Difference between revisions of "2004 AMC 10B Problems/Problem 22"
m (→Solution 2) |
(→Solution 1) |
||
Line 6: | Line 6: | ||
==Solution 1== | ==Solution 1== | ||
+ | <asy> | ||
+ | import geometry; | ||
+ | |||
+ | unitsize(0.6 cm); | ||
+ | |||
+ | pair A, B, C, D, E, F, I, O; | ||
+ | |||
+ | A = (5^2/13,5*12/13); | ||
+ | B = (0,0); | ||
+ | C = (13,0); | ||
+ | I = incenter(A,B,C); | ||
+ | D = (I + reflect(B,C)*(I))/2; | ||
+ | E = (I + reflect(C,A)*(I))/2; | ||
+ | F = (I + reflect(A,B)*(I))/2; | ||
+ | O = (B + C)/2; | ||
+ | |||
+ | draw(A--B--C--cycle); | ||
+ | draw(incircle(A,B,C)); | ||
+ | draw(I--D); | ||
+ | draw(I--E); | ||
+ | draw(I--F); | ||
+ | draw(I--O); | ||
+ | |||
+ | label("$A$", A, N); | ||
+ | label("$B$", B, SW); | ||
+ | label("$C$", C, SE); | ||
+ | dot("$D$", D, S); | ||
+ | dot("$E$", E, NE); | ||
+ | dot("$F$", F, NW); | ||
+ | dot("$I$", I, N); | ||
+ | dot("$O$", O, S); | ||
+ | </asy> | ||
This is obviously a right triangle. Pick a coordinate system so that the right angle is at <math>(0,0)</math> and the other two vertices are at <math>(12,0)</math> and <math>(0,5)</math>. | This is obviously a right triangle. Pick a coordinate system so that the right angle is at <math>(0,0)</math> and the other two vertices are at <math>(12,0)</math> and <math>(0,5)</math>. | ||
Revision as of 20:25, 31 July 2018
Contents
[hide]Problem
A triangle with sides of 5, 12, and 13 has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles?
Solution 1
This is obviously a right triangle. Pick a coordinate system so that the right angle is at and the other two vertices are at and .
As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at .
The radius of the inscribed circle can be computed using the well-known identity , where is the area of the triangle and its perimeter. In our case, and . Thus, . As the inscribed circle touches both legs, its center must be at .
The distance of these two points is then .
Solution 2
We directly apply Euler’s Theorem, which states that if the circumcenter is and the incenter , and the inradius is and the circumradius is , then
We notice that this is a right triangle, and hence has area . We then find the inradius with the formula , where denotes semiperimeter. We easily see that , so .
We now find the circumradius with the formula . Solving for gives .
Substituting all of this back into our formula gives: So,
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.