2009 AIME I Problems/Problem 12

Solution 3

As in Solution $2$ , let $P$ and $Q$ be the intersections of $\omega$ with $BI$ and $AI$ respectively.

Recall that the distance from a point outside a circle to that circle is the same along both tangent lines to the circle drawn from the point.

Recall also that the length of the altitude to the hypotenuse of a right-angle triangle is the geometric mean of the two segments into which it cuts the hypotenuse.

Let $x = \overline{AD} = \overline{AQ}$ . Let $y = \overline{BD} = \overline{BP}$ . Let $z = \overline{PI} = \overline{QI}$ . The semi-perimeter of $ABI$ is $x + y + z$ . Since the lengths of the sides of $ABI$ are $x + y$ , $y + z$ and $x + z$ , the square of its area by Heron's formula is $(x+y+z)xyz$ .

The radius $r$ of $\omega$ is $\overline{CD}/2$ . Therefore $r^2 = xy/4$ . As $\omega$ is the in-circle of $ABI$ , the area of $ABI$ is also $r(x+y+z)$ , and so the square area is $r^2(x+y+z)^2$ .

Therefore $(x+y+z)xyz = r^2(x+y+z)^2 = \frac{xy(x+y+z)^2}{4}$ Dividing both sides by $xy(x+y+z)/4$ we get: $4z = (x+y+z),$ and so $z = (x+y)/3$ . The semi-perimeter of $ABI$ is therefore $\frac{4}{3}(x+y)$ and the whole perimeter is $\frac{8}{3}(x+y)$ . Now $x + y = \overline{AB}$ , so the ratio of the perimeter of $ABI$ to the hypotenuse $\overline{AB}$ is $8/3$ and our answer is $8+3=\boxed{011}$

and $8+3=\boxed{011}$ .

2009 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2009 AIME I Problems/Problem 12

Solution 3

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