1988 AIME Problem 7

Question

In triangle

ABC,

\tan\angle CAB=\frac{22}{7},

and the altitude from

A

divides

BC

into segments of length

3

and

17.

What is the area of triangle

ABC?

Answer

Let the altitude have length

h

and place its foot at the origin

Use cross product over dot product to express the tangent of the angle between the two side vectors

Let the altitude length be

h.

From

A,

vectors to the endpoints of

BC

may be taken as

(-3,-h)

and

(17,-h).

Therefore

\tan\angle CAB=\frac{20h}{h^2-51}=\frac{22}{7}.

Thus

11h^2-70h-561=0,

whose positive root is

h=11.

Since

BC=3+17=20,

the area is

\frac12(20)(11)=110.

Problem 7 in other years