2009 AIME I Problems/Problem 12
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.
First, by pythagorean theorem, . Now the area of is , so and the inradius of is .
Now from we find that so and similarly, .
Note , , and . So we have , . Now we can compute the area of in two ways: by heron's formula and by inradius times semiperimeter, which yields
The quadratic formula now yields . Plugging this back in, the perimeter of is so the ratio of the perimeter to is and our answer is
Solution 3
As in Solution , let and be the intersections of with and respectively.
Recall that the distance from a point outside a circle to that circle is the same along both tangent lines to the circle drawn from the point.
Recall also that the length of the altitude to the hypotenuse of a right-angle triangle is the geometric mean of the two segments into which it cuts the hypotenuse.
Let . Let . Let . The semi-perimeter of is . Since the lengths of the sides of are , and , the square of its area by Heron's formula is .
The radius of is . Therefore . As is the in-circle of , the area of is also , and so the square area is .
Therefore Dividing both sides by we get: and so . The semi-perimeter of is therefore and the whole perimeter is . Now , so the ratio of the perimeter of to the hypotenuse is and our answer is
Solution 4
Let , let , and let . Let . We find . Let , , and be the angles , , and respectively. Then , so The perimeter of is . The desired ratio is then We need to find . In , . We get Similarly, . Then Computing in two ways we get , so . Using this value of we get . Thus and .
Solution 5
This solution is not a real solution and is solving the problem with a ruler and compass.
Draw . Then, drawing the tangents and intersecting them, we get that is around and is around . We then find the ratio to be around . Using long division, we find that this ratio is approximately 2.666, which you should recognize as . Since this seems reasonable, we find that the answer is ~ilp
Solution 6
Denoting three tangents has length while lies on with .The area of is , so and the inradius of is .As we know that the diameter of the circle is the height of from to . Assume that and and . But we know that According to the basic computation, we can get that ; So we know that according to the tangent addition formula. Hence, it is not hard to find that the length of is . According to basic addition and division, we get the answer is which leads to ~bluesoul
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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