Difference between revisions of "Sharygin Olympiads, the best"
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Therefore point <math>P</math> lies on the circle with diameter <math>BQ</math> (except points <math>B</math> and <math>Q.)</math> | Therefore point <math>P</math> lies on the circle with diameter <math>BQ</math> (except points <math>B</math> and <math>Q.)</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==2024, Problem 21== | ||
+ | [[File:2024 21 0.png|350px|right]] | ||
+ | A chord <math>PQ</math> of the circumcircle of a triangle <math>ABC</math> meets the sides <math>BC, AC</math> at points <math>A', B',</math> respectively. The tangents to the circumcircle at <math>A</math> and <math>B</math> meet at point <math>X,</math> and the tangents at points <math>P</math> and <math>Q</math> meets at point <math>Y.</math> The line <math>XY</math> meets <math>AB</math> at point <math>C'.</math> | ||
+ | |||
+ | Prove that the lines <math>AA', BB',</math> and <math>CC'</math> concur. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | WLOG, <math>P \in \overset{\Large\frown} {AC}.</math> | ||
+ | Denote <math>\Omega = \odot ABC, Z = BB' \cap AA', D = AQ \cap BP.</math> | ||
+ | |||
+ | Point <math>D</math> is inside <math>\Omega.</math> | ||
+ | |||
+ | We use Pascal’s theorem for quadrilateral <math>APQB</math> and get <math>D \in XY.</math> | ||
+ | |||
+ | We use projective transformation which maps <math>\Omega</math> to a circle and that maps the point <math>D</math> to its center. | ||
+ | |||
+ | From this point we use the same letters for the results of mapping. Therefore the segments <math>AQ</math> and <math>BP</math> are the diameters of <math>\Omega, C'D \in XY || AP \implies C'</math> is the midpoint <math>AB.</math> | ||
+ | |||
+ | <math>AB|| PQ \implies AB || B'A' \implies C' \in CZ \implies</math> preimage <math>Z</math> lies on preimage <math>CC'.\blacksquare</math> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 11:45, 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, see diagram). Let Similarly,
Let
Therefore Similarly, if then
vladimir.shelomovskii@gmail.com, vvsss
2024, Problem 22
A segment is given. Let be an arbitrary point of the perpendicular bisector to be the point on the circumcircle of opposite to and an ellipse centered at touche
Find the locus of touching points of the ellipse with the line
Solution
Denote the midpoint the point on the line
In order to find the ordinate of point we perform an affine transformation (compression along axis which will transform the ellipse into a circle with diameter The tangent of the maps into the tangent of the Denote
So point is the fixed point ( not depends from angle
Therefore point lies on the circle with diameter (except points and
vladimir.shelomovskii@gmail.com, vvsss
2024, Problem 21
A chord of the circumcircle of a triangle meets the sides at points respectively. The tangents to the circumcircle at and meet at point and the tangents at points and meets at point The line meets at point
Prove that the lines and concur.
Proof
WLOG, Denote
Point is inside
We use Pascal’s theorem for quadrilateral and get
We use projective transformation which maps to a circle and that maps the point to its center.
From this point we use the same letters for the results of mapping. Therefore the segments and are the diameters of is the midpoint
preimage lies on preimage
vladimir.shelomovskii@gmail.com, vvsss