AMC12 2015 B

AMC12 2015 B · Q19

AMC12 2015 B · Q19. It mainly tests Circle theorems, Coordinate geometry.

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 $AB=12$。在三角形外部构造正方形 $ABXY$ 和 $ACWZ$。点 $X,Y,Z,W$ 位于一个圆上。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 $XY$ and $ZW$, and $OX = OW$. Note that segments $XY$ and $AB$ have the same perpendicular bisector and segments $ZW$ and $AC$ have the same perpendicular bisector, from which it follows that $O$ lies on the perpendicular bisectors of segments $AB$ and $AC$; that is, $O$ is the circumcenter of $\triangle ABC$. Because $\angle C = 90^\circ$, $O$ is the midpoint of hypotenuse $AB$. 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$X、Y、Z、W$所在圆的圆心。则$O$在弦段$XY$与$ZW$的垂直平分线上，且$OX=OW$。注意到线段$XY$与$AB$有相同的垂直平分线，线段$ZW$与$AC$也有相同的垂直平分线，因此$O$在$AB$与$AC$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$AB$的中点。令$a=\frac{1}{2}BC$、$b=\frac{1}{2}CA$。则$a^2+b^2=6^2$，且$12^2+6^2=OX^2=OW^2=b^2+(a+2b)^2$。联立解得$a=b=3\sqrt{2}$。因此$\triangle ABC$的周长为$12+2a+2b=12+12\sqrt{2}$。

Topics