2023 AMC 10B Problems/Problem 25
Revision as of 14:23, 15 November 2023 by Technodoggo (talk | contribs)
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?
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]](http://latex.artofproblemsolving.com/0/f/a/0fa13003b42a94bd7790e1a246ef96e98ea136e0.png)
Let 
and 
be the circumradius of the big and smaller pentagon. Let 
be the apothem of the smaller pentagon.
From the diagram:
\cos{36^\circ} &= \dfrac{a_s}{r_s} = \dfrac{\phi}{2} = \dfrac{\sqrt{5}+1}{4}\\
a_s &= \dfrac{r_b}{2}\\
\text{Area of small pentagon} &= (\dfrac{r_s}{r_b})^2 \text{Area of big pentagon}\\
&=(\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
~Technodoggo
