🧠:[<think>Okay, so I need to solve this geometry problem. Let me try to visualize it first. There's a right triangle ABC with a right angle at C. The side CA is 4 units long. Then there's a semicircular arc over AC, and P is the midpoint of that arc. Then we connect BP, and this line divides the figure APCB into two parts. We need to find the absolute difference in the areas of these two parts. Hmm, let me break this down step by step.First, let me sketch the triangle. Since it's a right triangle at C, I can imagine point C at the origin (0,0), point A at (0,4) because CA is 4 units. Then point B must be somewhere along the x-axis, right? But wait, in a right triangle at C, sides AC and BC are the legs. Wait, but in the problem, they mention CA=4, so maybe CA is one leg, and BC is the other. But we need to know the length of BC or AB? The problem doesn't specify. Hmm, maybe I need to figure that out. Wait, maybe the semicircle is on AC, so the diameter is AC? Let me check.The problem says "the semicircular arc over AC". So the semicircle has AC as its diameter. Then the semicircle is drawn outward from the triangle, I suppose, since the triangle is right-angled at C, so the semicircle would be on the opposite side of the triangle. So points A and C are endpoints, and the semicircle is drawn above the triangle? Wait, maybe. Let me confirm: if the triangle is right-angled at C, then sides AC and BC are perpendicular. If the semicircular arc is over AC, meaning AC is the diameter, then the semicircle would lie outside the triangle. So the arc is from A to C, passing through a point P which is the midpoint of this semicircle. So P is the point that is 90 degrees from the diameter AC on the semicircle. Wait, but in a semicircle, the midpoint of the arc would be the point where the angle is 90 degrees from the diameter. So if AC is the diameter, then point P is the midpoint of the semicircle, so it's the point that forms a right angle with AC. Wait, but AC is the diameter, so the semicircle is the set of all points where angle APC is a right angle? Wait, no. Any triangle inscribed in a semicircle with the diameter as its base is a right triangle. So if P is on the semicircle with diameter AC, then angle APC is 90 degrees. But P is the midpoint of the arc, so it should be the point where the semicircle is at the top relative to AC. Let me think. If AC is vertical (since in my coordinate system, A is at (0,4) and C is at (0,0)), then the semicircle with diameter AC would be a semicircle that extends to the left or right? Wait, AC is vertical from (0,0) to (0,4). So the semicircle with diameter AC would be either to the left or right of AC. But since the triangle is ABC with right angle at C, and if we have points A(0,4), C(0,0), then point B must be on the x-axis at (b,0) for some b. Then the semicircle on AC would either be on the left side (negative x-direction) or the right side (positive x-direction). But since the problem mentions connecting BP, where P is on the semicircle, which is likely not overlapping with the triangle. So probably, the semicircle is drawn on the "free" side, i.e., not overlapping with the triangle. Since triangle ABC is right-angled at C with CA=4 and CB unknown, the semicircle on AC would be on the side opposite to B. So if B is on the x-axis at (b,0), then the semicircle on AC would be on the left side (negative x-direction). So point P is the midpoint of that semicircle. Therefore, coordinates of P can be determined.Let me assign coordinates to the points to make this easier. Let me place point C at (0,0), point A at (0,4). Then the semicircle with diameter AC would have its center at the midpoint of AC, which is (0,2), and radius 2. Since AC is vertical, the semicircle would extend either to the left or right. Since the triangle ABC is right-angled at C, point B is somewhere on the x-axis, say at (b,0). So the triangle is in the first quadrant if b is positive. The semicircle on AC would then be on the left side (negative x-direction) because the right side is occupied by the triangle. Therefore, the semicircle is centered at (0,2) with radius 2, extending from A(0,4) to C(0,0) on the left side. The midpoint of the arc would be the point 90 degrees from AC on the semicircle. Since AC is vertical, moving 90 degrees along the semicircle from A would be to the left and down, but since it's a semicircle over AC, the midpoint should be at (-2,2). Wait, let me calculate that.The center of the semicircle is at (0,2). The semicircle spans from (0,4) to (0,0) on the left side. To find the midpoint of the arc, which is 180 degrees from A to C. The midpoint would be 90 degrees from both A and C. Since the semicircle is on the left side, the midpoint P would be located at (0 - 2, 2), which is (-2,2). Because starting from A(0,4), moving 180 degrees along the semicircle (which is a half-circle) to C(0,0). The midpoint is at 90 degrees from A, which would be 2 units to the left (since the radius is 2). So coordinates of P are (-2,2). Is that right?Wait, parametric equations for the semicircle: center at (0,2), radius 2, on the left side (negative x-direction). The parametric equations would be x = -2 cos θ + 0, y = 2 sin θ + 2, where θ goes from 0 to π. At θ=0, we get x=-2(1)+0=-2, y=2(0)+2=2. Wait, no: wait, if center is at (0,2), radius 2, then the semicircle on the left would be points (x,y) such that (x - 0)^2 + (y - 2)^2 = 4, with x ≤ 0. So parametrizing it, starting from point A(0,4): when θ=90 degrees (π/2 radians), we reach (0,4). Wait, maybe I need to adjust the parameter. Let me think.Alternatively, in polar coordinates, with the center at (0,2). If we consider the angle from the positive y-axis, then starting at A(0,4), moving 180 degrees around the semicircle to C(0,0). The midpoint of the arc would be at 90 degrees from the start. So starting at A(0,4), moving 90 degrees along the semicircle would be to point (-2,2). Similarly, moving another 90 degrees would reach C(0,0). Yes, that makes sense. So point P is at (-2,2).So coordinates:- A(0,4)- C(0,0)- B(b,0)- P(-2,2)Now, we need to connect BP. The line BP divides the figure APCB into two parts. The figure APCB is composed of the triangle ABC and the semicircle APC? Wait, the problem says "the figure APCB". Let's see: APCB is a quadrilateral? Or is it the combination of the triangle and the semicircle?Wait, APCB would be the polygon formed by points A, P, C, B, connected in that order? So A to P to C to B to A? But P is on the semicircle, so the figure APCB includes the area of the triangle ABC and the area of the semicircle? Or is it just the quadrilateral? Wait, the problem says "the figure APCB". Let me think.If we connect the points A, P, C, B in order, then APCB is a quadrilateral that includes the triangle ABC and the region from A to P to C. But since P is on the semicircle outside the triangle, the figure APCB would be a four-sided figure with vertices at A(0,4), P(-2,2), C(0,0), and B(b,0). So connecting these points in order: A to P is a line from (0,4) to (-2,2), then P to C is a line from (-2,2) to (0,0), then C to B is a line from (0,0) to (b,0), then B back to A? Wait, but that would make a quadrilateral, but the original triangle is ABC. Hmm, perhaps the figure APCB includes both the triangle ABC and the semicircle APC? Wait, the problem mentions "the semicircular arc AC", so maybe the figure APCB is the combination of the triangle ABC and the semicircle APC. Then connecting BP divides this combined figure into two parts. The areas of these two parts would then be calculated, and we need the absolute difference.Alternatively, maybe APCB is the polygon formed by the points A, P, C, B. Let me confirm. The problem states: "Connect BP, the line segment BP divides the figure APCB into two parts." So figure APCB is divided by BP into two parts, so figure APCB is a single figure. So APCB is a four-sided figure, with vertices at A, P, C, B.But since P is on the semicircular arc AC, which is outside the triangle, the figure APCB would consist of the triangle ABC and the region bounded by the semicircle from A to C through P. Wait, but how exactly? If we connect A to P, then P to C, then C to B, then B back to A, the figure would be a quadrilateral, but the side AP and PC are not part of the original triangle. So perhaps the figure APCB is the union of the triangle ABC and the semicircle APC, connected via point P. Then BP is a line cutting through this figure.But I need to clarify. Let me try to parse the problem again."In the right triangle triangle ABC, angle ACB=90 degrees, CA=4, P is the midpoint of the semicircular arc over arc AC, connect BP, the line segment BP divides the figure APCB into two parts. The absolute value of the difference in the areas of these two parts is blank."So figure APCB is the figure formed by points A, P, C, B connected in some way. Since P is the midpoint of the semicircular arc AC, which is a semicircle with diameter AC. So the semicircle is drawn on AC as diameter, outside the triangle. So the figure APCB is the region bounded by the triangle ABC and the semicircle APC. So the area of APCB would be the area of triangle ABC plus the area of the semicircle APC. But wait, when you connect BP, this line would divide the figure into two parts. But maybe the figure APCB is actually the union of the triangle and the semicircle, so the total area is area of triangle plus area of semicircle. Then BP divides this union into two regions, and we need the difference in their areas.Alternatively, figure APCB could be the quadrilateral formed by the points A, P, C, B, connected in that order. So APCB is a four-sided figure with sides AP, PC, CB, and BA. Wait, but BA is not mentioned. Wait, the problem says "connect BP", so maybe the figure is A-P-C-B-A, forming a quadrilateral. Then BP is a diagonal that divides it into two parts: triangle BPC and quadrilateral APB or something? Wait, but need to visualize.Alternatively, maybe APCB is a region that includes the triangle ABC and the semicircle APC. Then BP cuts through both the triangle and the semicircle, creating two regions whose areas we need to compare.Hmm, this is a bit confusing. Let me try to approach this step by step.First, I need to figure out the coordinates of all points.Let me fix the coordinate system:- Let’s place point C at (0,0).- Since CA=4 and angle ACB=90 degrees, point A is at (0,4).- Point B is somewhere on the x-axis; let's denote it as (b,0). We need to find the value of b.Wait, but the problem doesn't specify the length of BC or AB. So is the triangle not fixed? But the problem mentions a semicircular arc over AC, so maybe the position of B is determined by some other condition? Wait, maybe not. Maybe the answer is independent of the position of B? That seems unlikely. Wait, perhaps the semicircle is constructed on AC as diameter, and regardless of where B is, the difference in areas is the same? Hmm, not sure.Wait, but if the problem doesn't specify the length of BC or AB, then maybe there's a missing piece of information. Wait, let me check the original problem again.Original problem: "In the right triangle triangle ABC, angle ACB=90 degrees, CA=4, P is the midpoint of the semicircular arc over arc AC, connect BP, the line segment BP divides the figure APCB into two parts. The absolute value of the difference in the areas of these two parts is blank."Hmm, so only CA=4 is given. The other sides are not specified. Therefore, maybe the answer is the same regardless of the length of CB? That seems odd. Alternatively, maybe the semicircle is part of the figure, and the position of B is such that the difference in areas can be calculated without knowing CB.Alternatively, perhaps there's a figure provided (Figure 1) that the user is referring to, but since I don't have access to Figure 1, I have to infer the configuration.Wait, another thought: Since P is the midpoint of the semicircular arc AC, which is the diameter. The semicircle is drawn outside the triangle. Then the coordinates of P can be determined as (-2,2) as previously thought if the semicircle is on the left side. Then point B is on the x-axis at (b,0). So if we connect BP, which is a line from (b,0) to (-2,2). Then this line divides the figure APCB into two parts. The figure APCB would be the quadrilateral formed by points A(0,4), P(-2,2), C(0,0), and B(b,0). So connecting A-P-C-B-A. So this is a quadrilateral. Then BP is a diagonal that splits it into two triangles: triangle BPC and triangle BPA? Wait, but AP is another side. Wait, no. The figure APCB is a quadrilateral, and BP is a diagonal that divides it into two parts: triangle BPC and quadrilateral APBP? Wait, no, maybe triangle BPC and quadrilateral APB... Wait, no, when you draw BP in quadrilateral APCB, which is a four-sided figure, it should split into two polygons: one is triangle BPC, and the other is a three-sided figure APB? Wait, but APB is not a polygon unless connected. Wait, maybe it's triangle APB and quadrilateral APBC? Hmm, this is confusing.Alternatively, perhaps the figure APCB includes the semicircle. So APCB is the combination of the triangle ABC and the semicircle APC. Then BP is a line that cuts through both the triangle and the semicircle. The area difference would be between the part of the figure on one side of BP and the other side.But without seeing the figure, it's a bit challenging. Let me try to proceed with coordinates.Assuming:- C(0,0)- A(0,4)- P(-2,2)- B(b,0)We need to find the absolute difference in areas of the two parts divided by BP.First, let's figure out the area of the figure APCB. If it's the quadrilateral APCB, then its area can be calculated as the sum of the areas of triangle APC and triangle ABC. Wait, but APC is part of the semicircle. Wait, maybe not. Alternatively, the figure APCB is the quadrilateral, so we can compute its area using the shoelace formula.Coordinates of quadrilateral APCB:A(0,4), P(-2,2), C(0,0), B(b,0). Let's apply the shoelace formula.Shoelace formula:Area = 1/2 |sum_{i=1 to n} (x_i y_{i+1} - x_{i+1} y_i)|Order of points: A(0,4), P(-2,2), C(0,0), B(b,0), back to A(0,4).Compute the terms:First pair: (0,4) and (-2,2): 0*2 - (-2)*4 = 0 + 8 = 8Second pair: (-2,2) and (0,0): (-2)*0 - 0*2 = 0 - 0 = 0Third pair: (0,0) and (b,0): 0*0 - b*0 = 0 - 0 = 0Fourth pair: (b,0) and (0,4): b*4 - 0*0 = 4b - 0 = 4bSum of terms: 8 + 0 + 0 + 4b = 8 + 4bArea = 1/2 |8 + 4b| = 1/2 (8 + 4b) = 4 + 2b (since areas are positive)But the area of quadrilateral APCB is 4 + 2b. However, the triangle ABC has area 1/2 * base * height = 1/2 * b * 4 = 2b. So the area of quadrilateral APCB is 4 + 2b, which is the area of triangle ABC (2b) plus 4. Where does the 4 come from? Maybe the area of the semicircle? Wait, the semicircle with diameter AC has area 1/2 * π * r^2. Since AC is 4, the radius is 2, so area is 1/2 * π * 4 = 2π. But 2π is approximately 6.28, which is not 4. Hmm, so perhaps the figure APCB is not including the entire semicircle, but just the arc. Then the area would just be the quadrilateral. But then where does the 4 come from?Wait, in the shoelace formula, we calculated the area of quadrilateral APCB as 4 + 2b. But in reality, if APCB is just the quadrilateral, then that's its area. However, the problem mentions "the figure APCB", which might include the area under the semicircle. But in our calculation, the shoelace formula only accounts for the quadrilateral. So perhaps the figure is intended to be the combination of the triangle ABC and the semicircle APC. In that case, the total area would be area of triangle ABC (2b) plus area of semicircle APC (2π). Then BP divides this combined figure into two parts. However, without knowing where exactly BP cuts through the semicircle, it's hard to compute the areas. Alternatively, if APCB is the quadrilateral, then BP divides it into two parts, and the difference in their areas is |(4 + 2b)/2 - (4 + 2b)/2| = 0, which can't be. So that's not possible. Therefore, my initial assumption about the figure APCB must be wrong.Wait, another approach: Maybe the figure APCB is the region bounded by the arc APC and the line segments PC, CB, and BA. So it's a kind of a crescent shape plus the triangle. Wait, but BA is part of the triangle. Alternatively, APCB is the region bounded by the semicircle APC and the line segments PC, CB, and BP. No, BP is the line that divides it.Alternatively, since the problem states "the line segment BP divides the figure APCB into two parts", the figure APCB must be a single connected figure. If P is on the semicircle outside the triangle, then connecting BP would divide APCB into two regions: one part is the triangle BPC and the other part is the region bounded by BA, AP, and BP. But without knowing exactly, it's tricky.Wait, maybe it's better to calculate coordinates of intersection points and compute areas using integrals or something. But since this is a geometry problem, there must be a more straightforward way.Let me assume that B is at (4,0). Wait, but why? Because CA=4, maybe it's a square? But angle at C is 90 degrees. If CA=4 and CB=4, then it's an isosceles right triangle. But the problem doesn't specify that. Hmm, but the problem doesn't give BC's length. So maybe the answer is independent of BC? That seems strange.Wait, let me think again. If P is the midpoint of the semicircular arc AC, then AP = PC in terms of arc length, but in coordinates, P is at (-2,2). Then BP connects B(b,0) to P(-2,2). The line BP divides the figure APCB into two parts. The figure APCB is likely the quadrilateral APCB as calculated before with area 4 + 2b, but divided by BP into two regions. The area difference would be |Area1 - Area2|. Since the total area is 4 + 2b, if BP divides it into two regions, the difference would be |Area1 - (4 + 2b - Area1)| = |2Area1 - (4 + 2b)|. So unless Area1 is exactly half of the total area, the difference would be non-zero. However, the problem states that the absolute value of the difference is a fixed number, so maybe regardless of b, the difference is the same. But how?Alternatively, maybe the figure APCB is not the quadrilateral but includes the semicircle. Then the total area would be area of triangle ABC (2b) plus area of semicircle (2π). Then BP divides this into two parts. The difference in areas would be |(2b + 2π)/2 - (2b + 2π)/2| = 0, which is not possible. So perhaps not.Wait, maybe the semicircle is part of the figure, and BP intersects the semicircle at some point, creating a segment whose area can be calculated. But since P is already on the semicircle, BP connects B to P, so it's a chord of the semicircle? Wait, BP connects B to P, where P is on the semicircle. So BP starts at B, goes to P, which is on the semicircle. Then the line BP would intersect the semicircle only at P, and the triangle ABC at point B and somewhere else?Wait, maybe I need to parametrize the line BP and see where it intersects other parts of the figure.Coordinates:- B(b,0)- P(-2,2)The equation of line BP: Let's find the slope first.Slope m = (2 - 0)/(-2 - b) = 2 / (-2 - b) = -2/(b + 2)Equation: y - 0 = m(x - b)So y = (-2)/(b + 2) (x - b)This line will go from B(b,0) to P(-2,2). We need to see how this line divides the figure APCB.If the figure APCB is the quadrilateral A-P-C-B-A, then BP divides it into two parts: triangle B-P-C and quadrilateral A-P-B-A? Wait, not sure.Wait, the quadrilateral APCB can be divided by BP into two parts: triangle BPC and the quadrilateral APB. Wait, but APB is not a quadrilateral. If you connect A-P-B-A, that's a triangle. Wait, maybe not.Wait, the figure APCB is the polygon A-P-C-B-A. If we draw BP, which connects B to P, then BP divides the quadrilateral into two parts: triangle B-P-C and pentagon A-P-B-C-A? Wait, no. Wait, in the quadrilateral APCB, drawing BP would create two polygons: triangle BPC and quadrilateral APB. Wait, APB would be A-P-B-A, which is a triangle. Wait, maybe.Wait, original quadrilateral is A-P-C-B-A.When we connect BP, we split it into two parts:1. Triangle B-P-C2. The remaining area, which is the quadrilateral A-P-B-A. Wait, but A-P-B-A is a triangle (A-P-B).Wait, no. The original quadrilateral has vertices A, P, C, B. If you connect B to P, then you have two polygons: triangle B-P-C and the quadrilateral A-P-C-B-P? No, that's overlapping.Alternatively, splitting the quadrilateral APCB along BP gives two regions: one is triangle BPC, and the other is the region consisting of triangle APB and the quadrilateral APBC? Hmm, this is confusing.Alternatively, perhaps it's better to use the shoelace formula to compute the areas of the two parts created by BP.The two parts would be:1. Polygon B-P-C: a triangle with points B(b,0), P(-2,2), C(0,0)2. Polygon A-P-B: a triangle with points A(0,4), P(-2,2), B(b,0)Wait, if we split the quadrilateral APCB along BP, then the two resulting figures are triangle BPC and the quadrilateral APB... Wait, no. The original quadrilateral is A-P-C-B-A. If we connect B-P, then the two resulting figures are:1. Triangle B-P-C2. Quadrilateral A-P-C-B-P? No, that doesn't make sense. Maybe it's A-P-B and A-P-C-B? Not sure.Alternatively, maybe the line BP divides the figure into a triangle and a quadrilateral. Let's check.Original quadrilateral APCB:Vertices: A(0,4), P(-2,2), C(0,0), B(b,0)After connecting BP, the two regions are:1. Triangle B-P-C (points B, P, C)2. Quadrilateral A-P-B-C (points A, P, B, C). Wait, but BC is already part of the original figure.Wait, no. When you split a quadrilateral with a diagonal, you get two triangles. But APCB is a quadrilateral, but it's not a convex quadrilateral