2009 AIME I Problems/Problem 12

Revision as of 17:30, 22 March 2009 by Chelseafc (talk | contribs) (Solution)

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 centre of the circle. P,Q be the two point of tangent such that P is on BI and Q is on AI

We know that AD:CD = CD:BD = 12:35.

Since the ratio doesn't change if two patterns are similar, 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 tanOAB = 35/24 and tanOBA = 6/35 , we have sin(OAB + OBA) = 1369/√(1801*1261), cos(OAB + OBA) = 630/√(1801*1261). Hence sinI = sin(2OAB + 2OBA) = (2*1369*630)/(1801*1261). let IP = IQ = x , then we have Area(IBC) = (2x + 1225*2 + 144*2)*210/2 = (x + 144)(x + 1225)*sinI/2

Then we get x + 1369 = 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 (Just assume it, if it's not, then the answer of the whole question will also be irrational, which is impossible to be expressed as m/n) that can be expressed as a/b such that (a,b) = 1, look at both side, we can know that a has to be 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 try if x = 1369/3 fits. Since 1369/3 + 1369 = 4*1369/3, and 3*1369*(x + 144)(x + 1225)/(1801*1261) = 3*1369*(1801/3)*(1261*4/3)/1801*1261 = 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 + (1369/3)*2 = 1369*(8/3), and BC is 1369. Hence m/n = 8/3, m + n = 11.

See also

2009 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
Invalid username
Login to AoPS