Difference between revisions of "2001 AIME II Problems/Problem 7"
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By the [[Two Tangent Theorem]], we find that <math>A_{1}Q = 60</math>, <math>A_{1}R = 90</math>. Using the similar triangles, <math>RA_{2} = 45</math>, <math>QA_{3} = 20</math>, so <math>A_{2}A_{3} = QR - RA_2 - QA_3 = 85</math>. Thus <math>(O_{2}O_{3})^{2} = (15 - 10)^{2} + (85)^{2} = 7250\implies n=\boxed{725}</math>. | By the [[Two Tangent Theorem]], we find that <math>A_{1}Q = 60</math>, <math>A_{1}R = 90</math>. Using the similar triangles, <math>RA_{2} = 45</math>, <math>QA_{3} = 20</math>, so <math>A_{2}A_{3} = QR - RA_2 - QA_3 = 85</math>. Thus <math>(O_{2}O_{3})^{2} = (15 - 10)^{2} + (85)^{2} = 7250\implies n=\boxed{725}</math>. | ||
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+ | ===Solution 3=== | ||
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+ | The radius of an incircle is $ r=A_t/semiperimeter | ||
== See also == | == See also == |
Revision as of 21:42, 2 October 2017
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 .
Solution 3
The radius of an incircle is $ r=A_t/semiperimeter
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.