2018 AMC 12B Problems/Problem 21
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 ?
Diagram
~MRENTHUSIASM
Solution
In this solution, let the brackets denote areas.
We place the diagram in the coordinate plane: Let and
Since is a right triangle with its circumcenter is the midpoint of from which Note that the circumradius of is
Let denote the semiperimeter of The inradius of is from which
Since is also tangent to both coordinate axes, its center is at and its radius is for some positive number Let be the point of tangency of and As and are both perpendicular to the common tangent line at we conclude that and are collinear. It follows that or Solving this equation, we have from which
Finally, we apply the Shoelace Theorem to Remark
Alternatively, we can use as the base and the distance from to as the height for
- By the Distance Formula, we have
- The equation of is so the distance from to is
Therefore, we get ~pieater314159 ~MRENTHUSIASM
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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