AMC10 2021 A

AMC10 2021 A · Q21

AMC10 2021 A · Q21. It mainly tests Triangles (properties), Circle theorems.

Let $ABCDEF$ be an equiangular hexagon. The lines $AB, CD,$ and $EF$ determine a triangle with area $192\sqrt{3}$, and the lines $BC, DE,$ and $FA$ determine a triangle with area $324\sqrt{3}$. The perimeter of hexagon $ABCDEF$ can be expressed as $m +n\sqrt{p}$, where $m, n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m + n + p$?

设$ABCDEF$是一个等角六边形。直线$AB$、$CD$和$EF$围成的三角形面积为$192\sqrt{3}$，直线$BC$、$DE$和$FA$围成的三角形面积为$324\sqrt{3}$。六边形$ABCDEF$的周长可表示为$m +n\sqrt{p}$，其中$m$、$n$和$p$是正整数，且$p$不被任何质数的平方整除。求$m + n + p$。

(A) 47 47

(B) 52 52

(D) 58 58

(E) 63 63

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let $P,Q,R,X,Y,$ and $Z$ be the intersections $\overleftrightarrow{AB}\cap\overleftrightarrow{CD},\overleftrightarrow{CD}\cap\overleftrightarrow{EF},\overleftrightarrow{EF}\cap\overleftrightarrow{AB},\overleftrightarrow{BC}\cap\overleftrightarrow{DE},\overleftrightarrow{DE}\cap\overleftrightarrow{FA},$ and $\overleftrightarrow{FA}\cap\overleftrightarrow{BC},$ respectively. The sum of the interior angles of any hexagon is $720^\circ.$ Since hexagon $ABCDEF$ is equiangular, each of its interior angles is $720^\circ\div6=120^\circ.$ By angle chasing, we conclude that the interior angles of $\triangle PBC,\triangle QDE,\triangle RFA,\triangle XCD,\triangle YEF,$ and $\triangle ZAB$ are all $60^\circ.$ Therefore, these triangles are all equilateral triangles, from which $\triangle PQR$ and $\triangle XYZ$ are both equilateral triangles. We are given that \begin{alignat*}{8} [PQR]&=\frac{\sqrt{3}}{4}\cdot PQ^2&&=192\sqrt3, \\ [XYZ]&=\frac{\sqrt{3}}{4}\cdot YZ^2&&=324\sqrt3, \end{alignat*} so we get $PQ=16\sqrt3$ and $YZ=36,$ respectively. By equilateral triangles and segment addition, we find the perimeter of hexagon $ABCDEF:$ \begin{align*} AB+BC+CD+DE+EF+FA&=AZ+PC+CD+DQ+YF+FA \\ &=(YF+FA+AZ)+(PC+CD+DQ) \\ &=YZ+PQ \\ &=36+16\sqrt{3}. \end{align*} Finally, the answer is $36+16+3=\boxed{\textbf{(C)} ~55}.$

设$P,Q,R,X,Y$和$Z$分别是 $\overleftrightarrow{AB}\cap\overleftrightarrow{CD}$、$\overleftrightarrow{CD}\cap\overleftrightarrow{EF}$、$\overleftrightarrow{EF}\cap\overleftrightarrow{AB}$、$\overleftrightarrow{BC}\cap\overleftrightarrow{DE}$、$\overleftrightarrow{DE}\cap\overleftrightarrow{FA}$ 和 $\overleftrightarrow{FA}\cap\overleftrightarrow{BC}$。任意六边形的内角和为$720^\circ$。由于六边形$ABCDEF$是等角的，其每个内角为$720^\circ\div6=120^\circ$。通过角度追踪，我们得出$\triangle PBC$、$\triangle QDE$、$\triangle RFA$、$\triangle XCD$、$\triangle YEF$ 和 $\triangle ZAB$的内角均为$60^\circ$。因此，这些三角形均为等边三角形，由此$\triangle PQR$ 和 $\triangle XYZ$均为等边三角形。已知 \begin{alignat*}{8} [PQR]&=\frac{\sqrt{3}}{4}\cdot PQ^2&&=192\sqrt3, \\ [XYZ]&=\frac{\sqrt{3}}{4}\cdot YZ^2&&=324\sqrt3, \end{alignat*} 从而得到$PQ=16\sqrt3$ 和 $YZ=36$。由等边三角形和线段加法，我们求得六边形$ABCDEF$的周长： \begin{align*} AB+BC+CD+DE+EF+FA&=AZ+PC+CD+DQ+YF+FA \\ &=(YF+FA+AZ)+(PC+CD+DQ) \\ &=YZ+PQ \\ &=36+16\sqrt{3}. \end{align*} 最后，答案为$36+16+3=\boxed{\textbf{(C)} ~55}$。

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