Difference between revisions of "2004 AMC 10B Problems/Problem 22"
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==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
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
![[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]](http://latex.artofproblemsolving.com/b/3/e/b3ee7391165284ef992a65e6ec0114c4beb26869.png)
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
![\[OI^2=R(R-2r)\]](http://latex.artofproblemsolving.com/1/e/e/1ee9dc3b52bdd0ea40931dc0ca792540fcb15ebc.png)
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:
![\[OI^2= \frac{65}{4}\]](http://latex.artofproblemsolving.com/f/a/a/faafcf2ce22effea97ba9366498d8dfe6840fb13.png)
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. 
