# Difference between revisions of "1996 AIME Problems/Problem 13"

## Problem

In triangle $ABC$, $AB=\sqrt{30}$, $AC=\sqrt{6}$, and $BC=\sqrt{15}$. There is a point $D$ for which $\overline{AD}$ bisects $\overline{BC}$, and $\angle ADB$ is a right angle. The ratio $\frac{[ADB]}{[ABC]}$ can be written in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

## Solution

$[asy] pointpen = black; pathpen = black + linewidth(0.7); pair B=(0,0), C=(15^.5, 0), A=IP(CR(B,30^.5),CR(C,6^.5)), E=(B+C)/2, D=foot(B,A,E); D(MP("A",A)--MP("B",B,SW)--MP("C",C)--A--MP("D",D)--B); D(MP("E",E)); MP("\sqrt{30}",(A+B)/2,NW); MP("\sqrt{6}",(A+C)/2,SE); MP("\frac{\sqrt{15}}2",(E+C)/2); D(rightanglemark(B,D,A)); [/asy]$

Let $E$ be the midpoint of $\overline{BC}$. Since $BE = EC$, then $\triangle ABE$ and $\triangle AEC$ share the same height and have equal bases, and thus have the same area. Similarly, $\triangle BDE$ and $BAE$ share the same height, and have bases in the ratio $DE : AE$, so $\frac{[BDE]}{[BAE]} = \frac{DE}{AE}$ (see area ratios). Now,

$\dfrac{[ADB]}{[ABC]} = \frac{[ABE] + [BDE]}{2[ABE]} = \frac{1}{2} + \frac{DE}{2AE}.$

By Stewart's Theorem, $AE = \frac{\sqrt{2(AB^2 + AC^2) - BC^2}}2 = \frac{\sqrt {57}}{2}$, and by the Pythagorean Theorem on $\triangle ABD, \triangle EBD$,

\begin{align*} BD^2 + \left(DE + \frac {\sqrt{57}}2\right)^2 &= 30 \\ BD^2 + DE^2 &= \frac{15}{4} \\ \end{align*}

Subtracting the two equations yields $DE\sqrt{57} + \frac{57}{4} = \frac{105}{4} \Longrightarrow DE = \frac{12}{\sqrt{57}}$. Then $\frac mn = \frac{1}{2} + \frac{DE}{2AE} = \frac{1}{2} + \frac{\frac{12}{\sqrt{57}}}{2 \cdot \frac{\sqrt{57}}{2}} = \frac{27}{38}$, and $m+n = \boxed{065}$.