necessarily. Wait, is APCB convex?Points:A(0,4), P(-2,2), C(0,0), B(b,0). If b is positive, then B is on the right side. P is on the left. So connecting these points, the quadrilateral would be a non-convex quadrilateral if b is positive. Therefore, drawing BP might split it into different parts.Wait, let's take specific coordinates. Let's assume that b is positive, say b=2 for concreteness. Then point B is at (2,0). Then APCB is the quadrilateral with points (0,4), (-2,2), (0,0), (2,0). Connecting BP is connecting (2,0) to (-2,2). Let's see if this line divides the quadrilateral into two parts.The quadrilateral APCB would have vertices:A(0,4) connected to P(-2,2), connected to C(0,0), connected to B(2,0), connected back to A(0,4). So the shape is like a kite but with one side dented. The line BP connects B(2,0) to P(-2,2). This line will cross through the interior of the quadrilateral. To compute the areas of the two resulting parts, we can use the shoelace formula for each part.First, let's compute the area of triangle BPC:Points B(2,0), P(-2,2), C(0,0)Area = 1/2 | (2*(2 - 0) + (-2)*(0 - 0) + 0*(0 - 2)) | = 1/2 | (4 + 0 + 0)| = 2.Then the area of quadrilateral APCB is 4 + 2b. If b=2, then the area is 4 + 4 = 8. So the area of triangle BPC is 2, and the remaining area is 8 - 2 = 6. The absolute difference is |2 - 6| = 4. But if I take b=2, the difference is 4. If I take another b, say b=1, let's see.If b=1, area of quadrilateral APCB is 4 + 2*1 = 6. Area of triangle BPC:Points B(1,0), P(-2,2), C(0,0)Area = 1/2 |1*(2 - 0) + (-2)*(0 - 0) + 0*(0 - 2)| = 1/2 |2 + 0 + 0| = 1.Then remaining area is 6 - 1 = 5. Difference |1 - 5| = 4. Hmm, same difference. Wait, interesting. Let's check with b=3.Area of quadrilateral APCB is 4 + 2*3 = 10.Area of triangle BPC:Points B(3,0), P(-2,2), C(0,0)Area = 1/2 |3*(2 - 0) + (-2)*(0 - 0) + 0*(0 - 2)| = 1/2 |6 + 0 + 0| = 3.Remaining area: 10 - 3 = 7. Difference |3 - 7| = 4. Again 4. Wait, so regardless of the value of b, the absolute difference is 4? So maybe the answer is 4? But why?Wait, that's interesting. When I calculated for specific values of b (2,1,3), the difference was always 4. So maybe the answer is 4 regardless of the position of B. But why is that? Is there a mathematical reason?Let me see. The area of the quadrilateral APCB is 4 + 2b. The area of triangle BPC is calculated as follows:Coordinates of B(b,0), P(-2,2), C(0,0). Using the shoelace formula:Area = 1/2 |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|Plugging in:1/2 |b*(2 - 0) + (-2)*(0 - 0) + 0*(0 - 2)|= 1/2 |2b + 0 + 0| = (1/2)(2b) = b.Wait, wait, that contradicts my previous calculations. Wait, when I took b=2, I calculated the area as 2, but according to this formula, it should be b=2. Wait, what's the issue here.Wait, maybe I made a mistake earlier. Let's recalculate the area of triangle BPC when b=2:Points B(2,0), P(-2,2), C(0,0)Using shoelace formula:Area = 1/2 |(2*(2 - 0) + (-2)*(0 - 0) + 0*(0 - 2))|= 1/2 |(4 + 0 + 0)| = 1/2 * 4 = 2. Correct.But according to the general formula I just derived, it's b. When b=2, area is 2, which matches. When b=1, area is 1, which matches. When b=3, area is 3, which matches. So in general, the area of triangle BPC is b.Then the remaining area is (4 + 2b) - b = 4 + b.Therefore, the absolute difference is |b - (4 + b)| = | -4 | = 4.Wait, that's brilliant! So regardless of the value of b, the absolute difference between the two areas is always 4. Therefore, the answer is 4. So boxed{4} is the answer.But let me verify once again.The area of quadrilateral APCB is 4 + 2b (from shoelace formula). The area of triangle BPC is b. Therefore, the area of the other part is (4 + 2b) - b = 4 + b. The absolute difference is |b - (4 + b)| = | -4 | = 4. So yes, the difference is always 4, regardless of b. Therefore, the answer is 4.Wow, that's elegant. Even though the position of B affects the individual areas, the difference remains constant. That's because the area of triangle BPC is b, and the remaining area is 4 + b, so their difference is exactly 4. Therefore, the absolute value of the difference is 4. So the answer is 4.Final Answerboxed{4}</think>To solve the problem, we start by considering the right triangle (triangle ABC) with (angle ACB = 90^circ) and (CA = 4). The semicircular arc (overparen{AC}) has (AC) as its diameter, and (P) is the midpoint of this arc. We need to find the absolute value of the difference in the areas of the two parts formed by the line segment (BP) dividing the figure (APCB).1. Coordinates Assignment: - Place (C) at ((0, 0)), (A) at ((0, 4)), and (B) at ((b, 0)) on the x-axis. - The semicircle with diameter (AC) has its center at ((0, 2)) and radius 2. The midpoint (P) of the semicircular arc is at ((-2, 2)).2. Area Calculation of Quadrilateral (APCB): - Using the shoelace formula for the quadrilateral (APCB) with vertices (A(0, 4)), (P(-2, 2)), (C(0, 0)), and (B(b, 0)): [ text{Area} = frac{1}{2} left| 0 cdot 2 + (-2) cdot 0 + 0 cdot 0 + b cdot 4 - (4 cdot (-2) + 2 cdot 0 + 0 cdot b + 0 cdot 0) right| ] Simplifying, the area is: [ frac{1}{2} left| 8 + 4b right| = 4 + 2b ]3. Area Calculation of Triangle (BPC): - Using the coordinates (B(b, 0)), (P(-2, 2)), and (C(0, 0)): [ text{Area} = frac{1}{2} left| b(2 - 0) + (-2)(0 - 0) + 0(0 - 2) right| = frac{1}{2} left| 2b right| = b ]4. Absolute Difference in Areas: - The remaining area after subtracting the area of triangle (BPC) from the area of quadrilateral (APCB) is: [ (4 + 2b) - b = 4 + b ] - The absolute difference between the areas of the two parts is: [ left| b - (4 + b) right| = left| -4 right| = 4 ]Thus, the absolute value of the difference in the areas of the two parts is (boxed{4}).,<think>Okay, so I need to solve this geometry problem. Let me try to visualize it first. There's a right triangle ABC with angle ACB being 90 degrees, and CA is 4 units long. Point P is the midpoint of the semicircular arc AC. Then, connecting BP divides the figure APCB into two parts. I need to find the absolute difference in the areas of these two parts.First, let me sketch this in my mind. Triangle ABC is right-angled at C. So, points A, B, and C form a right angle at C. CA is 4, which is one of the legs. Since it's a right triangle, the sides would be CA=4 (leg), CB (another leg), and AB (hypotenuse). But I don't know CB or AB yet. Wait, maybe I need more information. The problem mentions a semicircular arc over AC. So, maybe the semicircle is drawn on AC as the diameter? That makes sense. So, the semicircle would have AC as its diameter, and since P is the midpoint of the arc AC, that would be the point where the semicircle's arc is divided into two equal parts by P.So, the semicircle is constructed on AC as the diameter, with the arc facing outward from the triangle. Then point P is the midpoint of that arc. Connecting BP, which is a line from point B to point P, splits the figure APCB into two parts, and I need to find the difference in their areas.Hmm. Let me break this down step by step.First, maybe I need coordinates. Let me assign coordinates to the points to make calculations easier. Let's place point C at the origin (0,0). Then, since CA is 4 units, and angle C is 90 degrees, let's assume CA is along the x-axis. So point A is at (4, 0). Then, point B would be somewhere along the y-axis, since angle C is the right angle. Let me denote coordinates: C(0,0), A(4,0), and B(0,b), where b is the length of CB, which we don't know yet.Wait, but maybe we can figure out CB? Wait, the problem doesn't specify the length of CB or AB. Hmm, that's confusing. Is there any other information given? The problem says "in the right triangle triangle ABC", angle ACB is 90 degrees, CA=4. So, maybe the triangle isn't fixed? But then the semicircle is on AC, so the semicircle is fixed with diameter AC=4. Then point P is the midpoint of that semicircle's arc. Then BP is connecting B to P. The figure APCB is the quadrilateral formed by points A, P, C, B. So when BP is drawn, it divides APCB into two parts: one part is triangle BPC and the other is quadrilateral APBP? Wait, no. Wait, APCB is a four-sided figure. When you connect BP, it should split it into two parts: one part is triangle BPC and the other part is the quadrilateral APB? Hmm, maybe not. Let me think again.APCB is the figure formed by the points A, P, C, B. So connecting those in order: A to P, P to C, C to B, B to A? Wait, but the problem mentions the semicircular arc AC. So maybe APCB is the combination of the triangle ABC and the semicircle on AC? Wait, the wording is a bit unclear. The problem says "the figure APCB". So APCB is a figure formed by connecting points A, P, C, B. But P is a point on the semicircular arc AC. So perhaps the figure APCB includes the triangle ABC and the semicircle? Or is it just the quadrilateral?Wait, in the original problem statement: "in the right triangle ABC... P is the midpoint of the semicircular arc over AC, connect BP, the line segment BP divides the figure APCB into two parts." So figure APCB is the union of the triangle ABC and the semicircle AC with midpoint P. Then connecting BP divides this combined figure into two parts, and we need the absolute difference of their areas.Alternatively, maybe APCB is a quadrilateral formed by the points A, P, C, B. But since P is on the semicircle, which is outside the triangle, then APCB would be a four-sided figure with sides AP, PC, CB, and BA? Hmm, not sure.Wait, maybe I need to clarify this. Let me re-read the problem:"As shown in Figure 1, in the right triangle triangle ABC, angle ACB=90°, CA=4, P is the midpoint of the semicircular arc over AC, connect BP, the line segment BP divides the figure APCB into two parts. The absolute value of the difference in the areas of these two parts is ______."Since the original figure is the right triangle, and there's a semicircular arc over AC. So over AC, meaning AC is the diameter. Then P is the midpoint of that semicircle's arc. Then the figure APCB probably consists of the triangle ABC and the semicircle AC. So when you connect BP, the line BP would split this combined figure into two parts. So the two parts would be: one part is the area bounded by BP, BC, and the semicircle, and the other part is the area bounded by BP, BA, AP, and the semicircle. Hmm, maybe.Alternatively, since APCB is mentioned, maybe the figure is the quadrilateral formed by points A, P, C, B. But since P is on the semicircle, which is outside the triangle, AP would be an arc from A to P, and PC would be an arc from P to C? No, probably not. Probably the figure APCB is the combination of the triangle ABC and the semicircle, with AP and PC being straight lines? But that seems odd. Wait, no. If you have a semicircle on AC, then AP is the arc from A to P, and PC is the arc from P to C? But in that case, APCB would be a path going from A along the semicircle to P, then to C, then to B, then back to A? But that seems like a closed figure with a curved side.Alternatively, maybe APCB is the polygon A-P-C-B-A, where P is a point on the semicircle. So AP is a straight line from A to P, PC is a straight line from P to C, CB is from C to B, and BA is from B to A. But then connecting BP would split this quadrilateral into two parts: triangle BPC and quadrilateral APBP? Wait, that doesn't make sense.I think the key here is that the figure APCB is the union of the triangle ABC and the semicircle with diameter AC. Then, when BP is drawn, it divides this combined figure into two regions. The problem states that the line segment BP divides the figure into two parts, and we need the absolute difference between their areas.So, perhaps one region is the part of the triangle ABC that's on one side of BP, plus the part of the semicircle on that same side, and the other region is the remaining parts. But this is a bit vague. Alternatively, maybe the figure APCB is the area bounded by the triangle ABC and the semicircle AC. So APCB would be the area inside the triangle and inside the semicircle. But since the semicircle is drawn on AC, which is a side of the triangle, the semicircle would be outside the triangle. So the figure APCB would be the triangle ABC plus the semicircle on AC. Then BP connects point B to the midpoint of the semicircle's arc, and divides the figure into two parts. The areas of these two parts would be the area on one side of BP (including part of the triangle and part of the semicircle) and the area on the other side. Then, the absolute difference between these two areas is required.Alternatively, maybe the figure APCB is just the quadrilateral formed by the points A, P, C, and B. Since P is on the semicircle, which is outside the triangle, AP and PC are curves. But usually, figures are considered with straight lines unless specified. Maybe the problem is in Chinese, and the translation might have some nuances. Since the original problem mentions a semicircular arc, and P is the midpoint of that arc. So the figure APCB is probably the combination of the triangle ABC and the semicircle AC, with P being a point on the semicircle. Then connecting BP divides this combined figure into two parts: one part is the triangle BPC plus the region of the semicircle between P and C, and the other part is the triangle BPA plus the region of the semicircle between A and P. Wait, but the semicircle is only on AC, so the semicircle's area is fixed. But when you connect BP, how does it divide the figure?Alternatively, maybe the figure APCB is the area bounded by the semicircle from A to C through P and the triangle ABC. So the entire figure would be the union of the semicircle and the triangle. Then BP is a line cutting through this figure. The area on one side of BP would include parts of the triangle and parts of the semicircle, and the same for the other side. The difference between these two areas is what's needed.This is a bit confusing without the figure. Let me try to reconstruct it.1. Right triangle ABC with right angle at C, CA = 4. So legs are CA = 4 (along x-axis from C(0,0) to A(4,0)), CB unknown length (along y-axis from C(0,0) to B(0,b)), hypotenuse AB.2. Semicircular arc over AC: semicircle with diameter AC, located outside the triangle ABC. The arc from A to C passing through the top of the semicircle. The midpoint of this arc is point P. Since it's a semicircle, the midpoint of the arc AC would be the point diametrically opposite to the midpoint of AC. Wait, the midpoint of the arc on a semicircle with diameter AC would be the point that's 90 degrees from AC. Wait, in a semicircle, the midpoint of the arc AC is the point where the angle is 90 degrees to the diameter AC. Wait, actually, in a semicircle, any point on the arc forms a right angle with the diameter. But the midpoint of the arc would be the point that's equidistant from A and C along the arc. Since the semicircle is 180 degrees, the midpoint would be the point at 90 degrees from AC. So the coordinates of P can be found.Given that AC is the diameter from A(4,0) to C(0,0), the center of the semicircle is the midpoint of AC, which is (2,0), and the radius is 2. The semicircle is above AC (since it's a right triangle, and the semicircle is likely drawn outside). Therefore, the midpoint of the arc AC would be the point at (2,2), because starting from A(4,0), moving 180 degrees around the circle, the midpoint is at 90 degrees, which would be 2 units up from the center (2,0) along the y-axis. Wait, yes. The semicircle with center (2,0) and radius 2, the midpoint of the arc from A to C is at (2,2). So point P is (2,2).So coordinates:- C: (0,0)- A: (4,0)- B: (0,b) (unknown b)- P: (2,2)Then, connecting BP, which is the line from B(0,b) to P(2,2). The figure APCB is the combination of triangle ABC and the semicircle AC with midpoint P. Then BP divides this figure into two parts. The areas of these two parts will be the area on one side of BP and the other side. The problem asks for the absolute difference between these two areas.But since the figure includes both the triangle and the semicircle, we need to compute the areas divided by BP, considering both the triangle and the semicircle.But first, we need to find the value of b, the length of CB. Wait, the problem doesn't specify CB or AB. Hmm. Wait, the problem states it's a right triangle with CA=4, angle ACB=90°, so legs CA and CB, hypotenuse AB. But the length of CB isn't given. Is it possible that the answer is independent of CB? That seems unlikely. Maybe there's a missing piece of information here, or perhaps I need to look back at the problem.Wait, no, the problem is in Chinese, perhaps some details were lost in translation. Wait, original problem says: "in the right triangle ABC, angle ACB=90°, CA=4, P is the midpoint of the semicircular arc over AC, connect BP, the line segment BP divides the figure APCB into two parts. The absolute value of the difference in the areas of these two parts is ______."So maybe the semicircle is constructed on AC as diameter, outside the triangle, and the figure APCB is the union of triangle ABC and the semicircle. Then BP is drawn from B to P, and this line divides the figure into two regions. The problem is to find the absolute difference in their areas.But since the triangle's other leg CB isn't given, perhaps the difference is independent of CB? That seems strange. Alternatively, maybe the triangle is isoceles? But with CA=4 and angle at C, unless CB is also 4, which isn't stated. Wait, maybe the semicircle is drawn on AC as diameter, and point P is the midpoint, so coordinates (2,2). Then, regardless of where B is, the line BP will divide the figure. But without knowing where B is, how can we compute the areas? That must mean that perhaps the difference is the same regardless of the position of B? That seems counterintuitive.Wait, but the problem is from a competition, likely the answer is a fixed number, so maybe there's a way to show that the difference is equal to twice the area of some fixed component, regardless of the triangle's other leg. Alternatively, perhaps the problem implies that the triangle is a specific one, such as an isoceles right triangle? But CA=4, angle at C is 90°, so if it's isoceles, then CB=CA=4. But the problem doesn't specify that.Wait, maybe the semicircle is drawn on AC as diameter, and P is the midpoint of the arc AC. Then, regardless of the triangle's other leg, the difference in areas is always the same. But how?Alternatively, perhaps the problem is in 3D space, but no, it's a triangle, so 2D.Wait, perhaps the key is that P is the midpoint of the semicircle, so the line BP will pass through certain key points, making the areas subtract in a particular way.Alternatively, since the semicircle has a fixed area, and the triangle has an area dependent on CB. But the difference between the areas might involve cancellation.Wait, let's try to assign coordinates and see.Let me denote:- C(0,0)- A(4,0)- P(2,2)- B(0,b)We need to compute the areas divided by BP. The figure APCB is the union of triangle ABC and the semicircle AC. The total area is the area of ABC plus the area of the semicircle. The area of ABC is (CA * CB)/2 = (4 * b)/2 = 2b. The area of the semicircle is (π * r²)/2 = (π * 2²)/2 = 2π. So the total area of figure APCB is 2b + 2π.Now, BP divides this into two parts. Let's denote the two areas as S1 and S2. Then |S1 - S2| is required.To compute S1 and S2, we need to find the areas on either side of BP within the figure APCB. This involves computing the area of the part of the triangle ABC on one side of BP plus the part of the semicircle on that side, and similarly for the other side.Alternatively, since BP is a line from B(0,b) to P(2,2), we can find the equation of BP and then compute the areas accordingly.First, let's find the equation of line BP. The coordinates of B are (0, b) and P are (2, 2). The slope of BP is (2 - b)/(2 - 0) = (2 - b)/2. So the equation is:y - b = [(2 - b)/2](x - 0)=> y = [(2 - b)/2]x + bNow, we need to find where this line intersects the semicircle and the triangle ABC.But the semicircle is defined by the equation (x - 2)^2 + y^2 = 2^2, since it's centered at (2,0) with radius 2. Wait, no. Wait, the semicircle is on AC as diameter from A(4,0) to C(0,0), so the center is at (2,0), radius 2, and since it's a semicircle above AC (assuming it's the upper half), the equation is (x - 2)^2 + y^2 = 4, with y ≥ 0.But point P is (2,2), which is on this semicircle. The line BP goes from (0,b) to (2,2). So we need to check if this line intersects the semicircle again, besides at point P. Let's solve for intersections between BP and the semicircle.The parametric equations for BP can be parameterized as x = 2t, y = b + (2 - b)t, where t ranges from 0 to 1. Plugging into the semicircle equation:(x - 2)^2 + y^2 = 4(2t - 2)^2 + [b + (2 - b)t]^2 = 4Expand:(4t² - 8t + 4) + [b² + 2b(2 - b)t + (2 - b)² t²] = 4Combine terms:4t² -8t +4 + b² + 2b(2 - b)t + (2 - b)² t² -4 =0Simplify:(4 + (2 - b)^2) t² + (-8 + 2b(2 - b)) t + (4 + b² -4) =0Simplify each part:Coefficient of t²: 4 + (4 -4b + b²) = 8 -4b + b²Coefficient of t: -8 + 4b - 2b²Constant term: b²Therefore, the quadratic equation is:(8 -4b + b²) t² + (-8 +4b -2b²) t + b² =0This equation should have t=1 as a root since point P(2,2) is on both BP and the semicircle. Let's verify:Plug t=1:(8 -4b + b²)(1) + (-8 +4b -2b²)(1) + b² = (8 -4b + b² -8 +4b -2b² + b²) = 0. Yes, it works.Therefore, we can factor out (t -1). Let's perform polynomial division or factorization.Let’s write the quadratic as:At² + Bt + C =0, whereA = 8 -4b + b²,B = -8 +4b -2b²,C = b².Since t=1 is a root, factor as (t -1)(Dt + E) =0.Multiplying out:Dt² + (E - D)t - E =0Compare coefficients:Dt² = At² ⇒ D = A =8 -4b + b²(E - D)t = Bt ⇒ E - D = B ⇒ E = B + D = (-8 +4b -2b²) + (8 -4b + b²) = (-8 +8) + (4b -4b) + (-2b² +b²) = -b²- E = C ⇒ E = -C = -b², which matches the previous result.Thus, the quadratic factors as (t -1)( (8 -4b + b²) t -b² )=0.Therefore, the other root is given by (8 -4b + b²) t -b² =0 ⇒ t = b² / (8 -4b + b²)So the line BP intersects the semicircle at t=1 (point P) and t = b² / (8 -4b + b²). Let's denote this other intersection point as Q.But wait, BP is connecting B(0,b) to P(2,2). Since P is already on the semicircle, unless BP is tangent to the semicircle at P, which would mean only one intersection point. But we found another intersection point Q unless the quadratic has a double root at t=1. Let's check when the discriminant is zero.But since we already factored out (t -1), the other root is t = b² / (8 -4b + b²). Therefore, unless this t is equal to 1, there's another intersection point. Let's see when t=1:b² / (8 -4b + b²) =1 ⇒ b² =8 -4b +b² ⇒ 8 -4b=0 ⇒ b=2.So when b=2, the line BP is tangent to the semicircle at P. For other values of b, there's another intersection point Q.But in our problem, since P is the midpoint of the semicircular arc AC, and the line BP is drawn, so unless B is at a specific location, but the problem states it's a general right triangle. So unless the problem specifies that BP is tangent, which it doesn't, we have to consider the general case.Wait, but if b is variable, then the difference in areas would depend on b, but the problem doesn't specify b. Therefore, perhaps there is a specific value of b that makes the difference constant, or maybe the difference is always the same regardless of b.But since the answer is a fixed number, maybe the difference is equal to twice the area of the semicircle or something. Alternatively, maybe through some geometric properties, the difference is equal to the area of the semicircle minus the area of the triangle, but that seems arbitrary.Alternatively, perhaps there's a clever way to see that the difference in areas is equal to twice the area of triangle BPC or something.Alternatively, maybe coordinate geometry can solve this. Let's proceed with coordinates.First, let's note that the total area of the figure APCB is 2b + 2π.The line BP divides this into two regions. Let's compute the area of one region, say S1, which is the area below BP (assuming BP is going from B down to P), and S2 is the area above BP. Then |S1 - S2| is required.But actually, since the figure includes the semicircle and the triangle, the division by BP would split both the triangle and the semicircle into parts.To compute S1 and S2, we need to compute the area of the figure APCB on each side of BP.This seems complex, but let's try.First, the area of the triangle ABC is 2b, as before. The area of the semicircle is 2π.Now, the line BP will pass through the semicircle and the triangle. The intersection points are P and another point Q on BP inside the semicircle? Wait, but BP connects B(0,b) to P(2,2). If b > 2, then the line BP might pass through the semicircle only at P. If b < 2, then it might intersect the semicircle again somewhere else. Wait, but earlier when we solved for intersections, we found that besides P (t=1), there is another intersection at t = b² / (8 -4b + b²). Let's compute this value.Let’s denote t = b² / (8 -4b + b²). Let's see for different values of b:If b=4, then t = 16 / (8 -16 +16) = 16/8=2. But t=2 would be outside the range [0,1], since BP is from t=0 to t=1. Therefore, if b=4, the other intersection is at t=2, which is outside the segment BP, so only intersection at P.If b=0, then t=0 / (8 -0 +0)=0, which is point B itself. But B is at (0,0), which is point C? Wait, no. If b=0, then point B would coincide with C, but in the problem, B is a distinct point forming the triangle. So b must be positive.Wait, perhaps for b > 2, the other intersection is at t >1, so outside BP, so BP only intersects the semicircle at P. For b <2, t would be within [0,1], so BP intersects the semicircle at P and another point Q.But in our case, since P is the midpoint of the semicircle, and the triangle is right-angled at C, maybe there's a relation between b and the rest. Wait, the problem didn't specify the length of CB, so maybe the result is the same regardless of b. If that's the case, we can choose a specific value of b to simplify calculations. Maybe choosing b=2? Let me check.Suppose we set b=2. Then point B is at (0,2). Then line BP connects (0,2) to (2,2). That's a horizontal line y=2. So this line would intersect the semicircle at P(2,2) and maybe another point? Let's check.The semicircle equation is (x -2)^2 + y^2 =4, y ≥0. If y=2, then (x -2)^2 +4=4 ⇒ (x -2)^2=0 ⇒ x=2. So only point P. Therefore, BP is horizontal line y=2 from (0,2) to (2,2), tangent to the semicircle at P.In this case, the figure APCB is triangle ABC (with C(0,0), A(4,0), B(0,2)) and the semicircle on AC. The area of the triangle is (4*2)/2=4, semicircle area is 2π, total area 4+2π.Line BP is horizontal line y=2. The area above BP would be the part of the semicircle above y=2, which is zero, since the semicircle only goes up to y=2 (radius 2, centered at (2,0)). Wait, the semicircle from A(4,0) to C(0,0) with center at (2,0) and radius 2. The highest point is (2,2), which is P. So the semicircle is only up to y=2. Therefore, the line BP at y=2 coincides with the top of the semicircle. Therefore, the area above BP in the figure APCB would just be zero, and the area below BP would be the entire figure. But that doesn't make sense. Wait, no.Wait, if BP is the line y=2, then in the figure APCB (triangle ABC and semicircle AC), the area above BP would be the region of the triangle ABC above y=2, but since B is at (0,2) and the triangle is right-angled at C(0,0), the triangle's hypotenuse AB goes from (4,0) to (0,2). The line AB has equation y = (-1/2)x + 2.The area above BP (y=2) in the triangle ABC is zero because the highest point of the triangle is B(0,2). So the entire triangle is below or on y=2. The semicircle is also up to y=2. Therefore, the line BP divides the figure APCB into the semicircle (which is along y=2 at P) and the triangle. Wait, this seems like the line BP is along the top edge of the semicircle and passes through point B. So in this specific case, the area difference would be the area of the semicircle, since one part is the semicircle and the other is the triangle. But the semicircle area is 2π and the triangle area is 4, so the difference is |2π -4|. But the answer is supposed to be a fixed number, and if we choose b=2, we get this result. But if we choose another b, would we get the same difference?Wait, maybe the problem assumes that the triangle is isoceles, i.e., CA=CB=4, but the problem only states CA=4. However, in Chinese problems, sometimes figures are drawn with certain proportions unless specified otherwise. But since it's a right triangle at C, and CA=4, unless specified, CB can be any positive length. Therefore, maybe the answer is indeed |2π -4|, but I need to verify.Alternatively, perhaps the difference is always 4, regardless of the semicircle. Wait, but that seems unlikely. Let me check with another value of b.Suppose b=0. But that would collapse the triangle, so not possible. Let's take b=1.If b=1, then point B is at (0,1). Line BP connects (0,1) to (2,2). The equation is y = [(2-1)/2]x +1 = (1/2)x +1.This line will intersect the semicircle (x-2)^2 + y^2 =4 at points P(2,2) and another point Q.Using the earlier quadratic solution, t = b² / (8 -4b +b²) = 1 / (8 -4 +1) =1/5=0.2. Therefore, the intersection point Q is at t=0.2, which is x=2*0.2=0.4, y=1 + (2 -1)*0.2=1.2. So Q is at (0.4,1.2).Therefore, BP intersects the semicircle at Q(0.4,1.2) and P(2,2). Wait, but BP is from B(0,1) to P(2,2). The point Q is between B and P? At t=0.2, which is between 0 and1, yes. Therefore, BP crosses the semicircle at Q and P. Therefore, the area of the figure APCB on one side of BP would include the area of the semicircle from A to Q to P, and the area of the triangle from Q to C to B. The other side would include the semicircle from Q to P and the triangle from B to Q.This is getting complicated. Maybe integrating or using sector areas.Alternatively, maybe the difference in areas is equal to twice the area between BP and the triangle, or something similar. But this is too vague.Wait, another approach. Since the problem asks for the absolute difference between the two areas, maybe we can compute the area of one part minus the other, which could simplify due to symmetry or other properties.Alternatively, using the concept of area addition and subtraction. The total area is 2b + 2π. If we can compute one of the areas, say S1, then the difference would be |S1 - (2b + 2π - S1)| = |2S1 - 2b -2π|.But without knowing S1, this might not help.Wait, perhaps using Green's theorem or coordinate geometry to compute the area divided by BP.Let me try for the case when b=2. Earlier, we saw that if b=2, then BP is the line y=2, tangent to the semicircle at P. The figure APCB's total area is 4 + 2π. The line BP divides it into the semicircle (area 2π) and the triangle (area 4). Therefore, the difference is |2π -4|. However, the problem might be expecting a positive value, so 4 - 2π, but since it's absolute, it's |4 - 2π|. But 4 - 2π is negative (since π≈3.14, 2π≈6.28), so absolute value is 2π -4. But is this the case for all b?Wait, let's test with b=2. If the answer is 2π -4, then when b=2, yes. But for another b, say b=1, would this hold?Wait, if b=1, total area is 2*1 + 2π=2 +2π. The difference in areas would be |S1 - S2|. But without computing S1 and S2, it's hard to tell. But maybe the answer is always 2π -4 regardless of b. But that seems incorrect, unless there's some invariant.Alternatively, maybe the problem is in a configuration where BP is tangent to the semicircle, hence b=2. But the problem doesn't specify that. However, since P is the midpoint of the arc AC, and BP is drawn, maybe BP is tangent. Let me check.For BP to be tangent to the semicircle at P, the line BP must be perpendicular to the radius at P. The radius at P is from the center of the semicircle (2,0) to P(2,2), which is a vertical line. The tangent at P would be horizontal line y=2, which is indeed the case when BP is horizontal. For BP to be horizontal, point B must be at (0,2). Therefore, if the triangle is such that CB=2, then BP is tangent to the semicircle.But the problem doesn't specify that CB=2. So perhaps the problem assumes that the triangle is such that BP is tangent, making CB=2. Maybe in the original figure, this was the case. If that's true, then CB=2, so the triangle has legs 4 and 2, area 4, semicircle area 2π, total area 4 +2π. The line BP divides it into the semicircle (area 2π) and the triangle (area4), difference |2π -4|.But in the problem statement, it's just a general right triangle. So is there a way the difference is always 2π -4?Alternatively, perhaps the midpoint of the arc AC implies that BP is tangent. Let me verify.The midpoint of the arc AC on the semicircle is point P(2,2). The tangent at P is horizontal line y=2. If BP is this tangent line, then point B must lie on this tangent line. Since the tangent line at P is horizontal y=2, point B must be at (x,2). But in the triangle ABC, point B is at (0,b) because it's a right triangle at C. Therefore, to have B on the tangent line y=2, b=2. Therefore, only if b=2 is BP the tangent line. Therefore, in this case, the problem might be implying that BP is tangent to the semicircle, hence b=2.Therefore, if the problem's figure is such that BP is tangent to the semicircle at P, then CB=2, and the area difference is |2π -4|, which is 2π -4 since π>3.14 so 2π≈6.28>4.However, the problem says "the absolute value of the difference", so it should be 2π -4, which is approximately 2.28.But the problem might require an exact answer in terms of π, so the answer is 2π -4. But let's verify this with the case when b=2.If b=2, then yes, the difference is 2π -4.But the problem didn't specify that CB=2, so why would we assume that? Maybe the act of taking P as the midpoint of the arc AC causes BP to be tangent only when CB=2. Therefore, perhaps the problem is designed such that regardless of the triangle's dimensions, the difference is always 2π -4. But without further information, I think the problem expects us to assume that BP is tangent, hence CB=2, leading to the answer 2π -4.Alternatively, maybe there's a different approach using geometric transformations or symmetry.Let me think again.The figure APCB consists of triangle ABC and the semicircle on AC. The line BP divides this figure into two parts. The problem asks for the absolute difference in their areas.If we can show that one of the regions has area equal to the semicircle and the other equal to the triangle, then the difference would be |semicircle area - triangle area| = |2π - 2b|. But unless the triangle's area is equal to the semicircle's area, which would be when 2b = 2π ⇒ b=π, which is approximately 3.14, but there's no indication of that.Alternatively, maybe the line BP divides the figure such that one part is the semicircle and the other is the triangle, but this would only happen if BP is the boundary between them. However, in reality, BP cuts through both the semicircle and the triangle.Wait, going back to the case when b=2. If BP is the tangent at P, then the area above BP is just the semicircle, and the area below is the triangle. But actually, BP is part of the figure. Wait, no. The figure APCB includes both the semicircle and the triangle. So when you draw BP, which is from B to P, the two regions would be: one region is the semicircle plus the part of the triangle below BP, and the other region is the part of the triangle above BP. But if BP is tangent, then the area above BP in the triangle is zero, so one region is the semicircle and the entire triangle, and the other region is nothing? That doesn't make sense.Wait, no. If BP is part of the figure, then the figure is divided by BP into two parts: one part is the semicircle and the area of the triangle below BP, the other part is the area of the triangle above BP. But if BP is tangent, then there is no area above BP in the triangle, so one part is the entire figure, and the other part is nothing. That can't be.This is confusing. Maybe I need to use the shoelace formula or integrate to find the areas.Let me try coordinate geometry with b=2 as a specific case.Case when b=2:Points:C(0,0)A(4,0)B(0,2)P(2,2)Semicircle equation: (x-2)^2 + y^2 =4, y ≥0Line BP: from (0,2) to (2,2), which is y=2.The figure APCB is the union of triangle ABC and the semicircle. The line BP is the horizontal line y=2.The two regions divided by BP are:1. The region above BP: but y=2 is the highest point of the semicircle, so above BP there's nothing. The semicircle is up to y=2, and the triangle is below y=2. So the region above BP is just the line BP itself, which has no area. The region below BP is the entire figure. So the difference would be |0 - (4 + 2π)|=4 +2π, which contradicts. So this can't be right.Alternatively, perhaps the regions are the two parts formed by BP within the figure. But since BP is part of the figure, the figure is divided into two parts: one part is the semicircle APB (but APB is not a semicircle), wait.Wait, maybe the figure APCB is the polygon APCB, where AP is the arc from A to P, PC is the arc from P to C, CB is the line from C to B, and BA is the line from B to A. But in that case, AP and PC are arcs of the semicircle.But connecting BP would split this figure into two parts: the region bounded by arc AP, line PB, and line BA; and the region bounded by arc PC, line CB, and line BP. The area difference between these two regions is required.If that's the case, then we can compute the area of sector APB and subtract the area of triangle APB, and compare it to the area of sector PCB minus triangle PCB, but this is getting too vague.Alternatively, since APCB is a quadrilateral with two curved sides (AP and PC) and two straight sides (CB and BA), connecting BP would create two regions: one bounded by AP, BP, and BA, and the other bounded by PC, CB, and BP.To find their areas:Area1 = Area of sector APB + Area of triangle APBArea2 = Area of sector PCB + Area of triangle PCBBut I need to clarify.Wait, actually, AP and PC are arcs of the semicircle. So the figure APCB is the semicircle ACB combined with the triangle ABC? No.Wait, maybe APCB is the area bounded by the semicircle from A to C via P and the sides CB and BA. Then, connecting BP divides it into two regions: one is the sector APB and the other is the sector BPC combined with triangle BPC.But this is still unclear. Maybe it's better to use integration.Let me try for the general case with coordinates.The total area of figure APCB is area of triangle ABC (2b) plus area of semicircle (2π).Line BP divides this into two parts. To find the area of each part, we need to integrate or use geometric formulas.The line BP has equation y = [(2 - b)/2]x + b.This line intersects the semicircle at P(2,2) and another point Q, which we found earlier at t = b² / denominator.But this is complicated. Alternatively, we can use the principle of moments or symmetry.Wait, but maybe the answer is always 4. Because in the case when b=2, the difference is |2π -4|, but this is not 4. Alternatively, if the answer is 4, but in the case when b=2, 2π -4 ≈ 2.28, which is not 4.Alternatively, maybe the difference is twice the area of the triangle or something else.Wait, here's another thought. The midpoint of the arc AC on the semicircle is P, which has coordinates (2,2). The line connecting B to P. The area difference might be related to the power of point B with respect to the semicircle.The power of point B with respect to the semicircle is equal to the square of the tangent length from B to the semicircle. If BP is tangent, then the power is BP². Otherwise, it's equal to BQ * BP, where Q is the other intersection point.But the power of point B is also equal to (distance from B to center)^2 - radius^2. The center of the semicircle is (2,0), radius 2. So distance from B(0,b) to center (2,0) is √(4 + b²). Thus, power is (4 + b²) -4 = b².If BP is tangent, then BP² = power = b² ⇒ BP =b. But BP length is sqrt((2-0)^2 + (2 -b)^2) = sqrt(4 + (2 -b)^2). For this to equal b, we have:sqrt(4 + (2 -b)^2) =bSquare both sides:4 + (4 -4b +b²) =b² ⇒ 8 -4b=0 ⇒b=2. So again, when b=2, BP is tangent.Therefore, in this case, the power of point B is b²=4, which equals BP²=2² +0²=4? Wait, no. When b=2, BP is from (0,2) to (2,2), length is 2, so BP²=4, which matches the power.So in general, the power of point B is b²= BQ * BP, where Q is the other intersection point. But since we have BP = sqrt(4 + (2 -b)^2), and BQ = ?Wait, this might not be necessary.Alternatively, using coordinates, for the general case:The area difference can be computed as the absolute value of the difference between the area of the part of the semicircle above BP and the area of the part of the triangle below BP (or vice versa). But this is still complicated.Alternatively, since the answer is a constant, maybe we can use the case when b=2, compute 2π -4, and that's the answer.Given that in Chinese problems, often such problems assume certain configurations where the tangent is involved, and since when b=2, BP is tangent, and this results in a clean answer of 2π -4, which is a reasonable exact value, I think this is the intended answer.Therefore, the answer is 2π -4. Taking the absolute value, it's |2π -4| = 2π -4 since π >2.Therefore, the answer is boxed{2pi-4}.Final Answerboxed{2pi - 4}</think>Given a right triangle (triangle ABC) with (angle ACB = 90^circ) and (CA = 4), we need to find the absolute value of the difference in the areas of the two parts formed by the line segment (BP), where (P) is the midpoint of the semicircular arc (overparen{AC}).1. Coordinates and Setup: - Place (C) at the origin ((0, 0)), (A) at ((4, 0)), and (B) at ((0, b)). - The semicircle with diameter (AC) has its center at ((2, 0)) and radius 2. The midpoint (P) of the semicircular arc (overparen{AC}) is at ((2, 2)).2. Line Equation of (BP): - The coordinates of (B) are ((0, b)) and (P) are ((2, 2)). - The slope of (BP) is (frac{2 - b}{2}), and the equation of the line (BP) is (y = frac{2 - b}{2}x + b).3. Intersection with the Semicircle: - The semicircle equation is ((x - 2)^2 + y^2 = 4) with (y geq 0). - Solving for intersections between (BP) and the semicircle, we find that (P) is always one intersection point. For (B) at ((0, 2)), the line (BP) is tangent to the semicircle at (P).4. Area Calculation: - When (B) is at ((0, 2)), the triangle (ABC) has area (4) and the semicircle has area (2pi). - The line (BP) (horizontal line (y = 2)) divides the figure into the semicircle and the triangle. The absolute difference in areas is (|2pi - 4|).5. Conclusion: - The absolute value of the difference in the areas of the two parts is (2pi - 4).(boxed{2pi - 4})]

🔑:Ni.6.4.As shown in Figure 8, let A C and B P intersect at point D, and the point symmetric to D with respect to the circle center O is denoted as E. The line segment B P divides the figure A P C B into two parts, and the absolute value of the difference in the areas of these two parts is the area of triangle B E P, which is twice the area of triangle B O P. Thus,begin{array}{l}S_{triangle B P O}=frac{1}{2} P O cdot C O =frac{1}{2} times 2 times 2=2 .end{array}Therefore, the absolute value of the difference in the areas of these two parts is 4.

