2008 AIME II Problem 14

Attempt Problem 14 of the 2008 AIME II 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 2008 AIME II solutions, or check the answer key.

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14.

Let

a

and

b

be positive real numbers with

a \ge b.

Let

\rho

be the maximum possible value of

\frac{a}{b}

for which the system of equations

\begin{aligned} a^2 + y^2 &= b^2 + x^2 \\ &= (a - x)^2 + (b - y)^2 \end{aligned}

has a solution

(x, y)

satisfying

0 \le x \lt a

and

0 \le y \lt b.

Then

\rho^2

can be expressed as a fraction

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

Answer: 7

Concepts:equilateral triangle distance formula trigonometry optimization

Difficulty rating: 3270

Small hint:

The equal quantities are squared distances: with

D = (0,b),

E = (x,0),

F = (a,b-y)

in an

a \times b

rectangle, triangle

DEF

is equilateral

Big hint:

With

\theta = \angle ADE,

equal sides force

\frac{a}{b} = \frac{\cos(30^\circ - \theta)} {\cos\theta},

which increases in

\theta,

while

y \ge 0

caps

\theta

at

30^\circ

Solution:

Draw the rectangle with vertices

A = (0, 0),

B = (a, 0),

C = (a, b),

D = (0, b),

and let

E = (x, 0)

on

\overline{AB}

and

F = (a, b - y)

on

\overline{BC}.

Then

DE^2 = b^2 + x^2,

DF^2 = a^2 + y^2,

and

EF^2 = (a - x)^2 + (b - y)^2,

so the system says exactly that triangle

DEF

is equilateral, with the constraints keeping

E

and

F

on those two sides.

Let

\theta = \angle ADE,

so

x = b\tan\theta

and

DE = \frac{b}{\cos\theta}.

Since

\angle EDF = 60^\circ

and the corner angle at

D

is

90^\circ,

we get

\angle CDF = 30^\circ - \theta,

so

y = a\tan(30^\circ - \theta)

and

DF = \frac{a}{\cos(30^\circ - \theta)}.

Setting

DE = DF

gives

\begin{aligned} \frac{a}{b} &= \frac{\cos(30^\circ - \theta)}{\cos\theta} \\ &= \cos 30^\circ + \sin 30^\circ \tan\theta, \end{aligned}

which is increasing in

\theta.

The requirement

x \ge 0

gives

\theta \ge 0,

while

y \ge 0

forces

\theta \le 30^\circ.

The maximum is therefore at

\theta = 30^\circ,

where

\frac{a}{b} = \frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}} = \frac{2}{\sqrt{3}},

attained with

y = 0

and

x = \frac{b}{\sqrt{3}} \lt a.

Hence

\rho^2 = \frac{4}{3},

and

m + n = 4 + 3 = 7.