Difference between revisions of "2011 AMC 12B Problems/Problem 20"
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The question now becomes calculate the sum of distance from each vertices to the circumcenter. | 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 <math>XA = XB = XC</math> because <math>X</math> is the circumcenter. | + | We can calculate the distances with coordinate geometry. (Note that <math>XA = XB = XC</math> because <math>X</math> is the circumcenter.) |
Let <math>A = (5,12)</math>, <math>B = (0,0)</math>, <math>C = (14, 0)</math>, <math>X= (x_0, y_0)</math> | Let <math>A = (5,12)</math>, <math>B = (0,0)</math>, <math>C = (14, 0)</math>, <math>X= (x_0, y_0)</math> |
Revision as of 15:35, 5 February 2015
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 |
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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