🧠:[<think>Okay, let's see. I need to find the real number m such that when you take the union of sets A and B, which are given as A = {2, m} and B = {1, m²}, the result is {1, 2, 3, 9}. Hmm, so the union of A and B should have exactly those elements: 1, 2, 3, and 9. First, let me recall that the union of two sets includes all elements that are in either set. So, A ∪ B should contain all elements from A and all elements from B without repetition. The given union is {1, 2, 3, 9}, so that means the elements of A and B combined must be exactly these four numbers. Set A has elements 2 and m, and set B has elements 1 and m². So combining these, the union must include 1, 2, m, and m². But the union is {1, 2, 3, 9}. Therefore, m and m² must be 3 and 9, but I need to figure out which is which and ensure that m is a real number.So, m and m² have to be 3 and 9. Let's consider possible real numbers m such that m² is either 3 or 9. Wait, if m is 3, then m² would be 9. If m is √3, then m² would be 3. Alternatively, if m is -3, then m² is also 9. Similarly, if m is -√3, m² is 3. Let's check each possibility.First, let's suppose m is 3. Then set A would be {2, 3} and set B would be {1, 9}. Then their union is {1, 2, 3, 9}, which matches the given union. So m = 3 is a possible solution.Next, let's check m = √3. If m is √3 (approximately 1.732), then m² is 3. Then set A would be {2, √3} and set B would be {1, 3}. Then the union would be {1, 2, √3, 3}. But the given union is {1, 2, 3, 9}, so √3 is approximately 1.732, which is not equal to 9. Therefore, this doesn't work. So m can't be √3.What about m = -3? Then set A would be {2, -3} and set B would be {1, 9}. The union would then be {1, 2, -3, 9}, which is not the same as the given union {1, 2, 3, 9} because -3 is not equal to 3. Therefore, m = -3 is not a valid solution.Similarly, if m = -√3, then m² is 3. So set A would be {2, -√3} and set B would be {1, 3}. The union would be {1, 2, -√3, 3}, which again includes -√3 instead of 9. So that's not the solution either.Wait, another thought. What if m² is 9 and m is 3? Then that works as we saw earlier. What if m² is 3 and m is √3 or -√3? Then the union would have √3 or -√3 instead of 9, which is not in the given union. Therefore, m must be 3 or -3. But -3 leads to the union having -3 instead of 3, which isn't in the target union. So m must be 3.But let me verify once more. If m = 3:Set A = {2, 3}Set B = {1, 9}Union A ∪ B = {1, 2, 3, 9} which is exactly what's given. Perfect.Is there any other possible real number for m? Let's see. Let's suppose m is 9. Then m² would be 81. Then set A would be {2, 9} and set B would be {1, 81}. Then the union is {1, 2, 9, 81}, which does not match the given union. So m=9 is invalid.If m is 1, then m² is 1. Then set A would be {2,1} and set B would be {1,1} which is just {1}. So the union is {1, 2}, which is missing 3 and 9. So that's not right.If m is 2, then m² is 4. Then set A is {2, 2} which is just {2}, and set B is {1, 4}, so union is {1, 2, 4}, which is missing 3 and 9. Not correct.If m is 0, then m² is 0. Sets would be {2,0} and {1,0}, union is {0,1,2}, missing 3 and 9.Alternatively, if m is something else. Let's check if there are other possibilities. Suppose m is 3. As we saw, that works. If m is -3, union has -3 instead of 3, which is not allowed. So the only real number is 3. Wait, but maybe m is 3, but is there a case where m² is 3 and m is another number?Wait, if m² = 3, then m = √3 or -√3. But then m would be √3, which is approximately 1.732, not an integer. The union in that case would have √3 and 3, but the problem states that the union is {1,2,3,9}. So if m is √3, then the union would be {1, 2, √3, 3}, which is not the same as the given union because √3 is not 9, so that doesn't work. Similarly, m = -√3 would give union {1, 2, -√3, 3}, which includes a negative number and again not 9. So those don't work.Alternatively, if m² = 9, then m is 3 or -3. If m is 3, union is {1,2,3,9} which is correct. If m is -3, union is {1,2,-3,9} which is not correct. Therefore, only m=3 is valid.Wait, let me check again. The problem says that A ∪ B = {1, 2, 3, 9}. So all elements must be present. If m is 3, then A has 2 and 3, B has 1 and 9. Union is {1,2,3,9}. Perfect.If m is -3, then A is {2, -3} and B is {1, 9}. So union is { -3, 1, 2, 9}, but 3 is missing, and instead there's -3. So that's not matching.Therefore, the only solution is m=3.But wait, let me check if there's another possibility. Suppose m is both 3 and 9? No, m is a single real number. So m can't be two different numbers. Therefore, m must be 3. Wait another angle: The union has exactly four elements: 1,2,3,9. Therefore, A and B can't have any elements outside of these. So m must be either 1,2,3, or9. Because if m is something else, like 4, then 4 would be in A or B, and thus in the union, which is not allowed. Therefore, m must be one of 1,2,3,9.So let's check each possibility.If m=1: Then A={2,1}, B={1,1}={1}. Then union is {1,2}, which is missing 3 and 9. Not valid.If m=2: A={2,2}={2}, B={1,4}. Then union is {1,2,4}, missing 3 and 9. Not valid.If m=3: A={2,3}, B={1,9}. Union is {1,2,3,9}. Valid.If m=9: A={2,9}, B={1,81}. Union is {1,2,9,81}. Not valid because 81 is not in the given union.Therefore, only m=3 works.So the answer is 3. Wait, but just to confirm. Since the problem says "the real number m", and m=3 is a real number. So that's the answer.Final AnswerThe real number ( m ) is boxed{3}.</think>Given two sets ( A = {2, m} ) and ( B = {1, m^2} ), their union ( A cup B = {1, 2, 3, 9} ). We need to find the real number ( m ).1. The union of sets ( A ) and ( B ) must include all elements from both sets. Therefore, the elements ( m ) and ( m^2 ) must be 3 and 9, respectively, since the union is given as {1, 2, 3, 9}.2. We consider possible values for ( m ): - If ( m = 3 ), then ( m^2 = 9 ). This gives sets ( A = {2, 3} ) and ( B = {1, 9} ). Their union is ( {1, 2, 3, 9} ), which matches the given union. - If ( m = sqrt{3} ) or ( m = -sqrt{3} ), then ( m^2 = 3 ). However, this would result in the union containing ( sqrt{3} ) or ( -sqrt{3} ) instead of 9, which is not in the given union. - If ( m = -3 ), then ( m^2 = 9 ). This would result in the union containing -3 instead of 3, which is not in the given union. - Checking other possible values (1, 2, 9) for ( m ) results in unions that do not match the given union {1, 2, 3, 9}.3. After verifying all possibilities, the only valid solution is ( m = 3 ).Thus, the real number ( m ) is (boxed{3}).,<think>Okay, so I have this problem here. Let me read it again. It says: Given two sets A and B, where A is {2, m} and B is {1, m²}. Their union is {1, 2, 3, 9}. I need to find the real number m. Hmm, okay, let's break this down.First, let's recall what a union of two sets means. The union of sets A and B, denoted A ∪ B, is the set of elements that are in A, in B, or in both. So, all the elements from both sets combined, without duplicates. The problem states that A ∪ B is {1, 2, 3, 9}. So, the elements in A and B together must be exactly these four numbers: 1, 2, 3, and 9.Now, set A is {2, m} and set B is {1, m²}. So, the elements in A are 2 and m, and the elements in B are 1 and m². When we take the union, we should get all four elements: 1, 2, 3, 9. Therefore, the elements 2 and m from A and 1 and m² from B must combine to form {1, 2, 3, 9}. So, let me list the elements from both sets:From A: 2, mFrom B: 1, m²Combined, these should be 1, 2, 3, 9. So, m and m² must be 3 and 9 in some order. Because we already have 1 and 2 from the other elements. So, the union is 1, 2, m, m², which has to equal 1, 2, 3, 9. Therefore, m and m² must be 3 and 9. But which one is which? Let's check.If m is 3, then m² would be 9. Let's test that. If m = 3, then set A is {2, 3} and set B is {1, 9}. Then the union of A and B would be {1, 2, 3, 9}, which matches exactly the given union. So that works. Alternatively, what if m² is 3 and m is 9? Let's check. If m = 9, then m² would be 81, which is not in the union set {1, 2, 3, 9}. So that doesn't work because 81 isn't part of the union. So that's invalid. Alternatively, if m² is 3, then m would be sqrt(3) or -sqrt(3). But then m² would be 3, but m itself would be sqrt(3) ≈ 1.732 or -1.732, which is not in the union set {1, 2, 3, 9}. So, that's not possible either. Therefore, m cannot be sqrt(3) or -sqrt(3).Another angle: since m must be a real number, and m² is also part of the union, both m and m² must be elements of {1, 2, 3, 9}. So, possible candidates for m are 1, 2, 3, 9. Let's check these.If m = 1: Then set A is {2, 1}, set B is {1, 1} (since m² = 1). Then the union is {1, 2}, which is not {1, 2, 3, 9}. So invalid.If m = 2: Set A is {2, 2} (duplicate, so just {2}), set B is {1, 4}. Then the union is {1, 2, 4}, which is not the desired union. So invalid.If m = 3: As before, set A {2,3}, set B {1,9}, union is {1,2,3,9}. Perfect. That works.If m = 9: Then set A is {2,9}, set B is {1,81}. Union would be {1, 2, 9, 81}, which is not {1,2,3,9}. So invalid.So, m=3 is the only one that works. Let me verify again.Given m=3:A = {2, 3}B = {1, 3²} = {1, 9}A ∪ B = {1, 2, 3, 9} which matches exactly the given union. Therefore, m=3 is the correct answer.Wait, but the problem mentions that m is a real number. So, m could be negative? Let's check if m is negative.Suppose m = -3. Then m² = 9. So, set A would be {2, -3}, and set B would be {1, 9}. Then the union would be { -3, 1, 2, 9 }, which is not the same as {1,2,3,9} because -3 is not in the target union. Therefore, m cannot be -3.Similarly, if m is negative but m² is 3, which would lead to m being sqrt(3) or -sqrt(3), but as I checked before, m would not be in the union. Therefore, negative m's don't work here.Hence, m must be 3. So I think that's the answer.But let me just make sure there are no other possibilities. For example, suppose m is a real number not in the union. But wait, the union is supposed to contain all elements from A and B. So, since m is in set A, it must be part of the union. Therefore, m must be one of the elements in {1,2,3,9}. Similarly, m² must also be part of the union. Therefore, m² must be in {1,2,3,9}. Therefore, m must satisfy two conditions: m ∈ {1,2,3,9} and m² ∈ {1,2,3,9}.So let's check all elements of {1,2,3,9} for these conditions:- m=1: m²=1. So m² is in the union. But then union would be {1,2}, as above. Not valid.- m=2: m²=4. 4 is not in the union. So invalid.- m=3: m²=9. 9 is in the union. So valid.- m=9: m²=81. 81 not in union. Invalid.Therefore, only m=3 satisfies both m and m² being in the union {1,2,3,9}.Therefore, the answer must be 3.Final AnswerThe real number ( m ) is boxed{3}.</think>Given two sets ( A = {2, m} ) and ( B = {1, m^2} ), their union ( A cup B = {1, 2, 3, 9} ). We need to find the real number ( m ).1. The union of sets ( A ) and ( B ) must contain all elements from both sets. Therefore, the elements ( m ) and ( m^2 ) must be among the elements in the union set {1, 2, 3, 9}.2. We check possible values for ( m ) such that both ( m ) and ( m^2 ) are in the union set: - If ( m = 3 ), then ( m^2 = 9 ). This makes set ( A = {2, 3} ) and set ( B = {1, 9} ). The union ( A cup B = {1, 2, 3, 9} ), which matches the given union. - Other values such as ( m = 1 ), ( m = 2 ), or ( m = 9 ) do not satisfy the condition that both ( m ) and ( m^2 ) are in the union set {1, 2, 3, 9}. - Negative values like ( m = -3 ) result in elements not present in the union set.Thus, the real number ( m ) is (boxed{3}).]

🔑:Since A cup B = {1, 2, 3, 9}, it follows that both m and m^2 must be elements of the set {1, 2, 3, 9}.Let's solve for m in a systematic way:- Firstly, we see that 2 is already an element of A, so m cannot be 2.- Now we check if m can be 1, 3, or 9.1. If m=1, m^2=1^2=1. However, 1 is already accounted for by B, so this does not give us a new element. Hence, m neq 1.2. If m=3, m^2=3^2=9. Both numbers, 3 and 9, fit perfectly into the union without any repetition of elements already present. Therefore, m=3 is a valid solution.3. If m=9, m^2=9^2=81. However, 81 is not an element of A cup B, so m neq 9.By this process, the only valid solution for m that satisfies the given conditions is m=3.Therefore, the real number m is boxed{3}.

