1987 AMC 12 Problem 21

Question

There are two natural ways to inscribe a square in a given isosceles right triangle. If it is done as in Figure

1

below, then one finds that the area of the square is

441\text{ cm}^2.

What is the area (in

\text{cm}^2

) of the square inscribed in the same

\triangle ABC

as shown in Figure

2

below?

378

392

400

441

484

Use Figure

1

to determine the leg length of the isosceles right triangle

In Figure

2,

compare the side of the tilted square with the two equal leg segments cut off by its opposite side

The first square has side

21,

so the triangle’s legs have length

42.

Let

s

be the side of the tilted square. Its side on the hypotenuse is parallel to the opposite side joining the legs. That opposite side has endpoints

(0,k)

and

(k,0),

so

s=k\sqrt2.

The distance between the parallel lines

x+y=k

and

x+y=42

is also

s,

giving

s=\frac{42-k}{\sqrt2},\qquad k=\frac{s}{\sqrt2}.

Hence

s=14\sqrt2,

and its area is

s^2=392.

Thus the correct answer is B.

Problem 21 in other years