1987 AIME Problem 15

Question

Squares

S_1,

S_2

are inscribed in right triangle

ABC

as shown. Find

AC+CB

if

\operatorname{area}(S_1)=441

and

\operatorname{area}(S_2)=440.

Answer

Let the legs be

a

and

b

and use the first square to relate

ab

to

a+b

For the second square, use the altitude to the hypotenuse and similar cross-sections

Put

p=a+b,

q=ab,

and let the hypotenuse be

c.

Since

S_1

has side

21,

the standard leg-aligned-square relation gives

21=\frac{ab}{a+b},

so

q=21p.

Hence

c^2=p^2-2q=p(p-42).

The altitude to the hypotenuse is

h=\frac{q}{c}.

If the side of

S_2

is

t,

similarity gives

t=\frac{ch}{c+h}.

Substitution simplifies this to

t=\frac{21c}{p-21}.

Therefore

440=\frac{441p(p-42)}{(p-21)^2}.

Since

p(p-42)=(p-21)^2-441,

this becomes

440=441\left(1-\frac{441}{(p-21)^2}\right).

Thus

(p-21)^2=441^2,

and

p>42

gives

p=462.

Problem 15 in other years