🧠:[<think>Okay, let's see. So the problem is about a math group where each member borrowed exactly two science popularization books from a mini bookshelf that has 12 books in total. Each of these 12 books has been borrowed by exactly 3 members. The question is asking how many people are in the math group.Hmm, okay. Let me break this down. First, the total number of books is 12. Each book is borrowed by 3 members. So if I think about each book, there are 3 people associated with it. So maybe if I multiply the number of books by the number of people per book, that gives me the total number of borrowings? Wait, but each person borrows two books. So each person is associated with two books. So maybe the total number of borrowings is also equal to the number of people multiplied by 2. Since each person contributes two borrowings. So perhaps both expressions equal the total number of borrowings, which would mean:Number of people * 2 (since each person borrows 2 books) = Number of books * 3 (since each book is borrowed by 3 people)So, if I let the number of people be P, then:2P = 12 * 32P = 36Therefore, P = 18.Wait, so 18 people? Let me check if that makes sense. Each person borrows 2 books, so 18 people would borrow 36 books in total. But there are only 12 books, each borrowed 3 times, so 12 * 3 = 36. Yeah, that matches up. So the total number of borrowings is 36, and since each person does 2 borrowings, 36 divided by 2 is 18. So that seems correct.Is there another way to think about this? Maybe using combinations or something? Hmm, not sure. Let me see. Each book is borrowed by 3 people, so for each book, there are 3 people associated with it. So the total number of "book-person" pairs is 12 * 3 = 36. On the other hand, each person borrows 2 books, so each person is associated with 2 books. Therefore, the total number of "person-book" pairs is P * 2. Since these are the same pairs, just counted from different perspectives, they should be equal. So again, 2P = 36, so P = 18. That checks out.Is there a possibility of overcounting or something? For example, if two people borrow the same two books, but the problem states that each member has borrowed exactly two of them, but it doesn't say that the pairs have to be unique. However, since each book is borrowed by exactly 3 members, each book can be associated with 3 different people. Therefore, even if two people share a common book, as long as each book is only counted 3 times, it's okay.Let me think of a smaller example to verify. Suppose there are 2 books, each borrowed by 3 people. Each person borrows 2 books. Then total borrowings would be 2*3 = 6. Number of people would be 6 / 2 = 3. Wait, but with 2 books, each person has to borrow 2 books, which would mean each person has to borrow both books. So 3 people would each borrow both books, so each book is borrowed 3 times. That works. So in this case, 3 people. So the formula works here too.Another example: 3 books, each borrowed by 2 people. Each person borrows 1 book. Then total borrowings are 3*2 = 6. Each person borrows 1 book, so number of people is 6. But if each person can only borrow 1 book, then 6 people each take 1 book, but each book is borrowed by 2 people. So 3 books, 2 people per book, total borrowings 6, number of people 6. That works. But in this case, each person borrows 1 book, so total people would be 6. If we did 3 books each borrowed by 2 people, and each person borrows 2 books, then total borrowings would be 6, so number of people is 6 / 2 = 3. But each person would need to borrow 2 books. But with 3 books, each person can borrow 2 books. However, each book is only allowed to be borrowed by 2 people. So each of the 3 people would have to borrow 2 books, but we have 3 books. So person 1: book 1 and 2, person 2: book 1 and 3, person 3: book 2 and 3. Then each book is borrowed by 2 people. Yes, that works. So 3 people, 3 books, each book borrowed by 2 people, each person borrows 2 books. So the formula 2P = 3*2 => P = 3. Which is correct. So the formula seems to hold here as well.So applying this formula to the original problem, 12 books each borrowed by 3 people, so total borrowings 36, divided by 2 borrowings per person, gives 18 people. That seems solid. So the answer should be 18.Final AnswerThe number of people in the math group is boxed{18}.</think>The problem involves determining the number of people in a math group where each member borrowed exactly two science popularization books from a total of 12 books. Each book was borrowed by exactly 3 members. To solve this, we consider the total number of borrowings. Each book is borrowed by 3 members, so the total number of borrowings is calculated by multiplying the number of books by the number of times each book was borrowed:[12 text{ books} times 3 text{ members per book} = 36 text{ total borrowings}]Next, we know that each member borrowed exactly 2 books. Therefore, the total number of borrowings can also be expressed as the number of members multiplied by the number of books each member borrowed:[P text{ members} times 2 text{ books per member} = 2P text{ total borrowings}]Equating the two expressions for the total number of borrowings, we get:[2P = 36]Solving for (P):[P = frac{36}{2} = 18]Thus, the number of people in the math group is (boxed{18}).,<think>Okay, let me try to figure out this problem. So, there's a mini bookshelf with 12 science popularization books. Each member of the math group has borrowed exactly two of these books, and each book has been borrowed by exactly 3 members. We need to find out how many people are in the math group.Hmm, let me start by understanding the problem again. There are 12 books. Each person borrows 2 books. Each book is borrowed by 3 people. So, we need to relate the number of people to the number of books and the borrowing details.This seems like a problem that involves counting the total number of borrowings in two different ways. Maybe using the principle of double counting? Like, if we count the total number of times books have been borrowed from the perspective of the people and from the perspective of the books, we can set those equal and solve for the number of people.Let me formalize this. Let’s denote the number of people in the math group as ( P ). Each person borrows 2 books, so the total number of borrowings, if we consider each person's borrowings, would be ( P times 2 ).On the other hand, each of the 12 books is borrowed by 3 people. So, the total number of borrowings from the books' perspective is ( 12 times 3 ).Since both expressions represent the total number of borrowings, they must be equal. Therefore, we can set up the equation:[2P = 12 times 3]Let me compute the right side first. ( 12 times 3 = 36 ). So the equation becomes:[2P = 36]Solving for ( P ), we divide both sides by 2:[P = 36 / 2 = 18]So, that gives ( P = 18 ). Therefore, there are 18 people in the math group.Wait, let me double-check that to make sure I didn't make a mistake. Each person borrows 2 books, so 18 people would borrow 36 books in total. But since there are 12 books, each borrowed by 3 people, the total number of borrowings is indeed 12 x 3 = 36. So both sides match. 36 total borrowings from both perspectives.That seems consistent. So, 18 people. Let me just think if there's another way to approach this problem or if there's any assumption I might have missed.Another way to think about it: Each book is borrowed by 3 people, so each book contributes 3 borrowings. There are 12 books, so 3 x 12 = 36 borrowings. Each person does 2 borrowings, so number of people is 36 / 2 = 18. Yep, same answer.Is there any possibility that the answer could be different? Maybe if there was some overlap or something, but the problem states that each member borrows exactly two books, and each book is borrowed by exactly three members. So, there's no overlap issue here because we are just counting total borrowings. Each borrowing is a unique pair of a person and a book. Since each person has two books, each book is paired with three people. So, the total number of such pairs is 2P and 3B, where B is 12. So, 2P = 3*12 = 36. Therefore, P=18. Yep, that seems solid.I think that's the correct answer. There's no missing component here. The key was recognizing that we can count the total borrowings from two different angles and set them equal. This is a classic combinatorial problem where double-counting helps. So, yeah, the answer is 18.Final AnswerThe number of people in the math group is boxed{18}.</think>Given that there are 12 science popularization books, each member of the math group borrows exactly 2 books, and each book is borrowed by exactly 3 members, we need to determine the number of people in the math group.We start by noting that the total number of borrowings can be calculated from two perspectives: the number of books borrowed by each member and the number of members borrowing each book.1. From the perspective of the members: If there are ( P ) people in the math group, and each person borrows 2 books, the total number of borrowings is ( 2P ).2. From the perspective of the books: There are 12 books, each borrowed by 3 members, so the total number of borrowings is ( 12 times 3 = 36 ).Equating these two expressions for the total number of borrowings, we get:[2P = 36]Solving for ( P ):[P = frac{36}{2} = 18]Thus, the number of people in the math group is (boxed{18}).]

🔑:3. The class mini bookshelf has a total of 12 science popularization books. According to statistics, each member of the math group has borrowed exactly two of them, and each science popularization book has been borrowed by exactly 3 members of the math group. How many people are there in this math group?【Analysis】Since the mini bookshelf has a total of 12 science popularization books, and each book has been borrowed by exactly 3 members of the math group, the total number of borrowings is 12 times 3=36 (person-times). Let the math group have x people. Since each member of the math group has borrowed exactly two books, the total number of borrowings is 2 x person-times. Therefore, we have 2 x=36, and then solve it.【Solution】Let the math group have x people,Each science popularization book has been borrowed by exactly 3 members of the math group, so the total number of borrowings is 12 times 3=36 (person-times). Since each member of the math group has borrowed exactly two books, the total number of borrowings is 2 x person-times,So, 2 x=36,Solving for x, we get: x=18,Answer: This math group has 18 people.

🧠:[<think>Okay, let's see. So the problem is about Ginger saving time by running instead of walking to school, and we need to find the distance from her home to her school. The answer choices are given, so that might help later. Let me read the problem again to make sure I get all the details.Ginger walks at 4 km/h and runs at 6 km/h. She saves 3 3/4 minutes by running instead of walking from her home to her school. What is the distance in kilometers?Alright, so key points here: two speeds, 4 km/h walking and 6 km/h running. The time saved by running instead of walking is 3 3/4 minutes. And we need the distance. Let's start breaking this down.First, time saved is the difference between the time it takes her to walk and the time it takes her to run. So, time walking minus time running equals 3 3/4 minutes. But since the speeds are in km per hour, the time will be in hours. So, we need to convert that 3 3/4 minutes into hours to keep the units consistent. That seems important.Let me recall that 1 hour is 60 minutes, so 3 3/4 minutes is 3.75 minutes, which is 3.75/60 hours. Let me compute that. 3.75 divided by 60. Let's see, 3.75 divided by 60. Hmm, 3.75 divided by 60. Well, 60 goes into 3.75, 0.0625 times. Because 60 times 0.0625 is 3.75. So, 3.75 minutes is 0.0625 hours. Okay, so the time saved is 0.0625 hours.Now, let's denote the distance from her home to school as d kilometers. Then, the time taken to walk would be distance divided by speed, so d / 4 hours. Similarly, the time taken to run would be d / 6 hours. The difference between these two times is 0.0625 hours. So, the equation we have is:(d / 4) - (d / 6) = 0.0625Hmm, let me check that again. Time walking is longer than time running, so subtracting the running time from the walking time gives the time saved. Yes, that makes sense. So the equation is correct.Now, let's solve for d. First, find a common denominator for the fractions. The denominators are 4 and 6. The least common denominator is 12. So, converting both fractions:(d / 4) is equivalent to (3d / 12)(d / 6) is equivalent to (2d / 12)So, substituting back into the equation:(3d / 12) - (2d / 12) = 0.0625Which simplifies to:(3d - 2d) / 12 = 0.0625d / 12 = 0.0625Then, multiplying both sides by 12:d = 0.0625 * 12Let's compute that. 0.0625 * 12. Hmm, 0.0625 is 1/16, right? Because 1/16 is 0.0625. So 1/16 * 12 = 12/16 = 3/4. So, 3/4 of a kilometer. Wait, but 3/4 is one of the answer choices, option E. But let me verify because maybe I made a mistake.Wait, let's compute 0.0625 * 12 directly. 0.06 * 12 = 0.72, and 0.0025 * 12 = 0.03, so total is 0.72 + 0.03 = 0.75. So, 0.75 km, which is 3/4. So the answer is E) 3/4. But wait, the options given are (A) 7 1/2, (B) 3 3/4, (C) 1 7/8, (D) 1 1/4, (E) 3/4. So, 3/4 is option E.But wait, let me double-check. Maybe I messed up in the equation setup. Let's go through the steps again.Distance is d. Time to walk is d / 4 hours. Time to run is d / 6 hours. Time saved is (d / 4) - (d / 6) = 3.75 minutes, which is 3.75 / 60 hours = 0.0625 hours. Then:d / 4 - d / 6 = 0.0625Which is, as before, (3d - 2d)/12 = 0.0625 => d / 12 = 0.0625 => d = 0.75 km. Hmm, that seems right. So 0.75 km is 3/4 km, which is answer E. But the options are listed with some larger numbers. Let me check the problem again.Wait, perhaps I made a mistake converting minutes to hours. Let's verify that. 3 3/4 minutes is 3.75 minutes. To convert minutes to hours, we divide by 60. So 3.75 / 60. Let's compute that.3.75 divided by 60. Let's do 3.75 / 60:60 goes into 3.75 0.0625 times, as before. So that part is correct.Alternatively, 3 minutes is 3/60 = 0.05 hours, and 0.75 minutes (which is 45 seconds) is 0.75/60 = 0.0125 hours. So total is 0.05 + 0.0125 = 0.0625 hours. Correct.So, time difference is 0.0625 hours. Then solving the equation gives d = 0.75 km. So answer E. But wait, let's check the answer options again. The options are:(A) 7 1/2(B) 3 3/4(C) 1 7/8(D) 1 1/4(E) 3/4So 3/4 is E. That's one of the choices. So maybe that's correct. But let's think intuitively. If the distance was 3/4 km, then the time to walk is (3/4)/4 = 3/16 hours, which is 3/16 *60 = 11.25 minutes. Time to run is (3/4)/6 = 3/24 = 1/8 hours = 7.5 minutes. The difference is 11.25 -7.5 = 3.75 minutes, which matches the given time saved. So that checks out. So the answer is indeed E) 3/4.Wait, but just to be thorough, maybe the problem is presented differently. Let me check once again. So the problem says she saves 3 3/4 minutes by running instead of walking. So the calculation we did, converting 3.75 minutes to hours, setting up the equation, solving for d, verifying by plugging back in. All steps seem correct, and the answer checks out. So I think E is the correct answer. But why is this problem given in the list with options that include larger numbers? Maybe it's a trick question where people might miscalculate the units.Alternatively, perhaps some might set up the equation incorrectly. Let me see. If someone forgets to convert minutes to hours, they might do:(d /4 - d /6) = 3.75But then, if they set the equation as equal to 3.75 minutes instead of hours, then solving:d/4 - d/6 = 3.75Which would give:(3d -2d)/12 = 3.75d/12 = 3.75d = 3.75 *12 = 45. So 45 km, which is not among the options. So that's wrong. Therefore, unit conversion is critical here.Alternatively, maybe some might flip the time difference. Suppose they think running time minus walking time is positive? But that would give a negative distance, which is impossible. So the correct setup is walking time minus running time equals the time saved.Alternatively, maybe someone might miscalculate the fraction. Let me see:3 3/4 minutes is 3.75 minutes. Divided by 60 is 0.0625 hours. Then solving d/4 - d/6 = 0.0625.Calculating:The difference in times is d/4 - d/6 = (3d - 2d)/12 = d/12. So d/12 = 0.0625. So d = 0.0625 *12 = 0.75. So that's 3/4 km.So, yes, the answer is E.Alternatively, maybe there's a different approach? Let's see.Another way to think about it is to find the relative speed or something, but since it's about the time difference for the same distance, the basic approach is correct.Alternatively, using the concept that the difference in times is given. So, for distance d, time1 = d/4, time2 = d/6, difference = d/4 - d/6 = d*(1/4 -1/6) = d*(3/12 -2/12) = d*(1/12). So, d/12 = 0.0625. So, d = 0.0625*12 = 0.75. Same result.Therefore, the answer is E. So, I think that's correct. Wait, but looking at the answer choices, option C is 1 7/8, which is 1.875, and option D is 1.25. So, none of these are 0.75 except E.Alternatively, perhaps I made a mistake in the problem statement. Wait, let me check again. The problem says she saves 3 3/4 minutes by running instead of walking. So if she runs, she arrives 3 3/4 minutes earlier. So, yes, the time difference is 3.75 minutes. So converting that to hours is 3.75/60 = 0.0625. Then the equation is correct.Alternatively, if the problem had stated she saved 3 3/4 hours, but that's not the case. So, the answer is E. Therefore, I think the correct answer is E) 3/4.Final Answerboxed{E}</think>Ginger walks at 4 km/h and runs at 6 km/h. She saves (3 frac{3}{4}) minutes (which is 3.75 minutes) by running instead of walking from her home to her school. We need to find the distance from her home to her school.First, convert the time saved from minutes to hours:[3.75 text{ minutes} = frac{3.75}{60} text{ hours} = 0.0625 text{ hours}]Let (d) be the distance from her home to her school in kilometers. The time taken to walk is (frac{d}{4}) hours, and the time taken to run is (frac{d}{6}) hours. The difference between these times is the time saved:[frac{d}{4} - frac{d}{6} = 0.0625]To solve this equation, find a common denominator for the fractions:[frac{3d}{12} - frac{2d}{12} = 0.0625]Simplify the equation:[frac{d}{12} = 0.0625]Multiply both sides by 12:[d = 0.0625 times 12 = 0.75]Thus, the distance from her home to her school is (frac{3}{4}) kilometers. The correct answer is (boxed{E}).,<think>Okay, let's see. Ginger walks at 4 km/h and runs at 6 km/h. She saves 3 3/4 minutes by running instead of walking to school. We need to find the distance from her home to school. Hmm, the options are given, so maybe I can work through the problem step by step.First, I remember that time is equal to distance divided by speed. So, if the distance is the same whether she walks or runs, the time difference will be because of the different speeds. The time saved by running is 3 3/4 minutes, which I need to convert into hours because the speeds are in km per hour. Let me convert 3 3/4 minutes to hours. Since there are 60 minutes in an hour, 3.75 minutes is 3.75/60 hours. Let me calculate that: 3 divided by 60 is 0.05, and 0.75 divided by 60 is 0.0125, so total is 0.05 + 0.0125 = 0.0625 hours. So, 3.75 minutes is 0.0625 hours.Let me denote the distance as d. Then, the time it takes her to walk is d divided by her walking speed, which is 4 km/h. So, walking time is d/4 hours. Similarly, her running time is d/6 hours. The difference between these two times is the time saved, which is 0.0625 hours. So, the equation would be:d/4 - d/6 = 0.0625I need to solve this equation for d. Let me compute the left side. To subtract these fractions, I can find a common denominator, which is 12. So, converting both fractions:d/4 = 3d/12d/6 = 2d/12So, 3d/12 - 2d/12 = d/12Therefore, d/12 = 0.0625So, solving for d:d = 0.0625 * 12Let me compute that. 0.0625 * 12. Hmm, 0.0625 is 1/16, so 1/16 *12 = 12/16 = 3/4 = 0.75. Wait, that gives d = 0.75 km. But looking at the options, (E) is 3/4, which is 0.75. But that seems too small. Let me check my steps again.Wait, the time saved is 3.75 minutes, which I converted to 0.0625 hours. That seems right. Then, the equation is walking time minus running time equals time saved. So, d/4 - d/6 = 0.0625.Then, d/4 - d/6 is (3d - 2d)/12 = d/12. So, d/12 = 0.0625. Then, d = 0.0625 * 12. Let's calculate 0.0625 * 12. 0.0625 * 10 is 0.625, and 0.0625 * 2 is 0.125. So, total is 0.625 + 0.125 = 0.75. So, d is 0.75 km. That's answer choice E. But wait, the options include (C) 1 7/8, (D) 1 1/4, etc. So, 0.75 is option E. But maybe I made a mistake in converting the time?Wait, 3 3/4 minutes is 3.75 minutes. To convert minutes to hours, we divide by 60. 3.75 / 60. Let me compute that again. 3 divided by 60 is 0.05. 0.75 divided by 60 is 0.0125. So, total 0.0625 hours. That seems correct. Then, the equation is correct. So, d is indeed 0.75 km, which is option E.But let me verify. If the distance is 0.75 km, then walking time is 0.75 / 4 hours. 0.75 divided by 4 is 0.1875 hours. Convert that to minutes: 0.1875 * 60 = 11.25 minutes. Running time is 0.75 / 6 hours, which is 0.125 hours. 0.125 * 60 = 7.5 minutes. The difference is 11.25 - 7.5 = 3.75 minutes, which matches the given time saved. So, the calculation seems correct. Therefore, the answer is E) 3/4.Wait, but the options are listed as (A) 7 1/2, (B) 3 3/4, (C) 1 7/8, (D) 1 1/4, (E) 3/4. So, 0.75 km is option E. So, that's the answer. But in the problem statement, the options have (C) 1 7/8, which is 1.875 km, (D) 1 1/4 is 1.25 km. So, seems like maybe the answer is E. But I thought the answer would be 1.5 km or something, but maybe not.Alternatively, maybe there's a miscalculation. Let's check again. Walking speed 4 km/h, running speed 6 km/h. Time difference is 3.75 minutes. Let's use variables.Let distance be d. Time walking is d / 4. Time running is d / 6. The difference is d/4 - d/6 = d*(1/4 - 1/6) = d*(3/12 - 2/12) = d/12. This time difference is equal to 3.75 minutes, which is 3.75/60 hours = 0.0625 hours. So, d/12 = 0.0625. Therefore, d = 0.0625 * 12 = 0.75. So, same result.Alternatively, maybe the problem says she saves 3 3/4 minutes by running instead of walking, so running is faster, so walking time is more. So, walking time minus running time is 3.75 minutes. So, that's correct. So, the calculation is correct. So, answer is E. But why is option C 1 7/8? Hmm. Maybe the answer is different? Wait, perhaps I misread the problem? Let me check again."Ginger walks at 4 km/h and runs at 6 km/h. She saves 3 3/4 minutes by running instead of walking from her home to her school. What is the distance, in kilometres, from her home to her school?"No, the problem seems correctly understood. So, unless there is a trick here. Wait, maybe she doesn't run all the way? But the problem says "by running instead of walking", which implies she covers the entire distance by running instead of walking. So, the time saved is the difference between walking the entire distance and running the entire distance. So, the answer is 0.75 km, which is 3/4. So, E.Alternatively, maybe convert the time saved into minutes first, but no, we converted correctly into hours. Let me check with minutes.Alternatively, set up the equation in minutes. Walking time is (d / 4) hours, which is (d / 4)*60 minutes. Similarly, running time is (d /6)*60 minutes. The difference is 3.75 minutes. So:(d / 4)*60 - (d /6)*60 = 3.75Which is (15d - 10d) = 3.75So, 5d = 3.75d = 3.75 /5 = 0.75 km. Same answer. So, indeed, 0.75 km. So, answer is E. So, maybe the answer is E. I need to check the options again. Yes, option E is 3/4, which is 0.75. So, that must be the answer.Therefore, the answer is boxed{E}.Final Answerboxed{E}</think>Ginger walks at (4 mathrm{~km/h}) and runs at (6 mathrm{~km/h}). She saves (3 frac{3}{4}) minutes (which is (3.75) minutes) by running instead of walking from her home to her school. We need to find the distance from her home to her school.First, convert the time saved from minutes to hours:[3.75 text{ minutes} = frac{3.75}{60} text{ hours} = 0.0625 text{ hours}]Let (d) be the distance in kilometres. The time taken to walk is (frac{d}{4}) hours, and the time taken to run is (frac{d}{6}) hours. The difference in time is given by:[frac{d}{4} - frac{d}{6} = 0.0625]To solve this equation, find a common denominator for the fractions:[frac{3d}{12} - frac{2d}{12} = frac{d}{12}]Thus, we have:[frac{d}{12} = 0.0625]Solving for (d):[d = 0.0625 times 12 = 0.75]Therefore, the distance from her home to her school is (boxed{E}).]

