2016 AMC 12B Problem 17

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

Let

BH=x.

Then

AH^2=7^2-x^2=9^2-(8-x)^2,

which gives

x=2

and

AH=\sqrt{45}

By the angle bisector theorem,

\dfrac{AP}{PH}=\dfrac{CA}{CH}=\dfrac96

and

\dfrac{AQ}{QH}=\dfrac{BA}{BH}=\dfrac72;

then

PQ=AQ-AP

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

\begin{aligned} PQ &= AQ-AP \\ &= \left(\dfrac79-\dfrac35\right)AH \\ &= \dfrac{8}{45}\sqrt{45} \\ &= \dfrac{8}{15}\sqrt5. \end{aligned}

Thus, the correct answer is D.

Problem 17 in other years