AMC12 2013 B

AMC12 2013 B · Q19

AMC12 2013 B · Q19. It mainly tests Triangles (properties), Pythagorean theorem.

In triangle $ABC$, $AB=13$, $BC=14$, and $CA=15$. Distinct points $D$, $E$, and $F$ lie on segments $\overline{BC}$, $\overline{CA}$, and $\overline{DE}$, respectively, such that $AD\perp BC$, $DE\perp AC$, and $AF\perp BF$. The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

在三角形 $ABC$ 中，$AB=13$，$BC=14$，$CA=15$。不同的点 $D$、$E$、$F$ 分别在线段 $\overline{BC}$、$\overline{CA}$、$\overline{DE}$ 上，并满足 $AD\perp BC$，$DE\perp AC$，且 $AF\perp BF$。线段 $\overline{DF}$ 的长度可以表示为 $\frac{m}{n}$，其中 $m$ 与 $n$ 为互质的正整数。求 $m+n$。

(A) 18 18

(B) 21 21

(D) 27 27

(E) 30 30

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The Pythagorean Theorem applied to right triangles $ABD$ and $ACD$ gives $AB^2-BD^2=AD^2=AC^2-CD^2$; that is, $13^2-BD^2=15^2-(14-BD)^2$, from which it follows that $BD=5$, $CD=9$, and $AD=12$. Because triangles $AED$ and $ADC$ are similar, $\dfrac{AE}{12}=\dfrac{DE}{9}=\dfrac{12}{15}$, implying that $ED=\dfrac{36}{5}$ and $AE=\dfrac{48}{5}$. Because $\angle AFB=\angle ADB=90^\circ$, it follows that $ABDF$ is cyclic. Thus $\angle ABLD+\angle AFD=180^\circ$ from which $\angle ABD=\angle AFE$. Therefore right triangles $ABD$ and $AFE$ are similar. Hence $\dfrac{FE}{5}=\dfrac{\frac{48}{5}}{12}$, from which it follows that $FE=4$. Consequently $DF=DE-FE=\dfrac{36}{5}-4=\dfrac{16}{5}$.

答案（B）：将勾股定理应用于直角三角形 $ABD$ 和 $ACD$，得 $AB^2-BD^2=AD^2=AC^2-CD^2$；即 $13^2-BD^2=15^2-(14-BD)^2$，由此可得 $BD=5$，$CD=9$，且 $AD=12$。因为三角形 $AED$ 与 $ADC$ 相似， $\dfrac{AE}{12}=\dfrac{DE}{9}=\dfrac{12}{15}$，从而 $ED=\dfrac{36}{5}$，$AE=\dfrac{48}{5}$。因为 $\angle AFB=\angle ADB=90^\circ$，可知 $ABDF$ 为圆内接四边形。因此 $\angle ABLD+\angle AFD=180^\circ$，从而 $\angle ABD=\angle AFE$。于是直角三角形 $ABD$ 与 $AFE$ 相似。故 $\dfrac{FE}{5}=\dfrac{\frac{48}{5}}{12}$，由此得到 $FE=4$。因此 $DF=DE-FE=\dfrac{36}{5}-4=\dfrac{16}{5}$。

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