🔑:Suppose that the distance from Ginger's home to her school is d mathrm{~km}.Since there are 60 minutes in an hour, then 3 frac{3}{4} minutes (or frac{15}{4} minutes) is frac{15}{4} times frac{1}{60}=frac{1}{16} of an hour.Since Ginger walks at 4 mathrm{~km} / mathrm{h}, then it takes her frac{d}{4} hours to walk to school.Since Ginger runs at 6 mathrm{~km} / mathrm{h}, then it takes her frac{d}{6} hours to run to school.Since she saves frac{1}{16} of an hour by running, then the difference between these times is frac{1}{16} of an hour, sobegin{aligned}frac{d}{4}-frac{d}{6} & =frac{1}{16} frac{3 d}{12}-frac{2 d}{12} & =frac{1}{16} frac{d}{12} & =frac{1}{16} d & =frac{12}{16}=frac{3}{4}end{aligned}Therefore, the distance from Ginger's home to her school is frac{3}{4} mathrm{~km}.ANswer: (E)

🧠:[<think>Okay, let me try to solve this problem. So, we have the universal set U which is all real numbers R. Then set A is defined as all x such that x is greater than 0, so A = {x | x > 0}. Set B is defined as all x such that x is less than or equal to 1, so B = {x | x ≤ 1}. The question is asking for the intersection of A and the complement of B in U, which is written as A ∩ (∁ᵤ B). The options are given as A, B, C, D with different intervals. First, I need to recall what the complement of a set means. The complement of B with respect to U, which is ∁ᵤ B, should be all the elements in U (so all real numbers) that are not in B. Since B is {x | x ≤ 1}, the complement would be all x where x is not less than or equal to 1. That means x has to be greater than 1. So, ∁ᵤ B = {x | x > 1}.Now, the problem is asking for the intersection of A and this complement set. A is {x | x > 0}, and ∁ᵤ B is {x | x > 1}. The intersection of these two sets would be all x that satisfy both conditions. So, x has to be greater than 0 and also greater than 1. But if x is greater than 1, it's automatically greater than 0, right? So, the intersection should just be all x where x > 1. Let me double-check that. If we take numbers greater than 1, they are certainly in A because they're greater than 0, and they are also in the complement of B because they're not less than or equal to 1. So, their intersection is indeed x > 1. Looking back at the options, option D is {x | x > 1}, which would be the correct answer. Wait, but let me make sure I didn't mix up anything. Let's visualize this on a number line. Set A is everything to the right of 0, not including 0. Set B is everything to the left of 1, including 1. The complement of B is everything not in B, so that's everything to the right of 1, not including 1. So, the intersection of A (which is from 0 onwards) and the complement of B (from 1 onwards) would be from 1 onwards. But since A starts at 0 and the complement of B starts at 1, overlapping part is from 1 to infinity. So, yes, x > 1. Therefore, the answer is D. Let me check the other options to ensure none of them make sense. Option A is {x | 0 ≤ x < 1}. But that's the intersection of A and B, isn't it? Because A is x > 0 and B is x ≤ 1. So their intersection would be 0 < x ≤ 1, which is option B. Wait, but the question is about A and the complement of B, so that's different. So, complement of B is x > 1, so A ∩ (complement of B) is x > 1. So, definitely D. Another way to think about it: A ∩ ∁ᵤ B is the set of elements that are in A but not in B. Since B is x ≤ 1, not being in B means x > 1. So, elements in A that are greater than 1. So, that's x > 1. Therefore, D. Alternatively, using set operations: ∁ᵤ B = U B = R {x | x ≤ 1} = {x | x > 1}. Then, A ∩ {x | x > 1} = {x | x > 0} ∩ {x | x > 1} = {x | x > 1}. Yes, that's correct. So, the answer should be D.Final Answerboxed{D}</think>Given the universal set ( U = mathbb{R} ), set ( A = {x mid x > 0} ), and set ( B = {x mid x leq 1} ), we need to find ( A cap (complement_{U} B) ).1. Complement of ( B ) in ( U ): - The complement of ( B ) with respect to ( U ) is all real numbers not in ( B ). - Since ( B = {x mid x leq 1} ), the complement ( complement_{U} B ) is ( {x mid x > 1} ).2. Intersection of ( A ) and ( complement_{U} B ): - Set ( A ) is ( {x mid x > 0} ). - The intersection ( A cap {x mid x > 1} ) consists of all elements that are in both sets. - This intersection is the set of all ( x ) such that ( x > 1 ).Thus, the result of ( A cap (complement_{U} B) ) is ( {x mid x > 1} ), which corresponds to option D.[boxed{D}],<think>Okay, let me try to figure out this problem step by step. So, we have the universal set U, which is all real numbers, right? Then set A is defined as all x such that x is greater than 0. So, A is everything positive, not including zero. Then set B is all x less than or equal to 1. So, B includes everything up to and including 1. The question is asking for the intersection of A and the complement of B in U. Let me write that down to visualize it better.First, let's recall that the complement of a set in U consists of all elements in U that are not in the set. So, complement of B (denoted as ∁ᵤB) would be all real numbers not in B. Since B is {x | x ≤ 1}, its complement should be {x | x > 1}, right? Because if x is not less than or equal to 1, it must be greater than 1. That makes sense.Now, set A is {x | x > 0}, which is all positive real numbers. So, we need to find the intersection between A and ∁ᵤB. The intersection of two sets includes all elements that are in both sets. So, elements that are in both A and the complement of B.Let me break it down:- A = (0, ∞)- ∁ᵤB = (1, ∞)Therefore, the intersection of A and ∁ᵤB would be the overlap between (0, ∞) and (1, ∞). The overlap here is everything greater than 1, because that's where both sets coincide. So, the intersection should be (1, ∞), which in the given options is option D: {x | x > 1}.Wait, let me double-check. Maybe I made a mistake in the complement of B. Let's confirm. The complement of B in U (which is R) is all real numbers not in B. Since B is x ≤ 1, the complement is indeed x > 1. So, ∁ᵤB = (1, ∞). Then, intersecting that with A, which is (0, ∞), the overlapping part is from 1 to infinity. So, yes, {x | x > 1} is the correct answer.Looking at the options, D is {x | x > 1}, so that must be the right choice. Let me check the other options just to be sure.Option A is {x | 0 ≤ x < 1}. That's the interval from 0 to 1, including 0 but not 1. But this is the intersection of A with B, maybe? Because A is x > 0 and B is x ≤ 1. So, their intersection would be (0,1], which is not exactly option A, but option B is {x | 0 < x ≤ 1}. Wait, maybe that's confusing.But the question is about A intersect complement of B. So, complement of B is x > 1, as we discussed. So, A is x > 0. So, the intersection is x > 1. So, D is correct. The other options don't make sense. Option C is x < 0, which is the complement of A, but we're looking at A, so that's not possible. So yeah, D must be the answer.Final Answerboxed{D}</think>Given the universal set ( U = mathbb{R} ), set ( A = {x mid x > 0} ), and set ( B = {x mid x leq 1} ), we need to find ( A cap (complement_U B) ).1. Complement of ( B ) in ( U ): - ( B ) is the set of all real numbers ( x ) such that ( x leq 1 ). - The complement of ( B ), denoted ( complement_U B ), consists of all real numbers not in ( B ). Therefore, ( complement_U B = {x mid x > 1} ).2. Intersection of ( A ) and ( complement_U B ): - Set ( A ) is ( {x mid x > 0} ). - The complement of ( B ) is ( {x mid x > 1} ). - The intersection ( A cap (complement_U B) ) includes all elements that are in both ( A ) and ( complement_U B ). This is the set of all real numbers ( x ) such that ( x > 1 ).Thus, the final answer is (boxed{D}).]

🔑:To solve the problem step by step:1. Given the set A={x|x gt 0}, which contains all the numbers greater than 0.2. Given the set B={x|xleqslant 1}, which contains all the numbers less than or equal to 1.3. The complement of B in the universal set U=mathbb{R}, denoted as complement _{U}B, consists of all elements not in B. So, complement _{U}B={x|x gt 1} because it includes all real numbers greater than 1, excluding those less than or equal to 1.4. To find the intersection of A and complement _{U}B, denoted as A⋂(complement _{U}B), we look for elements that are common to both sets. Both sets include numbers greater than 1. Therefore, A⋂(complement _{U}B)={x|x gt 1}.Encapsulating the final answer: boxed{D}.

🧠:[<think>Okay, let's see. I need to find the perimeter of this tangential quadrilateral ABCD. So first, let me recall what a tangential quadrilateral is. If I remember correctly, a tangential quadrilateral is one where there is an incircle tangent to all four sides. That means the sum of the lengths of the two opposite sides are equal. So, AB + CD = BC + DA. That's a key property I might need to use here.The problem gives me some angles: angle DAB and angle ABC are both 120 degrees, and angle CDA is 90 degrees. Also, the length of AB is 1. The perimeter is AB + BC + CD + DA, so if I can find the other three sides, BC, CD, and DA, then I can sum them up with AB (which is 1) to get the perimeter.Let me try to visualize this quadrilateral. Let me sketch it out mentally. ABCD is a quadrilateral. Let's start with vertex A. Angle at A (DAB) is 120 degrees. So, sides AD and AB meet at A with 120 degrees. Then angle ABC is 120 degrees, so at vertex B, sides BA and BC meet at 120 degrees. Then angle CDA is 90 degrees, so at vertex D, sides CD and DA meet at 90 degrees. The fourth angle, which is at vertex C, angle BCD, isn't given. But since it's a quadrilateral, the sum of all angles should be 360 degrees. Let me check:Sum of given angles: 120 (DAB) + 120 (ABC) + 90 (CDA) = 330 degrees. So the remaining angle at C is 360 - 330 = 30 degrees. So angle BCD is 30 degrees.Hmm, that's useful. So angle at C is 30 degrees.Now, since it's a tangential quadrilateral, the incircle is tangent to all sides. Let me recall that in a tangential quadrilateral, the lengths of the sides can be expressed in terms of the tangent lengths from each vertex to the points of tangency with the incircle. Specifically, if we denote the tangent lengths from each vertex as follows: from A, the tangents to the incircle are of length x, from B it's y, from C it's z, and from D it's w. Then the sides can be written as:AB = x + yBC = y + zCD = z + wDA = w + xAnd because it's tangential, we have AB + CD = BC + DA. So x + y + z + w = y + z + w + x, which is trivially true, but perhaps more importantly, the semiperimeter (s) is equal to x + y + z + w. But maybe I need a different approach here.Alternatively, since we have angles given, perhaps using the Law of Sines or Cosines in some triangles. Let me see if I can divide the quadrilateral into triangles and apply trigonometric laws.Let me try to split the quadrilateral into two triangles by drawing a diagonal. Let's say diagonal BD. Then we have triangle ABD and triangle BCD. But maybe not. Alternatively, diagonal AC? Hmm, not sure. Alternatively, maybe split it into triangles ABC and ADC? Wait, no. Let me think.Alternatively, perhaps construct the inradius and use some area relations. In a tangential quadrilateral, the area is equal to the semiperimeter multiplied by the inradius (Area = r * s, where s is the semiperimeter). But since we don't know the inradius or the area, this might not help directly. Unless we can compute the area another way.Alternatively, let's assign coordinates to the points and use coordinate geometry. Maybe place point A at the origin and build coordinates from there. Let's try that.Let's place point A at (0,0). Since angle DAB is 120 degrees, and AB has length 1. Let's assume AB is along the x-axis. So point B is at (1,0). Now, angle at A is 120 degrees, so side AD makes a 120-degree angle with AB. So if AB is along the x-axis from A(0,0) to B(1,0), then AD would go from A(0,0) at an angle of 120 degrees. Let me compute the coordinates of D.The direction of AD is 120 degrees from the positive x-axis. Let’s denote the length of AD as 'a'. Then coordinates of D would be (a * cos(120°), a * sin(120°)). Cos(120°) is -0.5 and sin(120°) is √3/2. So D is at (-0.5a, (√3/2)a).Similarly, angle at B, which is angle ABC = 120 degrees. So from point B(1,0), side BC makes a 120-degree angle with AB. Since AB is along the x-axis from A to B, the direction from B to C is 120 degrees above the x-axis? Wait, no. Let me think. At vertex B, the sides BA and BC meet at 120 degrees. BA is from B to A, which is along the negative x-axis. So angle ABC is 120 degrees between BA and BC. So if BA is pointing to the left (negative x-direction), then BC would make a 120-degree angle upwards from BA. So the direction of BC is 60 degrees above the positive x-axis? Let me confirm.If angle at B is 120 degrees, with BA pointing to the left (negative x), then BC would form a 120-degree angle with BA. So the angle between BA (which is towards the left) and BC is 120 degrees. So if we take BA as 180 degrees, then BC would be 180 - 120 = 60 degrees above the positive x-axis. So the direction of BC is 60 degrees from the positive x-axis. So if the length of BC is 'b', then coordinates of point C would be (1 + b * cos(60°), 0 + b * sin(60°)). Cos(60°) is 0.5, sin(60°) is √3/2. So C is at (1 + 0.5b, (√3/2)b).Now, we also know that angle at D is angle CDA = 90 degrees. So angle at D between CD and DA is 90 degrees. Let's use this information to relate coordinates of points C and D.Point D is at (-0.5a, (√3/2)a). Point C is at (1 + 0.5b, (√3/2)b). So CD is the segment from C to D, and DA is from D to A(0,0). The angle at D between CD and DA is 90 degrees. So the vectors DC and DA should be perpendicular. Let's compute vectors DC and DA.Vector DC is C - D = [1 + 0.5b - (-0.5a), (√3/2)b - (√3/2)a] = [1 + 0.5b + 0.5a, (√3/2)(b - a)]Vector DA is A - D = [0 - (-0.5a), 0 - (√3/2)a] = [0.5a, - (√3/2)a]Since angle CDA is 90 degrees, vectors DC and DA are perpendicular, so their dot product is zero.Dot product of DC and DA:[1 + 0.5b + 0.5a] * [0.5a] + [(√3/2)(b - a)] * [ - (√3/2)a ] = 0Let me compute each term:First term: [1 + 0.5b + 0.5a] * 0.5a = 0.5a * (1 + 0.5b + 0.5a)Second term: [(√3/2)(b - a)] * [ - (√3/2)a ] = - (3/4)a(b - a)So combining these:0.5a(1 + 0.5b + 0.5a) - (3/4)a(b - a) = 0Let's factor out 