2011 AMC 12B Problems/Problem 20
Problem
Triangle has , and . The points , and are the midpoints of , and respectively. Let be the intersection of the circumcircles of and . What is ?
Solution
Answer: (C)
Let us also consider the circumcircle of .
Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of which is , Also, since . is cyclic, similarly, and are also cyclic. With this, we know that the circumcircles of , and all intercept at , so is .
The question now becomes calculate the sum of distance from each vertices to the circumcenter.
We can do it will coordinate geometry, note that because of being circumcenter.
Let , , ,
Then is on the line and also the line with slope and passes through .
So
and
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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All AMC 12 Problems and Solutions |