AMC10 2013 B

AMC10 2013 B · Q23

AMC10 2013 B · Q23. It mainly tests Triangles (properties), Similarity.

In triangle ABC, AB = 13, BC = 14, and CA = 15. Distinct points D, E, and F lie on segments BC, CA, and DE, respectively, such that AD ⊥BC, DE ⊥AC, and AF ⊥BF. The length of segment DF can be written as 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。不同点 D, E, F 分别位于线段 BC, CA 和 DE 上，使得 AD ⊥ BC, DE ⊥ AC, AF ⊥ BF。线段 DF 的长度可以写成 m/n，其中 m 和 n 互质正整数。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, \[ \frac{AE}{12}=\frac{DE}{9}=\frac{12}{15}, \] implying that $ED=\frac{36}{5}$ and $AE=\frac{48}{5}$. Because $\angle AFB=\angle ADB=90^\circ$, it follows that $ABDF$ is cyclic. Thus $\angle ABD+\angle AFD=180^\circ$ from which $\angle ABD=\angle AFE$. Therefore right triangles $ABD$ and $AFE$ are similar. Hence \[ \frac{FE}{5}=\frac{\frac{48}{5}}{12}, \] from which it follows that $FE=4$. Consequently $DF=DE-FE=\frac{36}{5}-4=\frac{16}{5}$.

答案（B）：将勾股定理应用于直角三角形 $ABD$ 和 $ACD$，得到 $AB^2-BD^2=AD^2=AC^2-CD^2$；即 $13^2-BD^2=15^2-(14-BD)^2$，由此可得 $BD=5$、$CD=9$、$AD=12$。因为三角形 $AED$ 与 $ADC$ 相似， \[ \frac{AE}{12}=\frac{DE}{9}=\frac{12}{15}, \] 从而 $ED=\frac{36}{5}$ 且 $AE=\frac{48}{5}$。又因为 $\angle AFB=\angle ADB=90^\circ$，可知 $ABDF$ 四点共圆。因此 $\angle ABD+\angle AFD=180^\circ$，从而 $\angle ABD=\angle AFE$。所以直角三角形 $ABD$ 与 $AFE$ 相似。于是 \[ \frac{FE}{5}=\frac{\frac{48}{5}}{12}, \] 进而 $FE=4$。因此 $DF=DE-FE=\frac{36}{5}-4=\frac{16}{5}$。

Topics