# Difference between revisions of "2016 AMC 12B Problems/Problem 17"

## Problem

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 $PQ$?

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$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ \frac{5}{8}\sqrt{3} \qquad \textbf{(C)}\ \frac{4}{5}\sqrt{2} \qquad \textbf{(D)}\ \frac{8}{15}\sqrt{5} \qquad \textbf{(E)}\ \frac{6}{5}$

## Solution 1

Get the area of the triangle by heron's formula: $$\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(12)(3)(4)(5)} = 12\sqrt{5}$$ Use the area to find the height AH with known base BC: $$Area = 12\sqrt{5} = \frac{1}{2}bh = \frac{1}{2}(8)(AH)$$ $$AH = 3\sqrt{5}$$ $$BH = \sqrt{AB^2 - AH^2} = \sqrt{7^2 - (3\sqrt{5})^2} = 2$$ $$CH = BC - BH = 8 - 2 = 6$$ Apply angle bisector theorem on triangle $ACH$ and triangle $ABH$, we get $AP:PH = 9:6$ and $AQ:QH = 7:2$, respectively. To find AP, PH, AQ, and QH, apply variables, such that $AP:PH = 9:6$ is $\frac{3\sqrt{5} - x}{x} = \frac{9}{6}$ and $AQ:QH = 7:2$ is $\frac{3\sqrt{5} - y}{y} = \frac{7}{2}$. Solving them out, you will get $AP = \frac{9\sqrt{5}}{5}$, $PH = \frac{6\sqrt{5}}{5}$, $AQ = \frac{7\sqrt{5}}{3}$, and $QH = \frac{2\sqrt{5}}{3}$. Then, since $AP + PQ = AQ$ according to the Segment Addition Postulate, and thus manipulating, you get $PQ = AQ - AP = \frac{7\sqrt{5}}{3} - \frac{9\sqrt{5}}{5}$ = $$\boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}$$

## Solution 2

Let the intersection of $BD$ and $CE$ be the point $I$. Then let the foot of the altitude from $I$ to $BC$ be $I'$. Note that $II'$ is an inradius and that $II' \cdot s = [ABC]$, where $s$ is the semiperimeter of the triangle.

Using Heron's Formula, we see that $II' \cdot 12 = \sqrt{12 \cdot 3 \cdot 4 \cdot 5} = 12\sqrt{5}$, so $II' = \sqrt{5}$.

Then since $II'$ and $AH$ are parallel, $\triangle CI'I \sim \triangle CHP$ and $\triangle BHQ \sim \triangle BI'I$.

Thus, $\frac{II'}{PQ + QH} = \frac{CI'}{CH}$ and $\frac{II'}{QH} = \frac{BI'}{BH}$, so $PQ = \frac{II' \cdot CH}{CI'} - \frac{II' \cdot BH}{BI'}$.

By the Dual Principle, $CI' = 5$ and $BI' = 3$. With the same method as Solution 1, $CH = 6$ and $BH = 2$. Then $PQ = \frac{8}{15} II' =$ $$\boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}$$

## Solution 3 (if you're running short on time)

$PQ$ lies on altitude $AH$, which we find to have a length of $3\sqrt{5}$ by Heron's Formula and dividing twice the area by $BC$. From H we can construct a segment $HX$ with $X$ on $CE$ such that $HX$ is parallel to $EB$. A similar construction gives $Y$ on $BD$ such that $HY$ is parallel to $DC$. We can hence generate a system of ratios that will allow us to find $PQ/AH$. Note that such a system will generate a rational number for the ratio $PQ/AH$. Thus, we choose the only answer that has a $\sqrt{5}$ term in it, giving us $\textbf{D}$.

 2016 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 18 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions