150, but 100 ÷ 30 is not an integer.

Similarly, 75 divides 150, but not 100.

So, no other two-digit numbers are common factors.

Therefore, the answer should be 3.

But let me think again. Wait, is 50 considered a two-digit number? Yes, because it's between 10 and 99 inclusive. So, 50 is two-digit.

So, 10, 25, 50. Three numbers.

Alternatively, maybe I can approach this problem by finding all two-digit numbers and checking which ones divide both 100 and 150. But that might take longer, but let me try for verification.

Two-digit numbers range from 10 to 99.

But instead of checking all of them, maybe I can find the common factors first.

But since we already did that, and cross-verified, maybe it's redundant.

Alternatively, think about the factors of 50, which are 1, 2, 5, 10, 25, 50. So, among these, the two-digit numbers are 10, 25, 50.

Therefore, the answer is 3.

But just to be thorough, let me check if there's a possibility that another number outside of the factors of 50 could divide both 100 and 150.

But since the GCD is 50, any common factor must divide 50. Therefore, there can't be any other common factors outside of the factors of 50.

Therefore, the answer is 3.

Wait, but let me check if 50 is considered a two-digit number. Yes, because it's 5 and 0, so two digits.

So, 10, 25, 50.

Therefore, the number of positive two-digit integers that are factors of both 100 and 150 is 3.

I think that's solid.

**Final Answer**
The number of positive two-digit integers that are factors of both 100 and 150 is \boxed{3}.

To determine how many **positive two-digit integers** are factors of **both 100 and 150**, we follow a structured approach based on the concept of **common factors** and the **greatest common divisor (GCD)**.

---

### Step 1: Understand the Problem

We are looking for **two-digit integers** that divide **both 100 and 150** without leaving a remainder. That is, we want the **common factors** of 100 and 150 that are **between 10 and 99 inclusive**.

---

### Step 2: Find the GCD of 100 and 150

The **common factors** of two numbers are exactly the **factors of their GCD**. So, we first compute the GCD of 100 and 150.

Using the **Euclidean algorithm**:

- $ 150 \div 100 = 1 $ remainder $ 50 $
- $ 100 \div 50 = 2 $ remainder $ 0 $

So, the GCD is **50**.

---

### Step 3: Find All Factors of the GCD (50)

Now, we find all the **factors of 50**:

$$
\text{Factors of } 50 = \{1, 2, 5, 10, 25, 50\}
$$

---

### Step 4: Identify the Two-Digit Factors

From the list of factors of 50, we select those that are **two-digit numbers**:

- 10
- 25
- 50

These are the only two-digit numbers that divide both 100 and 150.

---

### Step 5: Final Verification

Let’s verify that each of these numbers divides both 100 and 150:

- $ 100 \div 10 = 10 $, $ 150 \div 10 = 15 $ → ✅
- $ 100 \div 25 = 4 $, $ 150 \div 25 = 6 $ → ✅
- $ 100 \div 50 = 2 $, $ 150 \div 50 = 3 $ → ✅

All three numbers are valid.

---

### Final Answer

$$
\boxed{3}
$$

Solution: The factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, 100.  Of these, the $\boxed{3}$ factors 10, 25 and 50 have two digits and are factors of 150 as well.