Difference between revisions of "2018 AMC 12B Problems/Problem 21"
m (→Solution 2) |
|||
| Line 14: | Line 14: | ||
== Solution 2 == | == Solution 2 == | ||
| − | Notice that we can let <math>M = C</math>. If <math> | + | Notice that we can let <math>M = C</math>. If <math>O = \left(0, 0\right)</math>, then <math>C = \left(6, -\frac{5}{2}\right)</math> and <math>I = \left(4, -\frac{1}{2}\right)</math>. Using shoelace formula, we get <math>\left[COI\right] = \frac{7}{2}</math>. <math>\boxed{\textbf E.}</math> |
==See Also== | ==See Also== | ||
Revision as of 17:12, 25 February 2018
Contents
Problem
In 
with side lengths 
, 
, and 
, let 
and 
denote the circumcenter and incenter, respectively. A circle with center 
is tangent to the legs 
and 
and to the circumcircle of . What is the area of 
?
Solution 1
Let the triangle have coordinates 
Then the coordinates of the incenter and circumcenter are 
and 
respectively. If we let 
then 
satisfies
![\[\sqrt{(2.5-x)^2+(6-x)^2}+x=6.5\]](http://latex.artofproblemsolving.com/3/6/8/368140aaa27ee8a317c5a4f5e688856ee7955cdc.png)
![\[2.5^2-5x+x^2+6^2-12x+x^2=6.5^2-13x+x^2\]](http://latex.artofproblemsolving.com/0/8/a/08a8720dab4c3f61243ea3440cce09e98fa5cdd1.png)
![\[x^2=(5+12-13)x\]](http://latex.artofproblemsolving.com/4/9/6/49677b70023e00be75482bf672220d3904a201d6.png)
![\[x\neq 0\implies x=4.\]](http://latex.artofproblemsolving.com/f/1/4/f14e1187d8112a9ff56ae6baf339643d846f58a5.png)
Now the area of our triangle can be calculated with the Shoelace Theorem. The answer turns out to be 
Solution 2
Notice that we can let 
. If 
, then 
and 
. Using shoelace formula, we get ![$\left[COI\right] = \frac{7}{2}$](http://latex.artofproblemsolving.com/d/6/a/d6a8c4ab53da53b811de91d84afc34bdd97da44c.png)
. 
See Also
| 2018 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 20 |
Followed by Problem 22 |
| 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 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
