Difference between revisions of "2021 IMO Problems/Problem 3"
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<math>\hspace{13mm}E</math> is the point of the circle <math>\Omega_1.</math> | <math>\hspace{13mm}E</math> is the point of the circle <math>\Omega_1.</math> | ||
− | Let <math>T</math> be the point of intersection <math>\omega \cap \omega',</math> let <math>T'</math> be the point of intersection <math>\omega \cap \Omega.</math> Since the circles <math>\omega</math> and <math>\omega'</math> are inverse with respect to <math>\Omega_0,</math> then <math>T</math> lies on <math>\Omega_0,</math> and <math>P</math> lies on the perpendicular bisector of <math>DT.</math> The | + | Let Y be the radical center of the circles <math>\omega, \omega',</math> and <math>\omega_1.</math> Let <math>T</math> be the point of intersection <math>\omega \cap \omega',</math> let <math>T'</math> be the point of intersection <math>\omega \cap \Omega.</math> Since the circles <math>\omega</math> and <math>\omega'</math> are inverse with respect to <math>\Omega_0,</math> then <math>T</math> lies on <math>\Omega_0,</math> and <math>P</math> lies on the perpendicular bisector of <math>DT.</math> The power of a point <math>Y</math> with respect to the circles <math>\omega, \omega',</math> and <math>\Omega</math> are the same <math>(\boldsymbol{Lemma\hspace{3mm}2}),</math> so <math>DL\cdot DT = DL \cdot DT' \implies</math> the points <math>T</math> and <math>T'</math> coincide. |
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The centers of the circles <math>\omega</math> and <math>\Omega</math> (<math>O_1</math> and <math>O_2</math>) are located on the perpendicular bisector <math>DT'</math>, the point <math>P</math> is located on the perpendicular bisector <math>DT</math> and, therefore, the points <math>P, O_1,</math> and <math>O_2</math> lie on a line, that is, the lines <math>BC, EF,</math> and <math>O_1 O_2</math> are concurrent. | The centers of the circles <math>\omega</math> and <math>\Omega</math> (<math>O_1</math> and <math>O_2</math>) are located on the perpendicular bisector <math>DT'</math>, the point <math>P</math> is located on the perpendicular bisector <math>DT</math> and, therefore, the points <math>P, O_1,</math> and <math>O_2</math> lie on a line, that is, the lines <math>BC, EF,</math> and <math>O_1 O_2</math> are concurrent. | ||
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In the given problem <math>EF</math> and <math>BC</math> are antiparallel with respect to the sides of an angle <math>A,</math> quadrangle <math>BCEF</math> is concyclic. | In the given problem <math>EF</math> and <math>BC</math> are antiparallel with respect to the sides of an angle <math>A,</math> quadrangle <math>BCEF</math> is concyclic. | ||
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+ | <math>\boldsymbol{Lemma \hspace{3mm}2}</math> | ||
+ | |||
+ | Let the point <math>Y</math> be the radical center of the circles <math>\omega, \omega', \omega_1.</math> It has the same degree <math>\nu</math> with respect to these circles. | ||
+ | The common chords of the pairs of circles <math>A'B, AC, DT</math> intersect at this point. | ||
+ | |||
+ | <math>Y</math> has degree <math>\nu</math> with respect to <math>\Omega_1</math> since <math>A'B</math> is the radical axis of <math>\omega', \omega_1, \Omega_1.</math> | ||
+ | |||
+ | <math>Y</math> has degree <math>\nu</math> with respect to <math>\Omega_1</math> since <math>XE</math> containing <math>Y</math> is the radical axis of <math>\Omega</math> and <math>\Omega_1.</math> | ||
+ | |||
+ | Hence <math>Y</math> has degree <math>\nu</math> with respect to <math>\omega, \omega', \Omega.</math> | ||
'''Shelomovskii, vvsss, www.deoma-cmd.ru''' | '''Shelomovskii, vvsss, www.deoma-cmd.ru''' |
Revision as of 15:30, 22 July 2022
Problem
Let be an interior point of the acute triangle with so that . The point on the segment satisfies , the point on the segment satisfies , and the point on the line satisfies . Let and be the circumcentres of the triangles and respectively. Prove that the lines , , and are concurrent.
Solution
By statement point is located on the bisector of Let be the intersection point of the tangent to the circle at the point and the line is inverse to with respect to the circle centered at with radius Then the pairs of points and and are inverse with respect to , so the points and are collinear. Quadrilaterals containing the pairs of inverse points and and and are inscribed, is antiparallel to with respect to angle .
Consider the circles centered at centered at and
Denote . Then is cyclic), is cyclic, is antiparallel),
is the point of the circle
Let Y be the radical center of the circles and Let be the point of intersection let be the point of intersection Since the circles and are inverse with respect to then lies on and lies on the perpendicular bisector of The power of a point with respect to the circles and are the same so the points and coincide.
The centers of the circles and ( and ) are located on the perpendicular bisector , the point is located on the perpendicular bisector and, therefore, the points and lie on a line, that is, the lines and are concurrent.
Let be bisector of the triangle , point lies on The point on the segment satisfies . The point is symmetric to with respect to The point on the segment satisfies Then and are antiparallel with respect to the sides of an angle and Proof
Symmetry of points and with respect bisector implies Corollary
In the given problem and are antiparallel with respect to the sides of an angle quadrangle is concyclic.
Let the point be the radical center of the circles It has the same degree with respect to these circles. The common chords of the pairs of circles intersect at this point.
has degree with respect to since is the radical axis of
has degree with respect to since containing is the radical axis of and
Hence has degree with respect to
Shelomovskii, vvsss, www.deoma-cmd.ru
Video solution
https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems]