AMC12 2025 A

AMC12 2025 A · Q16

AMC12 2025 A · Q16. It mainly tests Triangles (properties), Pythagorean theorem.

Triangle $\triangle ABC$ has side lengths $AB = 80$, $BC = 45$, and $AC = 75$. The bisector of $\angle B$ and the altitude to side $\overline{AB}$ intersect at point $P$. What is $BP$?

三角形 $\triangle ABC$ 的边长 $AB = 80$，$BC = 45$，$AC = 75$。$\angle B$ 的角平分线与侧边 $\overline{AB}$ 的高线交于点 $P$。$BP$ 等于多少？

(A) 18 18

(B) 19 19

(D) 21 21

(E) 22 22

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $CD \perp AB$ with foot $D$. Right triangles $ACD$ and $BCD$ give $AC^2 = AD^2+CD^2$, $BC^2 = BD^2+CD^2$, $AC^2-BC^2 = AD^2-BD^2 =(AD-BD)(AD+BD)$. Since $AD+BD = AB = 80$ and $AC^2-BC^2 = 75^2-45^2 = 3600$, we get the equation $3600 = 80(AD-BD)$. This equation simplifies to $45 = AD - BD$. We can solve the system of equations $AD + BD = 80$ and $AD - BD = 45$ easily via elimination, and we get $AD = \frac{125}{2}$, $BD = \frac{35}{2}$. $CD^2 = AC^2-AD^2 = 75^2-\left(\frac{125}{2}\right)^2 = \frac{6875}{4}$, $CD = \frac{25\sqrt{11}}{2}$. By Angle Bisector Theorem, $\frac{DP}{PC} = \frac{DB}{BC} = \frac{\frac{35}{2}}{45} = \frac{7}{18}$, $PC = CD-DP$ thus, $18DP = 7(CD-DP)$, $25DP = 7CD$, $DP = \left(\frac{7}{25}\right)CD = \left(\frac{7}{25}\right)\left(\frac{25\sqrt{11}}{2}\right) = \frac{7\sqrt{11}}{2}$. $BP^2 = BD^2+DP^2 = \left(\frac{35}{2}\right)^2+\left(\frac{7\sqrt{11}}{2}\right)^2 = \frac{1225}{4}+\frac{49(11)}{4} = \frac{1764}{4} = 441$, thus $BP = \boxed{\text{(D) }21}.$

设 $CD \perp AB$，垂足为 $D$。直角三角形 $ACD$ 和 $BCD$ 得 $AC^2 = AD^2+CD^2$，$BC^2 = BD^2+CD^2$，$AC^2-BC^2 = AD^2-BD^2 =(AD-BD)(AD+BD)$。因为 $AD+BD = AB = 80$，且 $AC^2-BC^2 = 75^2-45^2 = 3600$，得方程 $3600 = 80(AD-BD)$。化简得 $45 = AD - BD$。解方程组 $AD + BD = 80$ 和 $AD - BD = 45$，得 $AD = \frac{125}{2}$，$BD = \frac{35}{2}$。$CD^2 = AC^2-AD^2 = 75^2-\left(\frac{125}{2}\right)^2 = \frac{6875}{4}$，$CD = \frac{25\sqrt{11}}{2}$。由角平分线定理，$\frac{DP}{PC} = \frac{DB}{BC} = \frac{\frac{35}{2}}{45} = \frac{7}{18}$，$PC = CD-DP$，从而 $18DP = 7(CD-DP)$，$25DP = 7CD$，$DP = \left(\frac{7}{25}\right)CD = \left(\frac{7}{25}\right)\left(\frac{25\sqrt{11}}{2}\right) = \frac{7\sqrt{11}}{2}$。$BP^2 = BD^2+DP^2 = \left(\frac{35}{2}\right)^2+\left(\frac{7\sqrt{11}}{2}\right)^2 = \frac{1225}{4}+\frac{49(11)}{4} = \frac{1764}{4} = 441$，故 $BP = \boxed{\text{(D) }21}$。

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