2009 AIME I Problems/Problem 12

Problem

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 $m + n$ .

Solution

Let $O$ be center of the circle and $P$ , $Q$ be the two points of tangent such that $P$ is on $BI$ and $Q$ is on $AI$ . We know that $AD:CD = CD:BD = 12:35$ .

Since the ratios between corresponding lengths of two similar diagrams are equal, we can let $AD = 144, CD = 420$ and $BD = 1225$ . Hence $BP = 144, AQ = 1225, AB = 1369$ and the radius $r = OD = 210$ .

Since we have $\tan OAB = \frac {35}{24}$ and $\tan OBA = \frac{6}{35}$ , we have $\sin {(OAB + OBA)} = \frac {1369}{\sqrt {(1801*1261)}},$ $\cos {(OAB + OBA)} = \frac {630}{\sqrt {(1801*1261)}}$ .

Hence $\sin I = \sin {(2OAB + 2OBA)} = \frac {2*1369*630}{1801*1261}$ . let $IP = IQ = x$ , then we have Area $(IBC)$ = $(2x + 1225*2 + 144*2)*\frac {210}{2}$ = $(x + 144)(x + 1225)* \sin {\frac {I}{2}}$ . Then we get $x + 1369 = \frac {3*1369*(x + 144)(x + 1225)}{1801*1261}$ .

Now the equation looks very complex but we can take a guess here. Assume that $x$ is a rational number (If it's not then the answer to the problem would be irrational which can't be in the form of $\frac {m}{n}$ ) that can be expressed as $\frac {a}{b}$ such that $(a,b) = 1$ . Look at both sides; we can know that $a$ has to be a multiple of $1369$ and not of $3$ and it's reasonable to think that $b$ is divisible by $3$ so that we can cancel out the $3$ on the right side of the equation.

Let's see if $x = \frac {1369}{3}$ fits. Since $\frac {1369}{3} + 1369 = \frac {4*1369}{3}$ , and $\frac {3*1369*(x + 144)(x + 1225)}{1801*1261} = \frac {3*1369* \frac {1801}{3} * \frac {1261*4}{3}} {1801*1261} = \frac {4*1369}{3}$ . Amazingly it fits!

Since we know that $3*1369*144*1225 - 1369*1801*1261 < 0$ , the other solution of this equation is negative which can be ignored. Hence $x = 1369/3$ .

Hence the perimeter is $1225*2 + 144*2 + \frac {1369}{3} *2 = 1369* \frac {8}{3}$ , and $BC$ is $1369$ . Hence $\frac {m}{n} = \frac {8}{3}$ , $m + n = 11$ .

2009 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2009 AIME I Problems/Problem 12

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