'a' since a ≠ 0 (as it's a side length):a[0.5(1 + 0.5b + 0.5a) - (3/4)(b - a)] = 0So the expression inside the brackets must be zero:0.5(1 + 0.5b + 0.5a) - (3/4)(b - a) = 0Let me expand the terms:First term: 0.5 * 1 + 0.5 * 0.5b + 0.5 * 0.5a = 0.5 + 0.25b + 0.25aSecond term: - (3/4)b + (3/4)aSo combining all terms:0.5 + 0.25b + 0.25a - 0.75b + 0.75a = 0Combine like terms:0.5 + (0.25a + 0.75a) + (0.25b - 0.75b) = 0Which simplifies to:0.5 + a - 0.5b = 0So:a - 0.5b = -0.5Multiply both sides by 2:2a - b = -1Thus, b = 2a + 1So we have a relation between a and b: b = 2a + 1Okay, that's one equation. Now, we need another equation to solve for a and b.Since ABCD is a tangential quadrilateral, we know that AB + CD = BC + DA.Given AB = 1, BC = b, DA = a, and CD is the length between points C and D.Let me compute CD.Coordinates of C: (1 + 0.5b, (√3/2)b)Coordinates of D: (-0.5a, (√3/2)a)So distance CD:√[(1 + 0.5b - (-0.5a))² + ((√3/2)b - (√3/2 a))²]Simplify the terms inside the square roots:x-coordinate difference: 1 + 0.5b + 0.5ay-coordinate difference: (√3/2)(b - a)Thus, CD = √[(1 + 0.5b + 0.5a)^2 + ( (√3/2)(b - a) )^2 ]Let me compute this expression:First term inside sqrt:(1 + 0.5b + 0.5a)^2 + ( (√3/2)(b - a) )^2Let me expand the first square:= [1 + 0.5b + 0.5a]^2 = 1 + (0.5b + 0.5a)^2 + 2*(0.5b + 0.5a)Wait, no. Wait, (x + y)^2 = x^2 + 2xy + y^2. So:[1 + 0.5(b + a)]^2 = 1^2 + 2*1*0.5(b + a) + [0.5(b + a)]^2 = 1 + (b + a) + 0.25(b + a)^2Similarly, the second term:( (√3/2)(b - a) )^2 = (3/4)(b - a)^2Therefore, CD^2 = 1 + (b + a) + 0.25(b + a)^2 + (3/4)(b - a)^2Let me combine these terms:First, expand 0.25(b + a)^2 and 0.75(b - a)^2.Wait, 0.25(b + a)^2 + 0.75(b - a)^2Let me compute that:0.25[(b + a)^2] + 0.75[(b - a)^2]= 0.25(b² + 2ab + a²) + 0.75(b² - 2ab + a²)= 0.25b² + 0.5ab + 0.25a² + 0.75b² - 1.5ab + 0.75a²Combine like terms:b²: 0.25 + 0.75 = 1ab: 0.5ab - 1.5ab = -1aba²: 0.25 + 0.75 = 1So total: 1*b² - 1*ab + 1*a² = b² - ab + a²Therefore, CD^2 = 1 + (b + a) + (b² - ab + a²)So CD^2 = 1 + b + a + b² - ab + a²But CD is a length, so CD = sqrt(1 + b + a + b² - ab + a²)But since in the tangential quadrilateral, AB + CD = BC + DA.Given AB = 1, BC = b, DA = a, so:1 + CD = b + aThus, CD = b + a - 1Therefore, CD^2 = (b + a - 1)^2So:(b + a - 1)^2 = 1 + b + a + b² - ab + a²Let me expand the left side:(b + a - 1)^2 = (b + a)^2 - 2(b + a) + 1 = b² + 2ab + a² - 2b - 2a + 1Set equal to right side:b² + 2ab + a² - 2b - 2a + 1 = 1 + b + a + b² - ab + a²Simplify both sides:Left side: b² + a² + 2ab - 2b - 2a + 1Right side: 1 + b + a + b² - ab + a²Subtract right side from left side:(b² + a² + 2ab - 2b - 2a + 1) - (1 + b + a + b² - ab + a²) = 0Simplify term by term:b² - b² = 0a² - a² = 02ab - (-ab) = 2ab + ab = 3ab-2b - b = -3b-2a - a = -3a1 - 1 = 0So overall:3ab - 3b - 3a = 0Divide both sides by 3:ab - b - a = 0So:ab - a - b = 0Factor:ab - a - b = a(b - 1) - b = (a - 1)(b - 1) - 1 = 0? Wait, maybe not.Alternatively, add 1 to both sides:ab - a - b + 1 = 1Which factors as (a - 1)(b - 1) = 1But let me check:(a - 1)(b - 1) = ab - a - b + 1, so yes, that's correct. So:(a - 1)(b - 1) = 1But earlier, we had from the angle condition: b = 2a + 1So substitute b = 2a + 1 into (a - 1)(b - 1) = 1:(a - 1)(2a + 1 - 1) = (a - 1)(2a) = 2a(a - 1) = 1Thus:2a(a - 1) = 1Expand:2a² - 2a - 1 = 0So quadratic equation:2a² - 2a - 1 = 0Solving for a:a = [2 ± sqrt(4 + 8)] / 4 = [2 ± sqrt(12)] / 4 = [2 ± 2*sqrt(3)] / 4 = [1 ± sqrt(3)] / 2Since a is a length, it must be positive. So:[1 + sqrt(3)] / 2 ≈ (1 + 1.732)/2 ≈ 1.366, which is positive.[1 - sqrt(3)] / 2 ≈ (1 - 1.732)/2 ≈ -0.366, which is negative. So we discard this.Therefore, a = [1 + sqrt(3)] / 2Then b = 2a + 1 = 2*[1 + sqrt(3)]/2 + 1 = [1 + sqrt(3)] + 1 = 2 + sqrt(3)So now, we have a = (1 + sqrt(3))/2, b = 2 + sqrt(3)Now, we can compute CD = b + a - 1 (from AB + CD = BC + DA, CD = BC + DA - AB = b + a - 1)Compute CD:b + a - 1 = (2 + sqrt(3)) + (1 + sqrt(3))/2 - 1Convert to common denominator:= (4 + 2sqrt(3))/2 + (1 + sqrt(3))/2 - 2/2= [4 + 2sqrt(3) + 1 + sqrt(3) - 2]/2= [3 + 3sqrt(3)]/2So CD = (3 + 3sqrt(3))/2Therefore, now we have all sides:AB = 1BC = b = 2 + sqrt(3)CD = (3 + 3sqrt(3))/2DA = a = (1 + sqrt(3))/2So perimeter P = AB + BC + CD + DALet's compute each term:AB = 1BC = 2 + sqrt(3)CD = (3 + 3sqrt(3))/2DA = (1 + sqrt(3))/2Sum them up:1 + (2 + sqrt(3)) + (3 + 3sqrt(3))/2 + (1 + sqrt(3))/2Combine the terms:First, convert all to halves to add easily:1 = 2/22 + sqrt(3) = (4 + 2sqrt(3))/2(3 + 3sqrt(3))/2 remains the same(1 + sqrt(3))/2 remains the sameSo total:2/2 + (4 + 2sqrt(3))/2 + (3 + 3sqrt(3))/2 + (1 + sqrt(3))/2Now add numerators:2 + 4 + 2sqrt(3) + 3 + 3sqrt(3) + 1 + sqrt(3) all over 2Combine like terms:Constants: 2 + 4 + 3 + 1 = 10sqrt(3) terms: 2sqrt(3) + 3sqrt(3) + sqrt(3) = 6sqrt(3)So total numerator: 10 + 6sqrt(3)Therefore, perimeter P = (10 + 6sqrt(3))/2 = 5 + 3sqrt(3)So the perimeter is 5 + 3√3.Let me check if this makes sense. Let's verify the steps again to ensure no errors.First, we placed the quadrilateral in coordinate system, found coordinates of points D and C in terms of a and b, then used the perpendicularity condition at D to get an equation between a and b (b = 2a + 1). Then, using the tangential quadrilateral condition AB + CD = BC + DA, we derived another equation leading to (a - 1)(b - 1) = 1. Substituted b = 2a + 1 into this, solved for a, got a positive value, then found b. Then computed CD and DA, summed all sides. Calculations seem correct.Let me verify the final perimeter:AB = 1BC = 2 + √3 ≈ 3.732CD = (3 + 3√3)/2 ≈ (3 + 5.196)/2 ≈ 4.098DA = (1 + √3)/2 ≈ (1 + 1.732)/2 ≈ 1.366Sum: 1 + 3.732 + 4.098 + 1.366 ≈ 10.1965 + 3√3 ≈ 5 + 5.196 ≈ 10.196, which matches. So seems correct.Therefore, the perimeter is 5 + 3√3.Final AnswerThe perimeter of quadrilateral (ABCD) is boxed{5 + 3sqrt{3}}.</think>Given a tangential quadrilateral (ABCD) with (angle DAB = angle ABC = 120^circ), (angle CDA = 90^circ), and (|AB| = 1), we need to find the perimeter of (ABCD).1. Key Properties and Setup: - Tangential quadrilateral: (AB + CD = BC + DA). - Angles: (angle DAB = 120^circ), (angle ABC = 120^circ), (angle CDA = 90^circ), and (angle BCD = 30^circ) (since the sum of angles in a quadrilateral is (360^circ)).2. Coordinate System: - Place (A) at ((0,0)) and (B) at ((1,0)). - Coordinates of (D) and (C) are derived using trigonometric relationships and angles.3. Vectors and Perpendicularity: - Vectors (DC) and (DA) are perpendicular, leading to the equation (2a - b = -1) (where (a = |AD|) and (b = |BC|)).4. Tangential Quadrilateral Condition: - Using (AB + CD = BC + DA), derive the equation ((a - 1)(b - 1) = 1).5. Solving for (a) and (b): - Substitute (b = 2a + 1) into ((a - 1)(b - 1) = 1) and solve the quadratic equation, yielding (a = frac{1 + sqrt{3}}{2}) and (b = 2 + sqrt{3}).6. Calculating Side Lengths: - (AB = 1) - (BC = 2 + sqrt{3}) - (CD = frac{3 + 3sqrt{3}}{2}) - (DA = frac{1 + sqrt{3}}{2})7. Perimeter Calculation: - Summing all sides: (AB + BC + CD + DA = 1 + (2 + sqrt{3}) + frac{3 + 3sqrt{3}}{2} + frac{1 + sqrt{3}}{2}) - Simplifying: (frac{10 + 6sqrt{3}}{2} = 5 + 3sqrt{3})The perimeter of quadrilateral (ABCD) is (boxed{5 + 3sqrt{3}}).,<think>Okay, so I need to find the perimeter of this tangential quadrilateral ABCD. Let me recall what a tangential quadrilateral is. It's a quadrilateral that has an incircle tangent to all four sides. That means the sum of the lengths of the two opposite sides are equal. So, AB + CD = BC + AD. That's a key property I can use here.Given angles: angle DAB is 120 degrees, angle ABC is 120 degrees, and angle CDA is 90 degrees. Also, the length of AB is 1. So, AB = 1. The perimeter would be AB + BC + CD + DA. Since AB is 1, I need to find BC, CD, and DA.Let me try to visualize this quadrilateral. Let me start by drawing a rough sketch. Let's denote the vertices in order: A, B, C, D. So, angle at A (DAB) is 120 degrees, angle at B (ABC) is 120 degrees, angle at D (CDA) is 90 degrees. The fourth angle, angle at C (BCD), I don't know yet. But since it's a quadrilateral, the sum of all angles should be 360 degrees. Let me check that.Sum of given angles: 120 (at A) + 120 (at B) + 90 (at D) + angle at C = 360. So angle at C is 360 - (120 + 120 + 90) = 30 degrees. So angle BCD is 30 degrees.Hmm, okay. So angles at A, B, C, D are 120°, 120°, 30°, 90° respectively. Now, since it's a tangential quadrilateral, there is an incircle tangent to all sides. Let me denote the points of tangency. Let's say the incircle touches AB at point P, BC at Q, CD at R, and DA at S. Then, the lengths from the vertices to the points of tangency satisfy certain equalities.Specifically, if we denote the tangent lengths as follows: from A to P and S, which should be equal; from B to P and Q; from C to Q and R; from D to R and S. Let me denote the tangent lengths as x, y, z, w. Wait, usually, in a tangential quadrilateral, if we let the tangent lengths from each vertex to the points of tangency be equal. So, if from A, the two tangent segments are equal, say length a. From B, the two segments are equal, length b. From C, length c. From D, length d. Then, the sides can be written as:AB = a + b,BC = b + c,CD = c + d,DA = d + a.And since AB + CD = BC + DA, substituting, we have (a + b) + (c + d) = (b + c) + (d + a), which simplifies to a + b + c + d = a + b + c + d, so it's always true. So, that's consistent.But how does that help me here? Maybe I can assign variables to these tangent lengths and use the angles to set up equations.Let me try to denote the tangent lengths as follows:From point A: the two tangent segments are both equal to x. So, AP = AS = x.From point B: the two tangent segments are both equal to y. So, BP = BQ = y.From point C: the two tangent segments are both equal to z. So, CQ = CR = z.From point D: the two tangent segments are both equal to w. So, DR = DS = w.Therefore, the sides can be written as:AB = AP + BP = x + y = 1 (given).BC = BQ + CQ = y + z.CD = CR + DR = z + w.DA = DS + AS = w + x.And the perimeter is AB + BC + CD + DA = (x + y) + (y + z) + (z + w) + (w + x) = 2x + 2y + 2z + 2w = 2(x + y + z + w). But since we need the perimeter, perhaps it's easier to compute x + y + z + w first, then multiply by 2.But how do I find these variables x, y, z, w?Alternatively, maybe I can use the fact that in a tangential quadrilateral, the area is equal to the product of the semiperimeter and the radius of the incircle. But I don't know the radius here. Maybe not helpful directly.Alternatively, perhaps using trigonometry. Let me consider the angles given and try to relate the sides.Let me think of the quadrilateral as being composed of four triangles with the inradius. Wait, maybe not. Alternatively, if I can split the quadrilateral into two triangles by drawing a diagonal, but since the angles are given, perhaps I can use the Law of Sines or Cosines in some triangles.Wait, let's try to model the quadrilateral step by step. Let me assign coordinates to the points to make it easier. Let me place point A at the origin (0,0). Then, since angle DAB is 120°, and side AB is length 1. Let me suppose that AB is along the x-axis. So point B is at (1,0). Then, angle at A is 120°, so the side AD makes a 120° angle with AB. Let me compute the coordinates of D.If AD has length, say, DA = s, then the coordinates of D would be (s*cos(120°), s*sin(120°)) = (s*(-1/2), s*(√3/2)).Similarly, angle ABC is 120°, so at point B, angle between AB and BC is 120°. Since AB is from (1,0) to (0,0), the direction from B to A is towards the left. So the angle of 120° at B would mean that BC is going off at 60° above the negative x-axis. Wait, perhaps I need to think carefully.At point B, angle ABC is 120°, so the angle between BA and BC is 120°. Since BA is from B to A, which is towards the left along the x-axis. So if BA is pointing to the left (negative x-direction), then BC is making a 120° angle with BA. So, starting from BA direction (left), turning 120° towards the upper side. That would put BC at 60° above the positive x-axis. Wait, maybe I need to draw this.Alternatively, using standard position: If we have point B at (1,0), and angle ABC is 120°, then the internal angle at B is 120°, so the external angle is 60°. The direction from B to A is along the negative x-axis. So, the direction of BC is 60° from BA. Hmm.Wait, perhaps coordinate system approach is the way to go here. Let's fix coordinates:Let me set point A at (0,0).AB is of length 1, so point B is at (1,0).At point A, angle DAB is 120°, so the side AD is at 120° from AB. Since AB is along the x-axis from A(0,0) to B(1,0), the angle of 120° at A would mean AD is going into the second quadrant. Let me assign coordinates to point D.Suppose AD has length 's'. Then, the coordinates of D would be s*cos(120°) along x-axis and s*sin(120°) along y-axis. Cos(120°) is -1/2, sin(120°) is √3/2. So D is at (-s/2, (s√3)/2).Similarly, at point B, angle ABC is 120°, so the side BC is making an angle of 120° with BA. Since BA is from B(1,0) to A(0,0), which is along the negative x-axis. So, the internal angle at B is 120°, meaning that the angle between BA and BC is 120°, so BC is at 60° above the positive x-axis. Let me denote the length of BC as 't'. Then, the coordinates of point C can be found.From point B(1,0), moving along a direction 60° above the x-axis (since angle with BA, which is negative x-axis, is 120°, so direction of BC is 60° from positive x-axis). Therefore, the coordinates of point C would be (1 + t*cos(60°), 0 + t*sin(60°)) = (1 + t*(1/2), 0 + t*(√3/2)).So point C is at (1 + t/2, (t√3)/2).Now, we have point D at (-s/2, (s√3)/2) and point C at (1 + t/2, (t√3)/2). Now, we need to connect points C and D, and also points D and A. Given that angle at D is 90°, angle CDA is 90°, so angle at D between CD and DA is 90°. Let me verify the coordinates.First, let's note that DA is from D(-s/2, (s√3)/2) to A(0,0). The vector DA is (s/2, -(s√3)/2). Similarly, CD is from C(1 + t/2, (t√3)/2) to D(-s/2, (s√3)/2). The vector CD is (-s/2 - (1 + t/2), (s√3)/2 - (t√3)/2) = (- (s/2 + 1 + t/2), (√3/2)(s - t)).Since angle at D is 90°, the vectors DA and DC must be perpendicular. Wait, angle CDA is 90°, so angle at D between CD and DA is 90°, so vectors DC and DA are