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| − | ==Problem==
| + | #redirect[[2023 AMC 12B Problems/Problem 25]] |
| − | A regular pentagon with area 1+\sqrt(5) is printed on paper and cut out. All five vertices are folded to the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?
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| − | | |
| − | | |
| − | ==Solution 1==
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| − | | |
| − | <asy>
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| − | unitsize(5cm);
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| − | | |
| − | // Define the vertices of the pentagons
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| − | pair A, B, C, D, E;
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| − | pair F, G, H, I, J;
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| − | | |
| − | // Calculate the vertices of the larger pentagon
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| − | A = dir(90);
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| − | B = dir(90 - 72);
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| − | C = dir(90 - 2*72);
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| − | D = dir(90 - 3*72);
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| − | E = dir(90 - 4*72);
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| − | | |
| − | // Draw the larger pentagon
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| − | draw(A--B--C--D--E--cycle);
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| − | | |
| − | pair O = (A+B+C+D+E)/5;
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| − | pair AA,OO;
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| − | real gap = 0.02;
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| − | AA = A+(0,0);
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| − | OO = O+(0,0);
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| − | | |
| − | draw(AA--OO, blue);
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| − | | |
| − | pair OOO, OAO;
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| − | OOO = O+(gap,0);
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| − | OAO = (O+A)/2 + (gap,0);
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| − | | |
| − | draw(OOO--OAO,green);
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| − | dot(O);
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| − | dot((O+A)/2);
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| − | | |
| − | label("$r_b$", (O+A)*.7, E,blue);
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| − | label("$a_s$", (O+A)*.2 +(0+0.18,0.05), E,green);
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| − | label("$r_s$", O+(-0.175,0.2), E,pink);
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| − | | |
| − | | |
| − | real scaleFactor = 1/1.618; // Adjust this value as needed
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| − | // Rotate the smaller pentagon by 180 degrees
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| − | F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180);
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| − | G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180);
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| − | H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180);
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| − | I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180);
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| − | J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180);
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| − | | |
| − | // Draw the smaller pentagon
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| − | | |
| − | draw(F--G--H--I--J--cycle,red);
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| − | | |
| − | draw(arc(O,(H+I)*.5*.6,H*.6));
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| − | label("$36^\circ$",O+(+0.05,0.15),NW);
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| − | draw(O--H,pink);
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| − | </asy>
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| − | | |
| − | Let <math>r_b</math> and <math>r_s</math> be the circumradius of the big and smaller pentagon. Let <math>a_s</math> be the apothem of the smaller pentagon.
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| − | | |
| − | From the diagram:
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| − | \cos{36^\circ} &= \dfrac{a_s}{r_s} = \dfrac{\phi}{2} = \dfrac{\sqrt{5}+1}{4}\\
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| − | a_s &= \dfrac{r_b}{2}\\
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| − | \text{Area of small pentagon} &= (\dfrac{r_s}{r_b})^2 \text{Area of big pentagon}\\
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| − | &=(\dfrac{a_s}{\cos{36^\circ} r_b})^2 (1+\sqrt{5})\\
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| − | &=(\dfrac{r_b}{\dfrac{\phi}{2} r_b})^2 (1+\sqrt{5})\\
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| − | &=(\dfrac{1}{2 \dfrac{\phi}{2}})^2 (1+\sqrt{5})\\
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| − | &=(\dfrac{2}{\sqrt{5}+1})^2 (1+\sqrt{5})\\
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| − | &=\dfrac{4}{\sqrt{5}+1} \\
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| − | &=\dfrac{4(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)} \\
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| − | &=\sqrt{5}-1
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| − | | |
| − | ~Technodoggo
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