Difference between revisions of "2013 USAMO Problems/Problem 1"
(Added a second solution) |
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The proof is complete. | The proof is complete. | ||
+ | ==Solution 3== | ||
+ | Use directed angles modulo <math>\pi</math>. | ||
+ | |||
+ | Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math> | ||
+ | |||
+ | Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath> | ||
+ | |||
+ | Now, it follows that (now not using directed angles) | ||
+ | <cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath> | ||
+ | using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>. | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:26, 20 May 2015
Contents
Problem
In triangle , points
lie on sides
respectively. Let
,
,
denote the circumcircles of triangles
,
,
, respectively. Given the fact that segment
intersects
,
,
again at
respectively, prove that
Solution 1
In this solution, all lengths and angles are directed.
Firstly, it is easy to see by that concur at a point
. Let
meet
again at
and
, respectively. Then by Power of a Point, we have
Thusly
But we claim that
. Indeed,
and
Therefore,
. Analogously we find that
and we are done.
courtesy v_enhance
Solution 2
Diagram Refer to the Diagram link.
By Miquel's Theorem, there exists a point at which intersect. We denote this point by
Now, we angle chase:
In addition, we have
Now, by the Ratio Lemma, we have
(by the Law of Sines in
)
(by the Law of Sines in
)
by the Ratio Lemma.
The proof is complete.
Solution 3
Use directed angles modulo .
Lemma.
Proof.
Now, it follows that (now not using directed angles)
using the facts that
and
,
and
are similar triangles, and that $\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}$ (Error compiling LaTeX. Unknown error_msg) equals twice the circumradius of the circumcircle of
.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.