2004 AIME II Problem 12

Question

Let

ABCD

be an isosceles trapezoid, whose dimensions are

AB = 6,

BC = 5 = DA,

and

CD = 4.

Draw circles of radius

3

centered at

A

and

B,

and circles of radius

2

centered at

C

and

D.

A circle contained within the trapezoid is tangent to all four of these circles. Its radius is

\frac{-k + m\sqrt{n}}{p},

where

k,

m,

n,

and

p

are positive integers,

n

is not divisible by the square of any prime, and

k

and

p

are relatively prime. Find

k + m + n + p.

Answer

Answer: 134

Concepts:tangent circles trapezoid Pythagorean Theorem radical

Difficulty rating: 3060

Small hint:

The trapezoid’s height is

\sqrt{24}.

Center the inner circle on the axis of symmetry; tangency means its center is at distance

x + 3

from

A

and

B,

and

x + 2

from

C

and

D

Big hint:

The center lies

\sqrt{x^2 + 6x}

above

\overline{AB}

and

\sqrt{x^2 + 4x}

below

\overline{CD},

and those two heights add to

\sqrt{24}

Solution:

Dropping perpendiculars from

C

and

D

shows each leg of length

5

spans a horizontal offset of

\frac{6 - 4}{2} = 1,

so the height of the trapezoid is

\sqrt{25 - 1} = \sqrt{24}.

By symmetry the inner circle’s center

O

lies on the vertical axis through the midpoints

E

of

\overline{AB}

and

F

of

\overline{CD}.

If its radius is

x,

external tangency gives

OA = x + 3

and

OC = x + 2,

so with

AE = 3

and

CF = 2,

\begin{aligned} OE &= \sqrt{(x+3)^2 - 9} \\ &= \sqrt{x^2 + 6x}, \\ OF &= \sqrt{(x+2)^2 - 4} \\ &= \sqrt{x^2 + 4x}. \end{aligned}

Since

OE + OF = \sqrt{24},

moving one radical across and squaring gives

\sqrt{24(x^2 + 4x)} = 12 - x,

and squaring again yields

24x^2 + 96x = 144 - 24x + x^2,

that is

23x^2 + 120x - 144 = 0.

The positive root is

\begin{aligned} x &= \frac{-120 + \sqrt{14400 + 13248}}{46} \\ &= \frac{-120 + 96\sqrt{3}}{46} \\ &= \frac{-60 + 48\sqrt{3}}{23}, \end{aligned}

so

k + m + n + p

= 60 + 48 + 3 + 23

= 134.