AMC12 2022 B

AMC12 2022 B · Q10

AMC12 2022 B · Q10. It mainly tests Triangles (properties), Distance / midpoint.

Regular hexagon $ABCDEF$ has side length $2$. Let $G$ be the midpoint of $\overline{AB}$, and let $H$ be the midpoint of $\overline{DE}$. What is the perimeter of $GCHF$?

正六边形$ABCDEF$边长为$2$。设$G$为$\overline{AB}$的中点，$H$为$\overline{DE}$的中点。$GCHF$的周长是多少？

(A) 4\sqrt3 4\sqrt3

(B) 8 8

(D) 4\sqrt7 4\sqrt7

(E) 12 12

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let the center of the hexagon be $O$. $\triangle AOB$, $\triangle BOC$, $\triangle COD$, $\triangle DOE$, $\triangle EOF$, and $\triangle FOA$ are all equilateral triangles with side length $2$. Thus, $CO = 2$, and $GO = \sqrt{AO^2 - AG^2} = \sqrt{3}$. By symmetry, $\angle COG = 90^{\circ}$. Thus, by the Pythagorean theorem, $CG = \sqrt{2^2 + \sqrt{3}^2} = \sqrt{7}$. Because $CO = OF$ and $GO = OH$, $CG = HC = FH = GF = \sqrt{7}$. Thus, the solution to our problem is $\sqrt{7} + \sqrt{7} + \sqrt{7} + \sqrt{7} = \boxed{\textbf{(D)}\ 4\sqrt7}$.

设六边形的中心为$O$。$\triangle AOB$、$\triangle BOC$、$\triangle COD$、$\triangle DOE$、$\triangle EOF$和$\triangle FOA$都是边长为2的等边三角形。因此，$CO = 2$，$GO = \sqrt{AO^2 - AG^2} = \sqrt{3}$。由对称性，$\angle COG = 90^{\circ}$。因此，由勾股定理，$CG = \sqrt{2^2 + \sqrt{3}^2} = \sqrt{7}$。因为$CO = OF$且$GO = OH$，$CG = HC = FH = GF = \sqrt{7}$。因此，我们的问题的解是$\sqrt{7} + \sqrt{7} + \sqrt{7} + \sqrt{7} = \boxed{\textbf{(D)}\ 4\sqrt7}$。

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