Question

In right $\triangle ABC$ with hypotenuse $\overline{AB},$ $AC = 12,$ $BC = 35,$ and $\overline{CD}$ is the altitude to $\overline{AB}.$ Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\omega.$ The ratio of the perimeter of $\triangle ABI$ to the length $AB$ can be expressed in the form $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Solution:

Because $\overline{CD} \perp \overline{AB}$ and $D$ is an endpoint of the diameter, $AB$ is tangent to $\omega$ at $D.$ Together with the tangent lines $AI$ and $BI,$ this makes $\omega$ the inscribed circle of triangle $ABI.$ Write $AD = y,$ $BD = z,$ and let $x$ be the tangent length from $I.$ The right-triangle altitude satisfies $CD^2 = AD \cdot BD,$ so the inradius of $ABI$ is $r = \frac{1}{2}\sqrt{yz}.$

With semiperimeter $s = x + y + z,$ the tangent lengths are exactly $s - AB = x,$ $s - BI = y,$ and $s - AI = z,$ so the area of $ABI$ equals both $rs$ and, by Heron's formula, $\sqrt{s \cdot xyz}.$ Equating and squaring, $\frac{s^2\,yz}{4} = s\,xyz, \qquad \text{so} \qquad s = 4x,$ which gives $AB = y + z = s - x = 3x.$

The perimeter is $2s = 8x,$ so its ratio to $AB$ is $\frac{8x}{3x} = \frac{8}{3}$ (independent of the given legs), and $m + n = 8 + 3 = 11.$

2009 AIME I Problem 12

Solution:

Problem 12 in Other Years