2009 AIME I Problem 12

Attempt Problem 12 of the 2009 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AIME I solutions, or check the answer key.

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12.

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

Answer: 11

Concepts:incircle, incenter, and inradius tangent line Heron’s Formula

Difficulty rating: 2990

Small hint:

AB

is tangent to

\omega

at

D,

so

\omega

is the inscribed circle of triangle

ABI

Big hint:

Let

x

be the tangent length from

I

and compute the area of

ABI

two ways:

rs

with

r = \frac{1}{2}\sqrt{AD \cdot BD},

and Heron’s formula

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,

so

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.