Difference between revisions of "2001 AIME II Problems/Problem 7"
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− | Let <math>P = (0,0)</math> be at the origin. Using the formula <math>A = rs</math> on <math>\triangle PQR</math>, where <math>r_{1}</math> is the [[inradius]] (similarly define <math>r_2, r_3</math> to be the radii of <math>C_2, C_3</math>), <math>s = \frac{PQ + QR + RP}{2} = 180</math> is the [[semiperimeter]], and <math>A = \frac 12 bh = 5400</math> is the area, we find <math>r_{1} = \frac As = 30</math>. Or, | + | Let <math>P = (0,0)</math> be at the origin. Using the formula <math>A = rs</math> on <math>\triangle PQR</math>, where <math>r_{1}</math> is the [[inradius]] (similarly define <math>r_2, r_3</math> to be the radii of <math>C_2, C_3</math>), <math>s = \frac{PQ + QR + RP}{2} = 180</math> is the [[semiperimeter]], and <math>A = \frac 12 bh = 5400</math> is the area, we find <math>r_{1} = \frac As = 30</math>. Or, the inradius could be directly by using the formula <math>\frac{a+b-c}{2}</math>, where <math>a</math> and <math>b</math> are the legs of the right triangle and <math>c</math> is the hypotenuse. (This formula should be used ''only for right triangles''.) Thus <math>ST, UV</math> lie respectively on the lines <math>y = 60, x = 60</math>, and so <math>RS = 60, UQ = 30</math>. |
Note that <math>\triangle PQR \sim \triangle STR \sim \triangle UQV</math>. Since the ratio of corresponding lengths of similar figures are the same, we have | Note that <math>\triangle PQR \sim \triangle STR \sim \triangle UQV</math>. Since the ratio of corresponding lengths of similar figures are the same, we have |
Revision as of 21:59, 11 February 2012
Problem
Let be a right triangle with , , and . Let be the inscribed circle. Construct with on and on , such that is perpendicular to and tangent to . Construct with on and on such that is perpendicular to and tangent to . Let be the inscribed circle of and the inscribed circle of . The distance between the centers of and can be written as . What is ?
Contents
[hide]Solution
Solution 1 (analytic)
Let be at the origin. Using the formula on , where is the inradius (similarly define to be the radii of ), is the semiperimeter, and is the area, we find . Or, the inradius could be directly by using the formula , where and are the legs of the right triangle and is the hypotenuse. (This formula should be used only for right triangles.) Thus lie respectively on the lines , and so .
Note that . Since the ratio of corresponding lengths of similar figures are the same, we have
Let the centers of be , respectively; then by the distance formula we have . Therefore, the answer is .
Solution 2 (synthetic)
We compute as above. Let respectively the points of tangency of with .
By the Two Tangent Theorem, we find that , . Using the similar triangles, , , so . Thus .
See also
2001 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |