Difference between revisions of "2011 AMC 12B Problems/Problem 20"
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===Solution 2 (Algebra)=== | ===Solution 2 (Algebra)=== | ||
− | Consider an additional circumcircle on <math>\triangle ADF</math>. After drawing the diagram, it is noticed that each triangle has side values: <math>7</math>, <math>\frac{15}{2}</math>, <math>\frac{13}{2}</math>. Thus they are congruent, and their respective circumcircles are. By inspection, we see that <math>XA</math>, <math>XB</math>, and <math>XC</math> are the | + | Consider an additional circumcircle on <math>\triangle ADF</math>. After drawing the diagram, it is noticed that each triangle has side values: <math>7</math>, <math>\frac{15}{2}</math>, <math>\frac{13}{2}</math>. Thus they are congruent, and their respective circumcircles are. By inspection, we see that <math>XA</math>, <math>XB</math>, and <math>XC</math> are the circumradii, and so they are congruent. Therefore, the solution can be found by calculating one of these circumdiameters and multiplying it by a factor of <math>3</math>. We can find the circumradius quite easily with the formula <math>\sqrt{(s)(s-a)(s-b)(s-c)} = \frac{abc}{4R}</math>, s.t. <math>s=\frac{a+b+c}{2}</math> and R is the circumradius. Since <math>s = \frac{21}{2}</math>: |
<cmath> \sqrt{(\frac{21}{2})(4)(3)(\frac{7}{2})} = \frac{\frac{15}{2}\cdot\frac{13}{2}\cdot 7}{4R} </cmath> | <cmath> \sqrt{(\frac{21}{2})(4)(3)(\frac{7}{2})} = \frac{\frac{15}{2}\cdot\frac{13}{2}\cdot 7}{4R} </cmath> |
Revision as of 01:57, 24 July 2017
Contents
[hide]Problem
Triangle has , and . The points , and are the midpoints of , and respectively. Let be the intersection of the circumcircles of and . What is ?
Solutions
Solution 1 (Coordinates)
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 intersect 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 (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. By inspection, we see that , , and are the circumradii, 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:
.
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.