2023 AIME I Problem 5

Attempt Problem 5 of the 2023 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AIME I solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

5.

Let

P

be a point on the circle circumscribing square

ABCD

that satisfies

PA \cdot PC = 56

and

PB \cdot PD = 90.

Find the area of

ABCD.

Answer

Answer: 106

Concepts:circle coordinate geometry trigonometric identity

Difficulty rating: 2400

Small hint:

Put the center at the origin with

A = (R, 0),

B = (0, R),

C = (-R, 0),

D = (0, -R),

and

P = (R\cos\theta, R\sin\theta).

Big hint:

Show

PA \cdot PC = 2R^2|\sin\theta|

and

PB \cdot PD = 2R^2|\cos\theta|;

a Pythagorean identity links the two products.

Solution:

Let the circle have center

O

and radius

R,

with

A = (R, 0),

B = (0, R),

C = (-R, 0),

D = (0, -R),

and

P = (R\cos\theta, R\sin\theta).

Then

PA^2 = 2R^2(1 - \cos\theta)

and

PC^2 = 2R^2(1 + \cos\theta),

so

PA \cdot PC = 2R^2|\sin\theta| = 56.

In the same way

PB \cdot PD = 2R^2|\cos\theta| = 90.

Squaring and adding,

4R^4 = 56^2 + 90^2 = 11236,

so

2R^2 = \sqrt{11236} = 106.

The square has diagonal

2R,

hence area

\frac{(2R)^2}{2} = 2R^2 = 106.