1990 AIME Problem 7

Question

A triangle has vertices

P=(-8,5),

Q=(-15,-19),

and

R=(1,-7).

The equation of the bisector of

\angle P

can be written in the form

ax+2y+c=0.

Find

a+c.

Answer

Find the unit vectors from

P

toward

Q

and

R

The internal angle-bisector direction is the sum of the two unit vectors

We have

\overrightarrow{PQ}=(-7,-24)

with length

25,

and

\overrightarrow{PR}=(9,-12)

with length

15.

The sum of their unit vectors is

\begin{aligned}v&=\left(-\frac7{25},-\frac{24}{25}\right)\\&\quad+\left(\frac35,-\frac45\right)\\&=\frac1{25}(8,-44),\end{aligned}

so the angle bisector has direction

(2,-11).

A normal vector is

(11,2).

Through

P=(-8,5),

its equation is

11(x+8)+2(y-5)=0,

or

11x+2y+78=0.

Hence

a+c=11+78=89.

Problem 7 in other years