2011 AMC 12B Problems/Problem 20
Contents
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 1
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 calculate the distances with coordinate geometry. (Note that because is the circumcenter.
Let , , ,
Then is on the line and also the line with slope and passes through .
So
and
Solution 2
Consider an additional circumcircle on . After drawing the diagram, it is noticed that each triangle has side values: , , . Thus they are congruent, and their respective circumcircles are. By inspection, we see that , , and are the circumdiameters, and so they are congruent. Therefore, the solution can be found by calculating one of these circumdiameters and multiplying it by a factor of . We can find the circumradius quite easily with the formula , s.t. and R is the circumradius. Since :
After a few algebraic manipulations:
.
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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