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

Revision as of 19:23, 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

Denote the distance from the center to one vertex of the pentagon be $a$ . The two pentagons are similar, thus the new pentagon's area could be calculated using the similarity ratio. The similarity ratio could be expressed as the ratio between the distance from the center to one of the sides.

$\begin{align*} ratio &= \frac{\frac{1}{2}d}{\cos36^\circ d} \\ &= \frac{1}{2\cos36^\circ} \end{align*}$

Therefore, our area ratio is $\frac{1}{4\cos^236^\circ}$ . Due to the golden ratio, $\cos36^\circ=\frac{1+\sqrt5}{4}$ .

Our area would be

$\begin{align*} A&=(\sqrt5+1)\cdot\frac{1}{4(\frac{1+\sqrt5}{4})^2}\\ &=(\sqrt5+1)\cdot\frac{4}{(1+\sqrt5)^2}\\ &=\frac{4}{\sqrt5+1}\\ &=\boxed{\textbf{(B)}~\sqrt{5}-1} \end{align*}$

~Solution by eric-z

2023 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ~Solution by eric-z
+==See Also==
+{{AMC12 box|year=2023|ab=B|num-b=24|after=Last Problem}}
+{{MAA Notice}}

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

Revision as of 19:23, 15 November 2023

Problem

Solution

See Also