1996 AIME Problem 13

Question

In triangle

ABC,

AB=\sqrt{30},

AC=\sqrt6,

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{\operatorname{Area}(\triangle ADB)}{\operatorname{Area}(\triangle ABC)}

can be written in the form

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m+n.

Answer

Answer: 65

Concepts:median (geometry)area ratio Pythagorean Theorem

Difficulty rating: 2380

Small hint:

Let

E

be the midpoint of

\overline{BC},

so

A,

E,

and

D

are collinear

Big hint:

Find

AE

with the median formula, then compare two right triangles sharing

BD

Solution:

Let

E

be the midpoint of

\overline{BC}.

Then

A,

E,

and

D

are collinear. The median formula gives

\begin{aligned}AE^2&=\frac{2AB^2+2AC^2-BC^2}{4}\\&=\frac{57}{4}.\end{aligned}

Since

AB^2>AE^2+BE^2,

angle

AEB

is obtuse, so the perpendicular foot

D

lies beyond

E.

Both

\triangle ABD

and

\triangle EBD

are right at

D.

Therefore

\begin{gathered}AB^2-BE^2\\=(AE+DE)^2-DE^2\\=AE^2+2AE\cdot DE.\end{gathered}

Substitution gives

30-\frac{15}{4}=\frac{57}{4}+2AE\cdot DE,

so

\frac{DE}{AE}=\frac{8}{19}.

Triangles

ABE

and

DBE

have bases

AE

and

DE

on the same line and share the altitude from

B,

while

[ABC]=2[ABE].

Hence

\begin{aligned}\frac{[ADB]}{[ABC]}&=\frac{[ABE]+[DBE]}{2[ABE]}\\&=\frac12\left(1+\frac8{19}\right)\\&=\frac{27}{38}.\end{aligned}

Thus

m+n=65.