Difference between revisions of "2012 USAJMO Problems/Problem 1"
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Note that (as in the first solution) the circumcircle of triangle <math>PRS</math> is tangent to <math>AB</math> at <math>P</math>. Similarly, since <math>\angle CQR = \angle QSR</math>, the circumcircle of triangle <math>QRS</math> is tangent to <math>AC</math> at <math>Q</math>. | Note that (as in the first solution) the circumcircle of triangle <math>PRS</math> is tangent to <math>AB</math> at <math>P</math>. Similarly, since <math>\angle CQR = \angle QSR</math>, the circumcircle of triangle <math>QRS</math> is tangent to <math>AC</math> at <math>Q</math>. | ||
− | Now, suppose these circumcircles are not the same circle. They already intersect at <math>R</math> and <math>S</math>, so they cannot intersect anymore. Thus, AS must touch the two circumcircles at points <math>M</math> and <math>N</math>, with <math>M</math> on the circumcircle of triangle <math>PRS</math>. By Power of a Point, <math> | + | Now, suppose these circumcircles are not the same circle. They already intersect at <math>R</math> and <math>S</math>, so they cannot intersect anymore. Thus, AS must touch the two circumcircles at points <math>M</math> and <math>N</math>, with <math>M</math> on the circumcircle of triangle <math>PRS</math>. By Power of a Point, <math>AQ^2 = AM \cdot AS</math> and <math>AP^2 = AN \cdot AS</math>. Hence, because <math>AP = AQ</math>, <math>AM = AN</math>, a contradiction because then, as they lie on the same line segment, M and N must be the same point! (Note line segment, not line.) Hence, the two circumcircles are the same circle. |
==See also== | ==See also== |
Revision as of 12:59, 7 April 2015
Contents
[hide]Problem
Given a triangle , let and be points on segments and , respectively, such that . Let and be distinct points on segment such that lies between and , , and . Prove that , , , are concyclic (in other words, these four points lie on a circle).
Solution
Since , the circumcircle of triangle is tangent to at . Similarly, since , the circumcircle of triangle is tangent to at .
For the sake of contradiction, suppose that the circumcircles of triangles and are not the same circle. Since , lies on the radical axis of both circles. However, both circles pass through and , so the radical axis of both circles is . Hence, lies on , which is a contradiction.
Therefore, the two circumcircles are the same circle. In other words, , , , and all lie on the same circle.
Solution 2
Note that (as in the first solution) the circumcircle of triangle is tangent to at . Similarly, since , the circumcircle of triangle is tangent to at .
Now, suppose these circumcircles are not the same circle. They already intersect at and , so they cannot intersect anymore. Thus, AS must touch the two circumcircles at points and , with on the circumcircle of triangle . By Power of a Point, and . Hence, because , , a contradiction because then, as they lie on the same line segment, M and N must be the same point! (Note line segment, not line.) Hence, the two circumcircles are the same circle.
See also
2012 USAJMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.