perpendicular. Therefore, the dot product of vectors DC and DA should be zero.Vector DC is from D to C: (1 + t/2 - (-s/2), (t√3)/2 - (s√3)/2) = (1 + t/2 + s/2, (√3/2)(t - s)).Vector DA is from D to A: (0 - (-s/2), 0 - (s√3)/2) = (s/2, - (s√3)/2).Wait, maybe I made a mistake in the direction. The angle at D is CDA, which is the angle between CD and DA. So vectors DC and DA. Wait, perhaps better to use vectors from point D: the angle is between CD and DA, so the vectors are CD and DA. So, CD is from C to D, and DA is from D to A. So vector CD is D - C, and vector DA is A - D. Then, the dot product of CD and DA should be zero.Vector CD = D - C = (-s/2 - (1 + t/2), (s√3)/2 - (t√3)/2) = (- (s/2 + 1 + t/2), (√3/2)(s - t)).Vector DA = A - D = (0 - (-s/2), 0 - (s√3)/2) = (s/2, - (s√3)/2).Dot product of CD and DA:[ - (s/2 + 1 + t/2) ] * (s/2) + [ (√3/2)(s - t) ] * [ - (s√3)/2 ]Let me compute each term:First term: [ - (s/2 + 1 + t/2) ] * (s/2) = - (s/2 + 1 + t/2) * s/2Second term: [ (√3/2)(s - t) ] * [ - (s√3)/2 ] = (√3/2)(s - t) * (- s√3/2 ) = (-3/4)s(s - t)So total dot product:- (s/2 + 1 + t/2) * s/2 - (3/4)s(s - t) = 0Let me expand the first term:- [ (s/2)(s/2) + (1)(s/2) + (t/2)(s/2) ] - (3/4)s(s - t) =- [ s²/4 + s/2 + st/4 ] - (3/4)s² + (3/4)st =- s²/4 - s/2 - st/4 - 3s²/4 + 3st/4 =Combine like terms:(- s²/4 - 3s²/4) + (-s/2) + (- st/4 + 3st/4)= (-s²) + (-s/2) + (2st/4)= -s² - s/2 + st/2Set equal to zero:-s² - s/2 + (st)/2 = 0Multiply both sides by -2:2s² + s - st = 0Factor s:s(2s + 1 - t) = 0Since s ≠ 0 (length of DA can't be zero), then:2s + 1 - t = 0So, t = 2s + 1Okay, so that's one equation relating t and s.Now, also, since ABCD is a tangential quadrilateral, the sum of the lengths of opposite sides are equal. So, AB + CD = BC + DA.Given AB = 1, CD is the length from C to D, BC is the length from B to C, which is t (since BC = y + z, but in our coordinate system, we set BC to be length t). Wait, no. Wait, in the coordinate system, BC is from B(1,0) to C(1 + t/2, (t√3)/2). Wait, that's not correct. Wait, in our earlier setup, when we set direction for BC, we considered BC as length t? Wait, perhaps confusion here.Wait, earlier, I considered BC as length t, but when we set up coordinates for C, we actually parameterized BC in terms of t. Wait, no. Wait, when we said from point B(1,0), moving along direction 60°, if we let the length of BC be t, then the coordinates of C would be (1 + t*cos60°, 0 + t*sin60°). Since cos60° = 0.5 and sin60° = √3/2, so that gives C as (1 + t/2, (t√3)/2). So yes, BC is length t. So BC = t.Similarly, DA is the length from D to A. Since D is at (-s/2, (s√3)/2), then DA is the distance from D to A(0,0):DA = √[ (-s/2 - 0)^2 + ((s√3)/2 - 0)^2 ] = √[ s²/4 + (3s²)/4 ] = √[s²] = s. So DA = s.Similarly, CD is the distance from C to D:C is at (1 + t/2, (t√3)/2), D is at (-s/2, (s√3)/2).So CD = √[ (1 + t/2 + s/2)^2 + ( (t√3)/2 - (s√3)/2 )^2 ]Let me compute this:First, the x-coordinate difference: (1 + t/2 + s/2)The y-coordinate difference: ( (t - s)√3 / 2 )So CD squared is:(1 + t/2 + s/2)^2 + [ ( (t - s)√3 / 2 ) ]^2= (1 + (t + s)/2 )^2 + (3/4)(t - s)^2Expand the first term:1 + (t + s) + ( (t + s)^2 )/4So, CD squared:1 + (t + s) + (t + s)^2 /4 + (3/4)(t - s)^2Now, expand (t + s)^2 /4 + (3/4)(t - s)^2:= [ (t² + 2ts + s²)/4 + 3(t² - 2ts + s²)/4 ]= [ (t² + 2ts + s²) + 3t² - 6ts + 3s² ] /4= (4t² -4ts + 4s²)/4= t² - ts + s²Therefore, CD squared is 1 + t + s + t² - ts + s²Hence, CD = √[1 + t + s + t² - ts + s²]But this seems complicated. However, since in the tangential quadrilateral, AB + CD = BC + DA, so 1 + CD = t + s.So, 1 + CD = t + sTherefore, CD = t + s -1But CD is also equal to √[1 + t + s + t² - ts + s²]Therefore, t + s -1 = √[1 + t + s + t² - ts + s²]Let me square both sides to eliminate the square root:(t + s -1)^2 = 1 + t + s + t² - ts + s²Left side:(t + s -1)^2 = t² + 2ts + s² - 2t - 2s + 1Right side:1 + t + s + t² - ts + s²Set equal:t² + 2ts + s² - 2t - 2s + 1 = 1 + t + s + t² - ts + s²Subtract 1 + t + s + t² - ts + s² from both sides:Left - Right:( t² + 2ts + s² - 2t - 2s + 1 ) - (1 + t + s + t² - ts + s² )= (0) + (2ts - (-ts)) + (0) + (-2t - t) + (-2s - s) + (1 -1)= 3ts - 3t - 3s = 0Thus:3ts - 3t - 3s = 0Divide both sides by 3:ts - t - s = 0Factor:t s - t - s = 0Add 1 to both sides:ts - t - s +1 =1Factor:(t -1)(s -1) =1Interesting. So (t -1)(s -1) =1.Earlier, from the perpendicularity condition, we had t = 2s +1.So, substitute t = 2s +1 into (t -1)(s -1)=1:(2s +1 -1)(s -1) =1(2s)(s -1)=12s(s -1)=12s² -2s -1=0Quadratic equation: 2s² -2s -1=0Solving for s:s = [2 ±√(4 +8)] /4 = [2 ±√12]/4 = [2 ±2√3]/4 = [1 ±√3]/2Since length cannot be negative, s must be positive. So:s = [1 +√3]/2 ≈ (1 +1.732)/2≈1.366, or s = [1 -√3]/2≈(1 -1.732)/2≈-0.366. Disregard negative.Thus, s = (1 +√3)/2Then, t = 2s +1 =2*(1 +√3)/2 +1= (1 +√3) +1=2 +√3So, t=2 +√3Therefore, BC = t = 2 +√3DA = s = (1 +√3)/2CD = t + s -1 = (2 +√3) + (1 +√3)/2 -1Compute that:= 2 +√3 + (1 +√3)/2 -1= (2 -1) + √3 + (1 +√3)/2= 1 + √3 + (1 +√3)/2Combine terms:= [1 + (1)/2] + [√3 + √3/2]= 3/2 + (3√3)/2= (3 + 3√3)/2Alternatively, CD = (3 + 3√3)/2Wait, but let me confirm with CD from coordinates:C is at (1 + t/2, (t√3)/2) = (1 + (2 +√3)/2, ( (2 +√3)√3 )/2 )Compute x-coordinate: 1 + (2 +√3)/2 = (2 + 2 +√3)/2 = (4 +√3)/2Wait, no:Wait, t = 2 +√3, so t/2 = (2 +√3)/2. Then, 1 + t/2 = 1 + (2 +√3)/2 = (2 + 2 +√3)/2 = (4 +√3)/2Similarly, y-coordinate: (t√3)/2 = (2 +√3)√3 /2 = (2√3 +3)/2Point D is at (-s/2, (s√3)/2) = ( - (1 +√3)/4, ( (1 +√3)√3 )/4 )Compute coordinates:-s/2 = - (1 +√3)/4(s√3)/2 = ( (1 +√3)√3 )/4 = (√3 +3)/4Thus, coordinates of D: (- (1 +√3)/4, (√3 +3)/4 )Coordinates of C: ( (4 +√3)/2 , (2√3 +3)/2 )So, CD distance squared:[ ( (4 +√3)/2 - (- (1 +√3)/4 ) ]² + [ ( (2√3 +3)/2 - (√3 +3)/4 ) ]²Compute x-coordinate difference:(4 +√3)/2 + (1 +√3)/4 = [ (8 + 2√3) +1 +√3 ] /4 = (9 + 3√3)/4Y-coordinate difference:(2√3 +3)/2 - (√3 +3)/4 = [ (4√3 +6) - (√3 +3) ] /4 = (3√3 +3)/4 = 3(√3 +1)/4Therefore, CD squared:[ (9 + 3√3)/4 ]² + [ 3(√3 +1)/4 ]²Compute first term:(9 +3√3)^2 /16 = (81 +54√3 +27)/16 = (108 +54√3)/16 = (54(2 +√3))/16 = (27(2 +√3))/8Second term:[ 3(√3 +1) ]² /16 =9( (√3 +1)^2 ) /16 =9(3 +2√3 +1)/16 =9(4 +2√3)/16 = (36 +18√3)/16 = (18(2 +√3))/16 = (9(2 +√3))/8Therefore, CD squared = (27(2 +√3) +9(2 +√3))/8 = (36(2 +√3))/8 = (9(2 +√3))/2Therefore, CD = √[ (9(2 +√3))/2 ] = (3)√[ (2 +√3)/2 ].Wait, but earlier we had CD = (3 +3√3)/2. Let me check if these are equal.Compute (3 +3√3)/2 squared:= (9 + 18√3 +27)/4 = (36 +18√3)/4 = (18 +9√3)/2Compare with CD squared from coordinates: (9(2 +√3))/2 = (18 +9√3)/2, which is the same. So, yes, CD = (3 +3√3)/2. Good.So now, the sides are:AB =1,BC =2 +√3,CD=(3 +3√3)/2,DA=(1 +√3)/2So perimeter is AB + BC + CD + DA =1 + (2 +√3) + (3 +3√3)/2 + (1 +√3)/2Let me compute term by term:1 + (2 +√3) = 3 +√3Then, adding (3 +3√3)/2 + (1 +√3)/2 = (3 +3√3 +1 +√3)/2 = (4 +4√3)/2 =2 +2√3So total perimeter: (3 +√3) + (2 +2√3) =5 +3√3So the perimeter is 5 +3√3.Wait, but let me check again:AB =1BC =2 +√3CD=(3 +3√3)/2 ≈(3 +5.196)/2≈4.098DA=(1 +√3)/2≈(1 +1.732)/2≈1.366Sum: 1 +2 +√3 + (3 +3√3)/2 + (1 +√3)/2Convert all to halves:2/2 + (4 +2√3)/2 + (3 +3√3)/2 + (1 +√3)/2Total: [2 +4 +2√3 +3 +3√3 +1 +√3]/2 = (10 +6√3)/2 =5 +3√3Yes, so perimeter is 5 +3√3.Wait, but let me check if this satisfies the property of tangential quadrilateral, i.e., AB + CD = BC + DA.AB + CD =1 + (3 +3√3)/2 = (2 +3 +3√3)/2 = (5 +3√3)/2BC + DA= (2 +√3) + (1 +√3)/2 = (4 +2√3 +1 +√3)/2 = (5 +3√3)/2Yes, equal. So the condition holds. Therefore, perimeter is 5 +3√3.But the problem says "calculate the perimeter of quadrilateral ABCD". So, the answer should be 5 +3√3, which is approximately 5 +5.196≈10.196. But we need an exact value.Alternatively, maybe there's a smarter way to get this without coordinates. Let me see.Alternatively, since the quadrilateral is tangential, so AB + CD = BC + DA. Also, given angles, maybe use the Law of Sines or Cosines on the triangles.Let me consider triangles ABD and BCD? Wait, maybe not. Wait, perhaps split the quadrilateral into two triangles by drawing diagonal BD.Wait, but with the given angles, maybe it's complex. Alternatively, using the fact that in a tangential quadrilateral, the lengths can be expressed in terms of the tangent segments, as we did earlier.We had AB = x + y =1,BC = y + z,CD = z + w,DA = w + x,and we also have angles given, so maybe we can use trigonometric relations in the triangles formed by the inradius.Wait, in a tangential quadrilateral, the inradius r can be related to the area by Area = r * s, where s is the semiperimeter. But since we don't know r or the area, perhaps not directly useful.Alternatively, maybe consider the triangles formed at each vertex with the inradius. For example, at vertex A, angle 120°, the inradius r is the height from the incenter to each side. The tangent lengths x and w (from A) can be related to r via trigonometric functions.Wait, perhaps for each angle, the tangent lengths can be expressed in terms of r and the angles.For instance, in triangle formed by the center of the incircle and the two tangent points adjacent to a vertex. Wait, maybe too vague.Alternatively, for each angle, the distance from the inradius to the vertex can be expressed in terms of r and the angle.Wait, let's consider vertex A with angle 120°. The inradius is r. The tangent lengths from A are both x (AP = AS =x). The distance from A to the incenter (I) can be expressed using trigonometry. The incenter is located at a distance of r/sin(α/2) from the vertex, where α is the angle at the vertex. So for angle 120°, the distance from A to I is r / sin(60°) = r / (√3/2) = 2r/√3.Similarly, the tangent lengths x can be related to r and the angle. In a tangential polygon, the tangent length from a vertex is equal to r / tan(α/2). Wait, is that true?Yes, in a tangential polygon, the length of the tangent from a vertex to the point of tangency is equal to r / tan(α/2), where α is the internal angle at that vertex.Wait, let me verify this formula.In a triangle, the tangent lengths are related to the inradius and angles. For a polygon, perhaps similar. Let's consider the tangent segment from vertex A. The angle at A is 120°, so half-angle is 60°. The tangent length x would be equal to r / tan(60°) = r / √3.Similarly, at vertex B, angle 120°, the tangent length y = r / tan(60°) = r / √3.At vertex C, angle 30°, so half-angle 15°, so tangent length z = r / tan(15°).At vertex D, angle 90°, half-angle 45°, so tangent length w = r / tan(45°) = r /1 = r.Therefore, the tangent lengths are:x = r / √3,y = r / √3,z = r / tan(15°),w = r.But in the quadrilateral, the sides are:AB = x + y = (r / √3) + (r / √3) = 2r / √3 =1 (given).So, 2r / √3 =1 => r = √3 / 2Then, compute other tangent lengths:z = r / tan(15°) = (√3 /2 ) / tan(15°)We know that tan(15°) = 2 -√3.Wait, tan(15°) = tan(45° -30°)= (1 - √3/3)/(1 + √3/3 )= (3 -√3)/(3 +√3)= multiply numerator and denominator by (3 -√3):= (9 -6√3 +3)/ (9 -3)= (12 -6√3)/6=2 -√3Yes, tan(15°)=2 -√3.Therefore, z = (√3 /2 ) / (2 -√3 )= (√3 /2 ) * (2 +√3)/ ( (2 -√3)(2 +√3) )= (√3 /2 )(2 +√3)/ (4 -3)= (√3 /2 )(2 +√3 )= (2√3 +3)/2Similarly, w = r = √3 /2Therefore, the sides:BC = y + z = (r / √3 ) + z = ( (√3 /2 ) / √3 ) + (2√3 +3)/2 = (1/2) + (2√3 +3)/2 = (1 +2√3 +3)/2 = (4 +2√3)/2 =2 +√3CD = z + w = (2√3 +3)/2 + √3 /2 = (2√3 +3 +√3)/2 = (3√3 +3)/2 =3(√3 +1)/2DA =w +x = √3 /2 + (√3 /2 ) / √3 = √3 /2 + (1/2 )= (√3 +1 )/2AB =1So perimeter is AB + BC + CD + DA=1 + (2 +√3) +3(√3 +1)/2 + (√3 +1)/2Compute term by term:1 +2 +√3 + [3(√3 +1) + (√3 +1)] /2=3 +√3 + [4(√3 +1)] /2=3 +√3 +2(√3 +1)=3 +√3 +2√3 +2=5 +3√3Same result as before. So perimeter is 5 +3√3. Therefore, the answer is boxed{5 + 3sqrt{3}}.Final AnswerThe perimeter of quadrilateral (ABCD) is boxed{5 + 3sqrt{3}}.</think>Given a tangential quadrilateral (ABCD) with (angle DAB = 120^circ), (angle ABC = 120^circ), (angle CDA = 90^circ), and (|AB| = 1), we need to find the perimeter.1. Properties of Tangential Quadrilaterals: - The sum of the lengths of opposite sides are equal: (AB + CD = BC + DA).2. Angle Calculation: - The sum of the internal angles of a quadrilateral is (360^circ). - Given angles: (angle DAB = 120^circ), (angle ABC = 120^circ), (angle CDA = 90^circ). - Remaining angle (angle BCD = 30^circ).3. Using Tangent Segments: - Denote the tangent lengths from each vertex to the points of tangency as (x, y, z, w). - For a tangential quadrilateral, the sides can be expressed as: - (AB = x + y = 1) - (BC = y + z) - (CD = z + w) - (DA = w + x)4. Trigonometric Relationships: - Using the inradius (r) and half-angle formulas, we find the tangent lengths: - (x = frac{r}{sqrt{3}}), (y = frac{r}{sqrt{3}}), (z = frac{r}{tan(15^circ)}), (w = r) - Solving for (r) using (AB = 1): - (2 cdot frac{r}{sqrt{3}} = 1 Rightarrow r = frac{sqrt{3}}{2})5. Calculating Tangent Lengths: - (z = frac{sqrt{3}/2}{2 - sqrt{3}} = frac{2sqrt{3} + 3}{2}) - (w = frac{sqrt{3}}{2})6. Side Lengths: - (BC = y + z = 2 + sqrt{3}) - (CD = z + w = frac{3(sqrt{3} + 1)}{2}) - (DA = w + x = frac{sqrt{3} + 1}{2})7. Perimeter Calculation: - Summing all sides: - (AB + BC + CD + DA = 1 + (2 + sqrt{3}) + frac{3(sqrt{3} + 1)}{2} + frac{sqrt{3} + 1}{2}) - Simplifying: (5 + 3sqrt{3})Thus, the perimeter of quadrilateral (ABCD) is (boxed{5 + 3sqrt{3}}).]

🔑:## First Solution.The fourth angle of the given quadrilateral is varangle B C D=30^{circ}. 1 pointLet E be the intersection of lines B C and D A.Let |D E|=x.The measures of angles varangle A B E and varangle B A E are 60^{circ}, so triangle A B E is equilateral, and thus |A E|=|B E|=|A B|=1.Triangle C D E is a right triangle, with acute angles varangle D C E=30^{circ} and varangle C E D=60^{circ}. Therefore, |C E|=2 x and |C D|=x sqrt{3}.Now, let's determine the lengths of all sides of quadrilateral A B C D:|A B|=1, quad|B C|=2 x-1, quad|C D|=x sqrt{3}, quad|D A|=x-1Since quadrilateral A B C D is tangential, we have |A B|+|C D|=|B C|+|D A|, i.e.,begin{gathered}1+x sqrt{3}=(2 x-1)+(x-1) (3-sqrt{3}) x=3end{gathered}Thus, x=frac{3}{3-sqrt{3}}=frac{3+sqrt{3}}{2}.The perimeter of quadrilateral A B C D isbegin{aligned}|A B|+|B C|+|C D|+|D A| & =1+(2 x-1)+x sqrt{3}+(x-1) & =(3+sqrt{3}) x-1 & =frac{1}{2}(3+sqrt{3})^{2}-1 & =5+3 sqrt{3}end{aligned}