Difference between revisions of "2001 AIME II Problems/Problem 7"
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− | The radius of an incircle is | + | The radius of an incircle is <math> r=A_t/semiperimeter </math>. The area of the triangle is equal to <math> (90)(120)/2 </math> or 5400 and the semiperimeter is equal to <math> (90+120+150)/2 </math> or 180. The radius therefore is equal to <math>5400/180</math> or 30. Thus using similar triangles the dimensions of the triangle circumscribing the circle with center <math>C_2</math> are equal to <math>120-2(30)</math> or <math> 60</math>, <math>1/2(90)</math> or <math>45</math>, and <math>1/2(150)</math> or <math>75</math>. The radius of the circle inscribed in this triangle with dimensions <math>45</math>x<math>60</math>x<math>75</math> is found using the formula mentioned at the very beginning. The radius of the incircle is equal to 15. Defining P as (0,0), C_2 is equal to (60+15,15) or (75,15). Also using similar triangles, the dimensions of the triangle circumscribing the circle with center <math>C_3</math> are equal to <math>90-2(30)</math>, <math>1/3(120)</math>, <math>1/3(150)</math> or <math> 30,40,50</math>. The radius of C_3 by using the formula mentioned at the beginning is 10. Using P as (0,0), <math>C_3</math> is equal to (<math>10, 60+10)</math> or <math>(10,70)</math>. Using the distance formula, the distance between <math>C_2</math> and <math>C_3</math>: <math>\sqrt{(75-10)^2 +(15-70)^2}</math> this equals <math>\sqrt{7250}</math> or <math>\sqrt{(725)(10)}</math>, thus n is <math>725</math>. |
== See also == | == See also == |
Revision as of 22:16, 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 . The area of the triangle is equal to or 5400 and the semiperimeter is equal to or 180. The radius therefore is equal to or 30. Thus using similar triangles the dimensions of the triangle circumscribing the circle with center are equal to or , or , and or . The radius of the circle inscribed in this triangle with dimensions xx is found using the formula mentioned at the very beginning. The radius of the incircle is equal to 15. Defining P as (0,0), C_2 is equal to (60+15,15) or (75,15). Also using similar triangles, the dimensions of the triangle circumscribing the circle with center are equal to , , or . The radius of C_3 by using the formula mentioned at the beginning is 10. Using P as (0,0), is equal to ( or . Using the distance formula, the distance between and : this equals or , thus n is .
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 |
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