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Difference between revisions of "2023 AMC 10B Problems/Problem 25"
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==Problem==
 
==Problem==
−
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?
+
A regular pentagon with area <math>1+\sqrt5</math> 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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</asy>
 
</asy>
  
−
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.
+
Let <math>r_b</math> and <math>r_s</math> be the circumradius of the big and small pentagon, respectively. Let <math>a_s</math> be the apothem of the smaller pentagon and <math>A_s</math> and <math>A_b</math> be the areas of the smaller and larger pentagon, respectively.  
  
 
From the diagram:
 
From the diagram:
 +
<cmath>\begin{align*}
 
     \cos{36^\circ} &= \dfrac{a_s}{r_s} = \dfrac{\phi}{2} = \dfrac{\sqrt{5}+1}{4}\\
 
     \cos{36^\circ} &= \dfrac{a_s}{r_s} = \dfrac{\phi}{2} = \dfrac{\sqrt{5}+1}{4}\\
 
     a_s &= \dfrac{r_b}{2}\\
 
     a_s &= \dfrac{r_b}{2}\\
−
     \text{Area of small pentagon} &= (\dfrac{r_s}{r_b})^2 \text{Area of big pentagon}\\
+
     A_s &= (\dfrac{r_s}{r_b})^2A_b\\
 
     &=(\dfrac{a_s}{\cos{36^\circ} r_b})^2 (1+\sqrt{5})\\
 
     &=(\dfrac{a_s}{\cos{36^\circ} r_b})^2 (1+\sqrt{5})\\
 
     &=(\dfrac{r_b}{\dfrac{\phi}{2} r_b})^2 (1+\sqrt{5})\\
 
     &=(\dfrac{r_b}{\dfrac{\phi}{2} r_b})^2 (1+\sqrt{5})\\
Line 73: Line 74:
 
     &=\dfrac{4(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)} \\
 
     &=\dfrac{4(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)} \\
 
     &=\sqrt{5}-1
 
     &=\sqrt{5}-1
−
 
+
\end{align*}</cmath>
 
~Technodoggo
 
~Technodoggo

Revision as of 13:27, 15 November 2023

Problem

A regular pentagon with area $1+\sqrt5$ 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?


Solution 1

[asy] unitsize(5cm); // Define the vertices of the pentagons pair A, B, C, D, E; pair F, G, H, I, J; // Calculate the vertices of the larger pentagon A = dir(90); B = dir(90 - 72); C = dir(90 - 2*72); D = dir(90 - 3*72); E = dir(90 - 4*72); // Draw the larger pentagon draw(A--B--C--D--E--cycle); pair O = (A+B+C+D+E)/5; pair AA,OO; real gap = 0.02; AA = A+(0,0); OO = O+(0,0); draw(AA--OO, blue); pair OOO, OAO; OOO = O+(gap,0); OAO = (O+A)/2 + (gap,0); draw(OOO--OAO,green); dot(O); dot((O+A)/2); label("$r_b$", (O+A)*.7, E,blue); label("$a_s$", (O+A)*.2 +(0+0.18,0.05), E,green); label("$r_s$", O+(-0.175,0.2), E,pink); real scaleFactor = 1/1.618; // Adjust this value as needed // Rotate the smaller pentagon by 180 degrees F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180); G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180); H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180); I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180); J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180); // Draw the smaller pentagon draw(F--G--H--I--J--cycle,red); draw(arc(O,(H+I)*.5*.6,H*.6)); label("$36^\circ$",O+(+0.05,0.15),NW); draw(O--H,pink); [/asy]

Let $r_b$ and $r_s$ be the circumradius of the big and small pentagon, respectively. Let $a_s$ be the apothem of the smaller pentagon and $A_s$ and $A_b$ be the areas of the smaller and larger pentagon, respectively.

From the diagram: \begin{align*} \cos{36^\circ} &= \dfrac{a_s}{r_s} = \dfrac{\phi}{2} = \dfrac{\sqrt{5}+1}{4}\\ a_s &= \dfrac{r_b}{2}\\ A_s &= (\dfrac{r_s}{r_b})^2A_b\\ &=(\dfrac{a_s}{\cos{36^\circ} r_b})^2 (1+\sqrt{5})\\ &=(\dfrac{r_b}{\dfrac{\phi}{2} r_b})^2 (1+\sqrt{5})\\ &=(\dfrac{1}{2 \dfrac{\phi}{2}})^2 (1+\sqrt{5})\\ &=(\dfrac{2}{\sqrt{5}+1})^2 (1+\sqrt{5})\\ &=\dfrac{4}{\sqrt{5}+1} \\ &=\dfrac{4(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)} \\ &=\sqrt{5}-1 \end{align*} ~Technodoggo

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