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Difference between revisions of "2023 AMC 10B Problems/Problem 25"

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\end{align*}</cmath>
 
\end{align*}</cmath>
 
<cmath>\boxed{\textbf{(B) }\sqrt{5}-1}</cmath>
 
<cmath>\boxed{\textbf{(B) }\sqrt{5}-1}</cmath>
−
~<math>\textbf{Techno}\textcolor{red}{doggo}</math>
+
~<math>T\raisebox{-0.5ex}{E}X\textbf{no}\textcolor{red}{doggo}</math> (Technodoggo)
  
 
==Solution 2==
 
==Solution 2==
 
Interestingly, we find that the pentagon we need is the one that is represented by the intersection of the diagonals. Through similar triangles and the golden ratio, we find that the side length ratio of the two pentagons is <math>\frac{\sqrt{5}-1}{2}</math> Thus, the answer is <math>\sqrt{5}+1 \cdot (\frac{\sqrt{5}-1}{2})^2 = \sqrt{5}-1</math>. <math>\boxed{\text{B}}</math>
 
Interestingly, we find that the pentagon we need is the one that is represented by the intersection of the diagonals. Through similar triangles and the golden ratio, we find that the side length ratio of the two pentagons is <math>\frac{\sqrt{5}-1}{2}</math> Thus, the answer is <math>\sqrt{5}+1 \cdot (\frac{\sqrt{5}-1}{2})^2 = \sqrt{5}-1</math>. <math>\boxed{\text{B}}</math>
 
~andliu766
 
~andliu766

Revision as of 17:57, 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*} \[\boxed{\textbf{(B) }\sqrt{5}-1}\] ~$T\raisebox{-0.5ex}{E}X\textbf{no}\textcolor{red}{doggo}$ (Technodoggo)

Solution 2

Interestingly, we find that the pentagon we need is the one that is represented by the intersection of the diagonals. Through similar triangles and the golden ratio, we find that the side length ratio of the two pentagons is $\frac{\sqrt{5}-1}{2}$ Thus, the answer is $\sqrt{5}+1 \cdot (\frac{\sqrt{5}-1}{2})^2 = \sqrt{5}-1$. $\boxed{\text{B}}$ ~andliu766

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