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

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 $CF = 2,$ $OE = \sqrt{(x+3)^2 - 9} = \sqrt{x^2 + 6x}, \qquad OF = \sqrt{(x+2)^2 - 4} = \sqrt{x^2 + 4x}.$

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 $x = \frac{-120 + \sqrt{14400 + 13248}}{46} = \frac{-120 + 96\sqrt{3}}{46} = \frac{-60 + 48\sqrt{3}}{23},$ so $k + m + n + p = 60 + 48 + 3 + 23 = 134.$

2004 AIME II Problem 12

Solution:

Problem 12 in Other Years