2009 AIME I Problems/Problem 12

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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 1

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 $AQ = 144, BP = 1225, AB = 1369$ 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 $\frac {3*1369*(x + 144)(x + 1225)}{1801*1261} = \frac {3*1369* \frac {1801}{3} * \frac {1261*4}{3}} {1801*1261} = \frac {4*13...$ . 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 $\frac {m}{n} = \frac {8}{3}$ , $m + n = 11$ .

Solution 2

As in Solution $1$ , let $P$ and $Q$ be the intersections of $\omega$ with $BI$ and $AI$ respectively.

First, by pythagorean theorem, $AB = \sqrt{12^2+35^2} = 37$ . Now the area of $ABC$ is $1/2*12*35 = 1/2*37*CD$ , so $CD=\frac{420}{37}$ and the inradius of $\triangle ABI$ is $r=\frac{210}{37}$ .

Now from $\triangle CDB \sim \triangle ACB$ we find that $\frac{BC}{BD} = \frac{AB}{BC}$ so $BD = BC^2/AB = 35^2/37$ and similarly, $AD = 12^2/37$ .

Note $IP=IQ=x$ , $BP=BD$ , and $AQ=AD$ . So we have $AI = 144/27+x$ , $BI = 1225/37+x$ . Now we can compute the area of $\triangle ABI$ in two ways: by heron's formula and by inradius times semiperimeter, which yields

$rs=210/37(37+x) = \sqrt{(37+x)(37-144/37)(37-1225/37)x}$ $210/37(37+x) = 12*35/37 \sqrt{x(37+x)}$ $37+x = 2 \sqrt{x(x+37)}$ $x^2+74x+1369 = 4x^2 + 148x$ $3x^2 + 74x - 1369 = 0$

The quadratic formula now yields $x=37/3$ . Plugging this back in, the perimeter of $ABI$ is $2s=2(37+x)=2(37+37/3) = 37(8/3)$ so the ratio of the perimeter to $AB$ is $8/3$ and our answer is $8+3=\boxed{011}$

2009 AIME I (Problems • 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

From AoPSWiki

Contents

Problem

Solution 1

Solution 2

See also