2023 AMC 12B Problems/Problem 25: Difference between revisions
No edit summary |
No edit summary |
||
| Line 4: | Line 4: | ||
<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> | <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>. | |||
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> | |||
\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> | |||
Revision as of 19:21, 15 November 2023
Problem
A regular pentagon with area 
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?
Solution
Denote the distance from the center to one vertex of the pentagon be 
.
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.
Therefore, our area ratio is 
. Due to the golden ratio, 
.
Our area would be
