AMC12 2023 B

AMC12 2023 B · Q25

AMC12 2023 B · Q25. It mainly tests Polygons, Geometry misc.

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?

一个面积为 $\sqrt{5}+1$ 的正五边形印在纸上并剪下。五边形的五个顶点被折向五边形中心，形成一个更小的五边形。新五边形的面积是多少？

(A) 4-\sqrt{5} 4-\sqrt{5}

(B) \sqrt{5}-1 \sqrt{5}-1

(D) \frac{\sqrt{5}+1}{2} \frac{\sqrt{5}+1}{2}

(E) \frac{2+\sqrt{5}}{3} \frac{2+\sqrt{5}}{3}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Since $A$ is folded onto $O$, $AM = MO$ where $M$ 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 $\left(\frac{OM}{ON}\right)^{2} = \left(\frac{\frac{OA}{2}}{OA\sin (\angle OAE)}\right)^{2} = \frac{1}{4\sin^{2}54}$ Remember that $\sin54 = \frac{1+\sqrt5}{4}$. $\cos54 = \sin36$ $4\cos^{3}18-3\cos18 = 2\sin18\cos18$ $4(1-\sin^{2}18)-3-2\sin18=0$ $4\sin^{2}18+2\sin18-1=0$ $\sin18 = \frac{-1+\sqrt5}{4}$ $\sin54 = \cos36 = 1-2\sin^{2}18 = \frac{1+\sqrt5}{4}$ $\sin^{2}54 =\frac{3+\sqrt5}{8}$ Let the inner pentagon be $Z$. $[Z] = \frac{1}{4\sin^{2}54}[ABCDE]$ $= \frac{2(1+\sqrt5)}{3+\sqrt5}$ $= \sqrt5-1$ So the answer is $\boxed{B}$

因为 $A$ 被折到 $O$，$AM = MO$，其中 $M$ 是 $AO$ 与折痕之间的交点。注意内五边形是正的，因此与原五边形相似，由于对称性。由于相似性，内五边形面积与外五边形面积之比可表示为 $\left(\frac{OM}{ON}\right)^{2} = \left(\frac{\frac{OA}{2}}{OA\sin (\angle OAE)}\right)^{2} = \frac{1}{4\sin^{2}54}$ 记住 $\sin54 = \frac{1+\sqrt5}{4}$。 $\cos54 = \sin36$ $4\cos^{3}18-3\cos18 = 2\sin18\cos18$ $4(1-\sin^{2}18)-3-2\sin18=0$ $4\sin^{2}18+2\sin18-1=0$ $\sin18 = \frac{-1+\sqrt5}{4}$ $\sin54 = \cos36 = 1-2\sin^{2}18 = \frac{1+\sqrt5}{4}$ $\sin^{2}54 =\frac{3+\sqrt5}{8}$ 设内五边形为 $Z$。 $[Z] = \frac{1}{4\sin^{2}54}[ABCDE]$ $= \frac{2(1+\sqrt5)}{3+\sqrt5}$ $= \sqrt5-1$ 因此答案为 $\boxed{B}$

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