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 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 <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>, | + | 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 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 <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>, such that <math>s=\frac{a+b+c}{2}</math> and <math>R</math> 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> | ||
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− | ===Solution 4 ( | + | ===Solution 4 (basically Solution 1 but without coordinates)=== |
− | + | Since Solution 1 has already proven that the circumcenter of <math>\triangle ABC</math> coincides with <math>X</math>, we'll go from there. Note that the radius of the circumcenter of any given triangle is <math>\frac{a}{2\sin{A}}</math>, and since <math>b=15</math> and <math>\sin{B}=\frac{12}{13}</math>, it can be easily seen that <math>XA = XB = XC = \frac{65}{8}</math> and therefore our answer is <cmath>3\cdot \frac{65}{8}=\boxed{\frac{195}{8}}.</cmath> | |
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− | Since Solution 1 has already proven that the circumcenter of <math>\triangle ABC</math> coincides with <math>X</math>, we'll go from there. Note that the radius of the circumcenter of any given triangle is <math>\frac{a}{2\sin{A}}</math>, and since <math>b= | ||
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=19|num-a=21|ab=B}} | {{AMC12 box|year=2011|num-b=19|num-a=21|ab=B}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 02:10, 28 June 2020
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 ?
Solutions
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 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 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. 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 , 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
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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