Difference between revisions of "Sharygin Olympiads, the best"
(Created page with "<i><b>Igor Fedorovich Sharygin</b></i> (13/02/1937 - 12/03/2004, Moscow) - Soviet and Russian mathematician and teacher, specialist in elementary geometry, popularizer of scie...") |
(→2024, Problem 23) |
||
Line 4: | Line 4: | ||
==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> | ||
+ | |||
+ | 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 01:45, 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,