2016 AIME I Problem 9

Question

Triangle

ABC

has

AB = 40,

AC = 31,

and

\sin A = \frac{1}{5}.

This triangle is inscribed in rectangle

AQRS

with

B

on

\overline{QR}

and

C

on

\overline{RS}.

Find the maximum possible area of

AQRS.

Answer

With

\beta = \angle BAQ

and

\gamma = \angle CAS,

the area is

40\cos\beta \cdot 31\cos\gamma,

and

\beta + \gamma = 90^\circ - A

Product-to-sum:

2\cos\beta\cos\gamma = \cos(\beta - \gamma)

+ \sin A,

maximized when

\beta = \gamma

Let

\beta = \angle BAQ

and

\gamma = \angle CAS,

so

\beta + \gamma = 90^\circ - A.

From the right triangles

AQB

and

ASC,

the sides of the rectangle are

AQ = 40\cos\beta

and

AS = 31\cos\gamma,

so its area is

\begin{aligned} 40 \cdot 31 \cos\beta\cos\gamma \\ &\tiny = 620\bigl(\cos(\beta - \gamma) + \cos(\beta + \gamma)\bigr) \\ &\tiny = 620\bigl(\cos(\beta - \gamma) + \sin A\bigr), \end{aligned}

using the product-to-sum identity and

\cos(90^\circ - A) = \sin A.

This is maximized when

\beta = \gamma,

which the constraint allows, giving area

620\left(1 + \frac{1}{5}\right) = 744.

Problem 9 in other years