Question

In $\triangle ABC$ shown in the figure, $AB=7,$ $BC=8,$ $CA=9,$ and $\overline{AH}$ is an altitude. Points $D$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB},$ respectively, so that $\overline{BD}$ and $\overline{CE}$ are angle bisectors, intersecting $\overline{AH}$ at $Q$ and $P,$ respectively. What is $PQ?$

$1$

$\dfrac58\sqrt3$

$\dfrac45\sqrt2$

$\dfrac{8}{15}\sqrt5$

$\dfrac65$

Solution:

Let $x=BH.$ Then $CH=8-x,$ and from the two right triangles $AH^2=7^2-x^2=9^2-(8-x)^2.$ This gives $x=2$ and $AH=\sqrt{45}.$ By the angle bisector theorem in $\triangle ACH,$ $\dfrac{AP}{PH}=\dfrac{CA}{CH}=\dfrac96,$ so $AP=\dfrac35 AH.$ Similarly in $\triangle ABH,$ $\dfrac{AQ}{QH}=\dfrac{BA}{BH}=\dfrac72,$ so $AQ=\dfrac79 AH.$ Then $PQ=AQ-AP=\left(\dfrac79-\dfrac35\right)AH=\dfrac{8}{45}\sqrt{45} =\dfrac{8}{15}\sqrt5.$

Thus, the correct answer is D.

2016 AMC 12B Problem 17

Solution:

Problem 17 in Other Years