Difference between revisions of "2018 AMC 12B Problems/Problem 21"
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== 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 18:12, 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
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
.
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 |
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