Difference between revisions of "2009 AIME I Problems/Problem 12"
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First, by pythagorean theorem, <math>AB = \sqrt{12^2+35^2} = 37</math>. Now the area of <math>ABC</math> is <math>1/2*12*35 = 1/2*37*CD</math>, so <math>CD=\frac{420}{37}</math> and the inradius of <math>\triangle ABI</math> is <math>r=\frac{210}{37}</math>. | First, by pythagorean theorem, <math>AB = \sqrt{12^2+35^2} = 37</math>. Now the area of <math>ABC</math> is <math>1/2*12*35 = 1/2*37*CD</math>, so <math>CD=\frac{420}{37}</math> and the inradius of <math>\triangle ABI</math> is <math>r=\frac{210}{37}</math>. | ||
− | Now from <math>\triangle CDB | + | Now from <math>\triangle CDB \sim \triangle ACB</math> we find that <math>\frac{BC}{BD} = \frac{AB}{BC}</math> so <math>BD = BC^2/AB = 35^2/37</math> and similarly, <math>AD = 12^2/37</math>. |
Now we have <math>AI = 144/27+x</math>, <math>BI = 1225/37+x</math>. Now we can compute the area of <math>\triangle ABI</math> in two ways: by heron's formula and by inradius times semiperimeter, which yields | Now we have <math>AI = 144/27+x</math>, <math>BI = 1225/37+x</math>. Now we can compute the area of <math>\triangle ABI</math> in two ways: by heron's formula and by inradius times semiperimeter, which yields |
Revision as of 00:32, 3 April 2009
Contents
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 1
Let be center of the circle and , be the two points of tangent such that is on and is on . We know that .
Since the ratios between corresponding lengths of two similar diagrams are equal, we can let and . Hence and the radius .
Since we have and , we have .
Hence . let , then we have Area = = . Then we get .
Now the equation looks very complex but we can take a guess here. Assume that 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 ) that can be expressed as such that . Look at both sides; we can know that has to be a multiple of and not of and it's reasonable to think that is divisible by so that we can cancel out the on the right side of the equation.
Let's see if fits. Since , and . Amazingly it fits!
Since we know that , the other solution of this equation is negative which can be ignored. Hence .
Hence the perimeter is , and is . Hence , .
Solution 2
As in Solution , let and be the intersections of with and respectively. Let .
First, by pythagorean theorem, . Now the area of is , so and the inradius of is .
Now from we find that so and similarly, .
Now we have , . Now we can compute the area of in two ways: by heron's formula and by inradius times semiperimeter, which yields
Solving yields . Plugging this back in, the perimeter of is so the ratio of the perimeter to is and our answer is
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 |