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 $m+n = \boxed{6}$ .

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

2006 iTest (Problems, Answer Key)
Preceded by: Problem U4	Followed by: Problem U6
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10

Page

Toolbox

Search

2006 iTest Problems/Problem U5

Problem

Solution

See Also