# 2006 iTest Problems/Problem 27

## Problem

Line $\ell$ passes through $A$ and into the interior of the equilateral triangle $ABC$. $D$ and $E$ are the orthogonal projections of $B$ and $C$ onto $\ell$ respectively. If $DE=1$ and $2BD=CE$, then the area of $ABC$ can be expressed as $m\sqrt n$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Determine $m+n$. $[asy] import olympiad; size(250); defaultpen(linewidth(0.7)+fontsize(11pt)); real r = 31, t = -10; pair A = origin, B = dir(r-60), C = dir(r); pair X = -0.8 * dir(t), Y = 2 * dir(t); pair D = foot(B,X,Y), E = foot(C,X,Y); draw(A--B--C--A^^X--Y^^B--D^^C--E); label("A",A,S); label("B",B,S); label("C",C,N); label("D",D,dir(B--D)); label("E",E,dir(C--E)); [/asy]$

## Solution

Let $X$ be the intercept of $CB$ and $ED$. By the Vertical Angle Theorem, $\angle CXE = \angle DXB$. Also, since both $CE$ and $BD$ are perpendicular to $AD$, $CE \parallel DB$. Thus, $\triangle CEX \sim \triangle BDX$ by AA Similarity. Since $2BD = CE$, $DX = \tfrac13$ and $EX = \tfrac23$.

Let $a$ be the side length of the triangle, so $BX = \tfrac{a}{3}$. By the Pythagorean Theorem, $DB = \tfrac{\sqrt{a^2 - 1}}{3}$. Also, $\angle ABC = 60^\circ$, so by the Law of Cosines, $AX = \sqrt{a^2 + \tfrac19 a^2 - \tfrac23 a^2 \cdot \tfrac12} = \tfrac{a\sqrt{7}}{3}$. $[asy] pair A=(0,0), X=(14/3,0), D=(5,0), B=(5,-1.732); draw(A--D--B--A); dot(A); label("A",A,NW); dot(D); label("D",D,NE); dot(B); label("B",B,SE); dot(X); label("X",X,NW); label("\frac{a\sqrt{7}}{3}",(7/3,0),N); label("\frac13",(14.5/3,0),N); label("\frac{\sqrt{a^2-1}}{3}",(5,-0.866),E); label("a",(2.5,-0.866), SW); [/asy]$

By using the Pythagorean Theorem again, we have \begin{align*} \left( \frac{a\sqrt{7} + 1}{3} \right)^2 + \left( \frac{\sqrt{a^2 - 1}}{3} \right)^2 &= a^2 \\ (a\sqrt{7} + 1)^2 + a^2 - 1 &= 9a^2 \\ 7a^2 + 2a\sqrt{7} + 1 + a^2 - 1 &= 9a^2 \\ 2a\sqrt{7} &= a^2 \\ a &= 2\sqrt{7} \end{align*} Thus, the area of the triangle is $\tfrac{28\sqrt{3}}{4} = 7\sqrt{3}$, so $m+n = \boxed{10}$.