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Difference between revisions of "2009 AIME I Problems/Problem 12"

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== Solution ==
 
== Solution ==
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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
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Let <math>O</math> be center of the circle and <math>P</math>,<math>Q</math> be the two points of tangent such that <math>P</math> is on <math>BI</math> and <math>Q</math> is on <math>AI</math>. We know that <math>AD:CD = CD:BD = 12:35</math>.
  
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We know that AD:CD = CD:BD = 12:35.
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Since the ratios between corresponding lengths of two similar diagrams are equal, we can let <math>AD = 144, CD = 420</math> and <math>BD = 1225</math>. Hence <math>BP = 144, AQ = 1225, AB = 1369</math> and the radius <math>r = OD = 210</math>.  
  
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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.  
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Since we have <math>\tan OAB = \frac {35}{24}</math> and <math>\tan OBA = \frac{6}{35}</math> , we have <math>\sin {(OAB + OBA)} = \frac {1369}{\sqrt {(1801*1261)}}, \cos {(OAB + OBA)} = \frac {630}{\sqrt {(1801*1261)}}</math>. Hence <math>\sin I = \sin {(2OAB + 2OBA)} = \frac {2*1369*630}{1801*1261}</math>. let <math>IP = IQ = x</math> , then we have Area<math>(IBC)</math> = <math>(2x + 1225*2 + 144*2)*\frac {210}{2}</math> = <math>(x + 144)(x + 1225)* \sin {\frac {I}{2}}</math>. Then we get <math>x + 1369 = \frac {3*1369*(x + 144)(x + 1225)}{1801*1261}</math>.
  
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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
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Now the equation looks very complex but we can take a guess here. Assume that <math>x</math> is a rational number
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+
(If it's not then the answer to the problem would be irrational which can't be in the form of m/n)  
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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.
 
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.
  

Revision as of 18:02, 22 March 2009

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 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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