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

Answer: 110

Concepts:triangle area trigonometry vector

Difficulty rating: 1870

Small Hint:

Let the altitude have length $h$ and place its foot at the origin

Big Hint:

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

Solution:

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 $\frac12(20)(11)=110.$

1988 AIME Problem 7

Answer: 110

Difficulty rating: 1870

Small Hint:

Big Hint:

Solution:

Problem 7 in Other Years