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

Revision as of 20:37, 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?

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

@@ Line 5: / Line 5: @@
 <math>\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}</math>
-==Solution==
-Denote the distance from the center to one vertex of the pentagon be <math>a</math>.
+==Solution 1==
-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.
-<cmath>
+[[File:Pentagon_2023_12B_Q25_dissmo.png]]
-\begin{align*}
-ratio &= \frac{\frac{1}{2}d}{\cos36^\circ d} \\
-&= \frac{1}{2\cos36^\circ}
-\end{align*}
-</cmath>
-Therefore, our area ratio is <math>\frac{1}{4\cos^236^\circ}</math>. Due to the golden ratio, <math>\cos36^\circ=\frac{1+\sqrt5}{4}</math>.
-Our area would be
-<cmath>
-\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*}
-</cmath>
-~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 20:37, 15 November 2023

Problem

Solution 1