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 (Coordinates)
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 intersect at , so is .
The question now becomes calculating the sum of the distance from each vertex 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 that passes through .
So
and
Solution 2 (Algebra)
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.
Let & be & 's circumcircles' respective centers. Since & are congruent, the distance & each are from are equal, so . The angle between & is , and since , is also . is a right triangle inscribed in a circle, so must be the diameter of . Using the same logic & reasoning, we could deduce that & are also circumdiameters.
Since the circumcircles are congruent, circumdiameters , , and 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 , such that and is the circumradius. Since :
After a few algebraic manipulations:
.
Solution 3 (Homothety)
Let be the circumcenter of and denote the length of the altitude from Note that a homothety centered at with ratio takes the circumcircle of to the circumcircle of . It also takes the point diametrically opposite on the circumcircle of to Therefore, lies on the circumcircle of Similarly, it lies on the circumcircle of By Pythagorean triples, Finally, our answer is
Solution 4 (basically Solution 1 but without coordinates)
Since Solution 1 has already proven that the circumcenter of coincides with , we'll go from there. Note that the radius of the circumcenter of any given triangle is , and since and , it can be easily seen that and therefore our answer is
Solution 5
Since is a midline of we have that with a side length ratio of
Consider a homothety of scale factor with on concerning point . Note that this sends to with By properties of homotheties, and are collinear. Similarly, we obtain that with all three points collinear. Let denote the circumcenter of It is well-known that and analogously However, there is only one perpendicular line to passing through , therefore, coincides with
It follows that where is the circumradius of and this can be computed using the formula from which we quickly obtain
Solution 6 (Trigonometry)
, , as the angles are on the same circle.
,
,
,
Therefore , and is the angle bisector of . By the angle bisector theorem , . In a similar fashion , where is the circumcircle of .
By the law of cosine, ,
By the extended law of sines, ,
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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