2015 AIME II Problem 15

Question

Circles

\mathcal{P}

and

\mathcal{Q}

have radii

1

and

4,

respectively, and are externally tangent at point

A.

Point

B

is on

\mathcal{P}

and point

C

is on

\mathcal{Q}

so that line

BC

is a common external tangent of the two circles. A line

\ell

through

A

intersects

\mathcal{P}

again at

D

and intersects

\mathcal{Q}

again at

E.

Points

B

and

C

lie on the same side of

\ell,

and the areas of

\triangle DBA

and

\triangle ACE

are equal. This common area is

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

Answer: 129

Concepts:homothety tangent circles coordinate geometry triangle area

Difficulty rating: 3500

Small hint:

The homothety at

A

taking

\mathcal{P}

to

\mathcal{Q}

sends

D

to

E,

so

AE = 4\,AD

and equal areas force

B

’s distance to

\ell

to be

4

times

C

’s

Big hint:

Set

B = (0, 0),

C = (4, 0),

A = \left(\frac{4}{5}, \frac{8}{5}\right)

and find the line through

A

whose distances to

B

and

C

have ratio

4 : 1,

same side

Solution:

Place line

BC

on the

x

-axis, so the centers are

P = (0, 1)

and

Q = (4, 4)

(their distance is

1 + 4 = 5

), with

B = (0, 0),

C = (4, 0),

and the tangency point

A = P + \frac{1}{5}(Q - P) = \left(\frac{4}{5}, \frac{8}{5}\right).

The homothety centered at

A

with ratio

-4

carries

\mathcal{P}

to

\mathcal{Q}

and

D

to

E,

so

AE = 4\,AD.

Since

[DBA] = \frac{1}{2} AD \cdot d(B, \ell)

and

[ACE] = \frac{1}{2} AE \cdot d(C, \ell),

the equal-area condition is

d(B, \ell) = 4\,d(C, \ell)

with

B

and

C

on the same side of

\ell.

Write

\ell

as

u\left(x - \frac{4}{5}\right) + v\left(y - \frac{8}{5}\right) = 0.

Its signed values at

B

and

C

are

-\frac{4u + 8v}{5}

and

\frac{16u - 8v}{5},

so the same-side ratio-

4

condition reads

-(4u + 8v) = 4(16u - 8v),

giving

24v = 68u,

i.e.

v = \frac{17}{6}u.

Taking

(u, v) = (6, 17),

the line is

6x + 17y = 32.

Then

d(B, \ell) = \frac{32}{\sqrt{325}},

and the center

P = (0, 1)

is at distance

\frac{15}{\sqrt{325}}

from

\ell,

so the chord gives

AD = 2\sqrt{1 - \frac{225}{325}} = \frac{20}{\sqrt{325}}.

The common area is

\begin{aligned} &\frac{1}{2} \cdot \frac{20}{\sqrt{325}} \cdot \frac{32}{\sqrt{325}} \\ &= \frac{320}{325} = \frac{64}{65}, \end{aligned}

so

m + n = 64 + 65 = 129.