Difference between revisions of "2023 AMC 12B Problems/Problem 25"

Revision as of 20:13, 15 November 2023

Problem

A regular pentagon with area $\sqrt{5}+1$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?

$\textbf{(A)}~4-\sqrt{5}\qquad\textbf{(B)}~\sqrt{5}-1\qquad\textbf{(C)}~8-3\sqrt{5}\qquad\textbf{(D)}~\frac{\sqrt{5}+1}{2}\qquad\textbf{(E)}~\frac{2+\sqrt{5}}{3}$

Solution 1

Let the original pentagon be $ABCDE$ centered at $O$ . The dashed lines represent the fold lines. WLOG, let's focus on vertex $A$ .

Since $A$ is folded onto $O$ , $AN = NO$ where $N$ is the intersection of $AO$ and the creaseline between $A$ and $O$ . Note that the inner pentagon is regular, and therefore similar to the original pentagon, due to symmetry.

Because of their similarity, the ratio of the inner pentagon's area to that of the outer pentagon can be represented by

$(\frac{ON}{OM})^{2} = (\frac{\frac{OA}{2}}{OA\sin (\angle OAE)})^{2} = \frac{1}{4\sin^{2}54}$

Option 1: Knowledge

Remember that $\sin54 = \frac{1+\sqrt5}{4}$ .

Option 2: Angle Identities

$\cos54 = \sin36$

$4\cos^{3}18-3\cos18 = 2\sin18\cos18$

$4(1-\sin^{2}18)-3-2\sin18=0$

$4\sin^{2}18+2\sin18-1=0$

$\sin18 = \frac{-1+\sqrt5}{4}$

$\sin54 = \cos36 = 1-2\sin^{2}18 = \frac{1+\sqrt5}{4}$

$\sin^{2}54 =\frac{3+\sqrt5}{8}$

Let the inner pentagon be $Z$ .

$[Z] = \frac{1}{4\sin^{2}54}[ABCDE]$

-Dissmo

@@ Line 25: / Line 25: @@
 <math>\cos54 = \sin36</math>
 <math>4\cos^{3}18-3\cos18 = 2\sin18\cos18</math>
 <math>4(1-\sin^{2}18)-3-2\sin18=0</math>
 <math>4\sin^{2}18+2\sin18-1=0</math>
 <math>\sin18 = \frac{-1+\sqrt5}{4}</math>
 <math>\sin54 = \cos36 = 1-2\sin^{2}18 = \frac{1+\sqrt5}{4}</math>
 <math>\sin^{2}54 =\frac{3+\sqrt5}{8}</math>

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Difference between revisions of "2023 AMC 12B Problems/Problem 25"

Revision as of 20:13, 15 November 2023

Problem

Solution 1