AMC10 2015 B

AMC10 2015 B · Q19

AMC10 2015 B · Q19. It mainly tests Triangles (properties), Circle theorems.

In $\triangle ABC$, $\angle C = 90^\circ$ and $AB = 12$. Squares $ABXY$ and $ACWZ$ are constructed outside of the triangle. The points $X, Y, Z$, and $W$ lie on a circle. What is the perimeter of the triangle?

在 $\triangle ABC$ 中，$\angle C = 90^\circ$ 且 $AB = 12$。在三角形外构造正方形 $ABXY$ 和 $ACWZ$。点 $X,Y,Z,W$ 位于同一个圆上。三角形的周长是多少？

(A) $12 + 9\sqrt{3}$ $12 + 9\sqrt{3}$

(B) $18 + 6\sqrt{3}$ $18 + 6\sqrt{3}$

(D) 30 30

(E) 32 32

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $O$ be the center of the circle on which $X$, $Y$, $Z$, and $W$ lie. Then $O$ lies on the perpendicular bisectors of segments $\overline{XY}$ and $\overline{ZW}$, and $OX = OW$. Note that segments $\overline{XY}$ and $\overline{AB}$ have the same perpendicular bisector and segments $\overline{ZW}$ and $\overline{AC}$ have the same perpendicular bisector, from which it follows that $O$ lies on the perpendicular bisectors of segments $\overline{AB}$ and $\overline{AC}$; that is, $O$ is the circumcenter of $\triangle ABC$. Because $\angle C = 90^\circ$, $O$ is the midpoint of hypotenuse $\overline{AB}$. Let $a = \frac{1}{2}BC$ and $b = \frac{1}{2}CA$. Then $a^2 + b^2 = 6^2$ and $12^2 + 6^2 = OX^2 = OW^2 = b^2 + (a + 2b)^2$. Solving these two equations simultaneously gives $a = b = 3\sqrt{2}$. Thus the perimeter of $\triangle ABC$ is $12 + 2a + 2b = 12 + 12\sqrt{2}$.

答案（C）：设$O$为点$X$、$Y$、$Z$、$W$所在圆的圆心。则$O$在弦$\overline{XY}$与$\overline{ZW}$的垂直平分线上，且$OX = OW$。注意到线段$\overline{XY}$与$\overline{AB}$有相同的垂直平分线，线段$\overline{ZW}$与$\overline{AC}$也有相同的垂直平分线，因此$O$也在线段$\overline{AB}$与$\overline{AC}$的垂直平分线上；也就是说，$O$是$\triangle ABC$的外心。由于$\angle C = 90^\circ$，$O$是斜边$\overline{AB}$的中点。令$a = \frac{1}{2}BC$，$b = \frac{1}{2}CA$。则$a^2 + b^2 = 6^2$，并且$12^2 + 6^2 = OX^2 = OW^2 = b^2 + (a + 2b)^2$。联立这两个方程可得$a = b = 3\sqrt{2}$。因此$\triangle ABC$的周长为$12 + 2a + 2b = 12 + 12\sqrt{2}$。

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