1992 AIME Problem 13

Question

Triangle

ABC

has

AB=9

and

BC:AC=40:41.

What’s the largest area that this triangle can have?

Answer

Set

BC=40t,

AC=41t,

and use the Law of Cosines with

AB=9

Express the area as a function of

\cos C

and maximize its square

Set

BC=40t,

AC=41t,

and

x=\cos C.

The Law of Cosines gives

81=t^2(3281-3280x),

while the area is

820t^2\sqrt{1-x^2}.

Hence it equals

820\cdot81\,\frac{\sqrt{1-x^2}}{3281-3280x}.

Differentiating its logarithm shows the maximum occurs at

x=\frac{3280}{3281}.

Then

\sqrt{1-x^2}=\frac{81}{3281}

and

3281-3280x=\frac{6561}{3281},

so the maximum area is

820.

Problem 13 in other years