2005 AIME II Problem 12

Attempt Problem 12 of the 2005 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 2005 AIME II solutions, or check the answer key.

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12.

Square

ABCD

has center

O,

AB = 900,

E

and

F

are on

\overline{AB}

with

AE \lt BF

and

E

between

A

and

F,

m\angle EOF = 45^\circ,

and

EF = 400.

Given that

BF = p + q\sqrt{r},

where

p,

q,

and

r

are positive integers and

r

is not divisible by the square of any prime, find

p + q + r.

Answer

Answer: 307

Concepts:trigonometric identity Vieta’s Formulas square (geometry)

Difficulty rating: 3060

Small hint:

Let

G

be the midpoint of

\overline{AB};

then

OG = 450,

and ray

OG

splits the

45^\circ

angle into

\alpha + \beta

Big hint:

From

\tan\alpha + \tan\beta = \frac{8}{9}

and

\tan(\alpha + \beta) = 1,

get

\tan\alpha\tan\beta = \frac{1}{9};

then solve a quadratic

Solution:

Let

G

be the midpoint of

\overline{AB},

so

OG \perp AB

and

OG = 450.

With

\alpha = \angle EOG

and

\beta = \angle FOG

on either side of ray

OG,

we have

EG = 450\tan\alpha,

FG = 450\tan\beta,

and

\alpha + \beta = 45^\circ.

From

EG + FG = EF = 400,

we get

\tan\alpha + \tan\beta = \frac{8}{9}.

The tangent addition formula gives

1 = \tan 45^\circ = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta},

so

\tan\alpha\tan\beta = 1 - \frac{8}{9} = \frac{1}{9}.

Hence

\tan\alpha

and

\tan\beta

are the roots of

9t^2 - 8t + 1 = 0,

namely

\frac{4 \pm \sqrt{7}}{9}.

Since

AE = 450 - EG

and

BF = 450 - FG,

the condition

AE \lt BF

means

EG \gt FG,

so

\tan\beta = \frac{4 - \sqrt{7}}{9}.

Then

BF = 450 - 450 \cdot \frac{4 - \sqrt{7}}{9}

= 250 + 50\sqrt{7},

and

p + q + r = 250 + 50 + 7 = 307.