AMC12 2002 B

AMC12 2002 B · Q20

AMC12 2002 B · Q20. It mainly tests Triangles (properties), Coordinate geometry.

Let $\triangle XOY$ be a right-angled triangle with $m\angle XOY = 90^{\circ}$. Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$, respectively. Given that $XN = 19$ and $YM = 22$, find $XY$.

设$\triangle XOY$为直角三角形，且$m\angle XOY=90^{\circ}$。设$M$与$N$分别为直角边$OX$与$OY$的中点。已知$XN=19$且$YM=22$，求$XY$。

(A) 24 24

(B) 26 26

(D) 30 30

(E) 32 32

Answer

Correct choice: (B)

正确答案：（B）

Solution

Error creating thumbnail: Unable to save thumbnail to destination Let $OM = x$, $ON = y$. By the Pythagorean Theorem on $\triangle XON, MOY$ respectively, \begin{align*} (2x)^2 + y^2 &= 19^2\\ x^2 + (2y)^2 &= 22^2\end{align*} Summing these gives $5x^2 + 5y^2 = 845 \Longrightarrow x^2 + y^2 = 169$. By the Pythagorean Theorem again, we have \[(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = \sqrt{4(169)} = \sqrt{676} = \boxed{\mathrm{(B)}\ 26}.\] Alternatively, we could note that since we found $x^2 + y^2 = 169$, segment $MN=13$. Right triangles $\triangle MON$ and $\triangle XOY$ are similar by Leg-Leg with a ratio of $\frac{1}{2}$, so $XY=2(MN)=\boxed{\mathrm{(B)}\ 26}$.

Error creating thumbnail: Unable to save thumbnail to destination 设$OM=x$，$ON=y$。分别对$\triangle XON,\ MOY$使用勾股定理， \begin{align*} (2x)^2+y^2&=19^2\\ x^2+(2y)^2&=22^2\end{align*} 将两式相加得$5x^2+5y^2=845 \Longrightarrow x^2+y^2=169$。再用一次勾股定理，有 \[(2x)^2+(2y)^2=XY^2 \Longrightarrow XY=\sqrt{4(x^2+y^2)}=\sqrt{4(169)}=\sqrt{676}=\boxed{\mathrm{(B)}\ 26}.\] 或者注意到，由于已求得$x^2+y^2=169$，线段$MN=13$。直角三角形$\triangle MON$与$\triangle XOY$由直角边-直角边相似，相似比为$\frac{1}{2}$，所以$XY=2(MN)=\boxed{\mathrm{(B)}\ 26}$。

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