2023 AMC 10B Problem 25


A regular pentagon with area 1+51+\sqrt5 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?

(A) 45(B) 51(C) 835(D) 5+12(E) 2+53\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}


Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.

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Category: AMC 10B
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