Difference between revisions of "Sharygin Olympiads, the best"
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==2024, Problem 23== | ==2024, Problem 23== | ||
| + | [[File:2023 22 2.png|350px|right]] | ||
A point <math>P</math> moves along a circle <math>\Omega.</math> Let <math>A</math> and <math>B</math> be fixed points of <math>\Omega,</math> and <math>C</math> be an arbitrary point inside <math>\Omega.</math> The common external tangents to the circumcircles of triangles <math>\triangle APC</math> and <math>\triangle BCP</math> meet at point <math>Q.</math> Prove that all points <math>Q</math> lie on two fixed lines. | A point <math>P</math> moves along a circle <math>\Omega.</math> Let <math>A</math> and <math>B</math> be fixed points of <math>\Omega,</math> and <math>C</math> be an arbitrary point inside <math>\Omega.</math> The common external tangents to the circumcircles of triangles <math>\triangle APC</math> and <math>\triangle BCP</math> meet at point <math>Q.</math> Prove that all points <math>Q</math> lie on two fixed lines. | ||
<i><b>Solution</b></i> | <i><b>Solution</b></i> | ||
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| + | Denote <math>A' = AC \cup \Omega, B' = BC \cup \Omega, \omega = \odot APC, \omega' = \odot BPC,</math> | ||
| + | <math>\theta = \odot ACB', \theta' = \odot BCA', O</math> is the circumcenter of <math>\triangle APC, O'</math> is the circumcenter of <math>\triangle BPC.</math> | ||
| + | |||
| + | Let <math>K</math> and <math>L</math> be the midpoints of the arcs <math>CB'</math> of <math>\theta.</math> | ||
| + | |||
| + | Let <math>K'</math> and <math>L'</math> be the midpoints of the arcs <math>CA'</math> of <math>\theta'.</math> These points not depends from position of point <math>P.</math> | ||
| + | |||
| + | Suppose, <math>P \in \overset{\Large\frown} {B'ABA'}.</math> | ||
| + | <cmath>\angle A'BC = 2 \alpha = \angle B'AC \implies \angle D'BC = \angle DAC = \alpha \implies \angle DOC = \angle D'O'C = 2 \alpha.</cmath> | ||
| + | <cmath>O'D' = O'C, OC = OD \implies \triangle OCD \sim \triangle O'D'C \implies OC||O'D'.</cmath> | ||
Revision as of 00:45, 24 March 2024
Igor Fedorovich Sharygin (13/02/1937 - 12/03/2004, Moscow) - Soviet and Russian mathematician and teacher, specialist in elementary geometry, popularizer of science. He wrote many textbooks on geometry and created a number of beautiful problems. He headed the mathematics section of the Russian Soros Olympiads. After his death, Russia annually hosts the Geometry Olympiad for high school students. It consists of two rounds – correspondence and final. The correspondence round lasts 3 months.
The best problems of these Olympiads will be published. The numbering contains the year of the Olympiad and the serial number of the problem. Solutions are often different from the original ones.
2024, Problem 23
A point 
moves along a circle 
Let 
and 
be fixed points of 
and 
be an arbitrary point inside The common external tangents to the circumcircles of triangles 
and 
meet at point 
Prove that all points 
lie on two fixed lines.
Solution
Denote 
is the circumcenter of 
is the circumcenter of 
Let 
and 
be the midpoints of the arcs 
of 
Let 
and 
be the midpoints of the arcs 
of 
These points not depends from position of point 
Suppose, 
![\[\angle A'BC = 2 \alpha = \angle B'AC \implies \angle D'BC = \angle DAC = \alpha \implies \angle DOC = \angle D'O'C = 2 \alpha.\]](http://latex.artofproblemsolving.com/b/1/2/b12b05c16f5b59b607b8d05d28f29f71ff5e877f.png)
![\[O'D' = O'C, OC = OD \implies \triangle OCD \sim \triangle O'D'C \implies OC||O'D'.\]](http://latex.artofproblemsolving.com/8/2/e/82edaa3da3fd43b297350ef6865e0440ebbac316.png)
