2009 AIME I Problems/Problem 12
Problem
In right with hypotenuse , , , and is the altitude to . Let be the circle having as a diameter. Let be a point outside such that and are both tangent to circle . The ratio of the perimeter of to the length can be expressed in the form , where and are relatively prime positive integers. Find .
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 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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