# 2006 iTest Problems/Problem U5

The following problem is from the Ultimate Question of the 2006 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.

## Problem

In triangle $ABC$, points $D$, $E$, and $F$ are the feet of the angle bisectors of $\angle A$, $\angle B$, $\angle C$ respectively. Let point $P$ be the intersection of segments $AD$ and $BE$, and let $p$ denote the perimeter of $ABC$. If $AP = 3PD$, $BE = 9$, and $CF = 9$, then the value of $\frac{AD}{p}$ can be expressed uniquely as $\frac{\sqrt{m}}{n}$ where $m$ and $n$ are positive integers such that $m$ is not divisible by the square of any prime. Find $m + n$.

## Solution

According to the Steiner-Lehmus Theorem, $AC = AB$. Thus, $\triangle CAD \cong \triangle BAD$ by SAS Congruency, so $AD \perp BC$ and $BD = CD$. $[asy] pair A=(4.593,12.990),B=(0,0),C=(9.186,0),D=(4.593,0),e=(7.348,5.196),F=(1.837,5.196); draw(B--C--A--B); draw(A--D); draw(B--e); draw(C--F); dot(D); label("D",D,S); dot(B); label("B",B,SW); dot(C); label("C",C,SE); dot(A); label("A",A,N); dot(e); dot("E",e,NE); dot(F); dot("F",F,NW); dot((4.593,3.248)); label("P",(3.5,3.248)); [/asy]$

Let $CP = y$ and $[CDP] = A$. Since $\triangle BDP$ and $\triangle CDP$ share an altitude and $\triangle BPC$ and $\triangle BPF$ share an altitude, we know that $[BDP] = A$ and $[BFP] = 2A \cdot \tfrac{9-y}{y}$. Also, since $[ABP] = 3 [BPD]$, $[AFP] = A \cdot (5 - \tfrac{18}{y})$. Similarly, $[PFB] = 2A \cdot \tfrac{9-y}{y}$ and $[AEP] = A \cdot (5 - \tfrac{18}{y})$.

Since $\triangle AEP$ and $\triangle ABP$ share an altitude, we have \begin{align*} \frac{y}{9-y} &= \frac{3A}{(5 - \frac{18}{y}) \cdot A} \\ \frac{y}{9-y} &= \frac{3}{(5 - \frac{18}{y})} \\ 5y - 18 &= 27-3y \\ 8y &= 45 \\ y &= \frac{45}{8}. \end{align*} Now let $CP = z$ and $DP = x$. By the Angle Bisector Theorem, $\tfrac{AC}{3x} = \tfrac{z}{x}$, so $AC = 3z$. By using the Pythagorean Theorem on $\triangle ADC$, we have $16x^2 + z^2 = 9z^2$, so $z^2 = 2x^2$. By using the Pythagorean Theorem on $\triangle PDC$, \begin{align*} x^2 + z^2 &= (\frac{45}{8})^2 \\ 3x^2 &= (\frac{45}{8})^2 \\ x^2 &= \frac{45}{8} \cdot \frac{1}{\sqrt{3}} \\ &= \frac{15\sqrt{3}}{8}. \end{align*} Therefore, $z = \tfrac{15\sqrt{6}}{8}$ the $AB = AC = 3 \cdot \tfrac{15\sqrt{6}}{8} = \tfrac{45\sqrt{6}}{8}$. The perimeter of the triangle equals $2 \cdot \tfrac{45\sqrt{6}}{8} + 2 \cdot \tfrac{15\sqrt{6}}{8} = 15\sqrt{6}$, and $AD = 4 \cdot \tfrac{15\sqrt{3}}{8} = \tfrac{15\sqrt{3}}{2}$. Thus, $\tfrac{AD}{p} = \tfrac{15\sqrt{3}}{2} \cdot \tfrac{1}{15\sqrt{6}} = \tfrac{\sqrt{2}}{4}$, and $m+n = \boxed{6}$.