| − | [[File:2021 IMO 3j.png|450px|right]] [[File:2021 IMO 3i.png|450px|right]] We prove that circles <math>ACD, EXD</math> and <math>\Omega_0</math> centered at <math>P</math> (the intersection point <math>BC</math> and <math>EF)</math> have a common chord. Let <math>P</math> be the intersection point of the tangent to the circle <math>\omega_2 = BDC</math> at the point <math>D</math> and the line <math>BC, A'</math> is inverse to <math>A</math> with respect to the circle <math>\Omega_0</math> centered at <math>P</math> with radius <math>PD.</math> Then the pairs of points <math>F</math> and <math>E, B</math> and <math>C</math> are inverse with respect to <math>\Omega_0</math>, so the points <math>F, E,</math> and <math>P</math> are collinear. Quadrilaterals containing the pairs of inverse points <math>B</math> and <math>C, E</math> and <math>F, A</math> and <math>A'</math> are inscribed, <math>FE</math> is antiparallel to <math>BC</math> with respect to angle <math>A</math> (see <math>\boldsymbol{Claim}</math>). Consider the circles <math>\omega = ACD</math> centered at <math>O_1, \omega' = A'BD,</math> <math>\omega_1 = ABC, \Omega = EXD</math> centered at <math>O_2 , \Omega_1 = A'BX,</math> and <math>\Omega_0.</math> Denote <math>\angle ACB = \gamma</math>. Then <math>\angle BXC = \angle BXE = \pi – 2\gamma,</math> <math>\angle AA'B = \gamma (AA'CB</math> is cyclic), <math>\angle AA'E = \pi – \angle AFE = \pi – \gamma (AA'EF</math> is cyclic, <math>FE</math> is antiparallel), <math>\angle BA'E = \angle AA'E – \angle AA'B = \pi – 2\gamma = \angle BXE \implies</math> <math>\hspace{13mm}E</math> is the point of the circle <math>\Omega_1.</math> Let the point <math>Y</math> be the radical center of the circles <math>\omega, \omega', \omega_1.</math> It has the same power <math>\nu</math> with respect to these circles. The common chords of the pairs of circles <math>A'B, AC, DT,</math> where <math>T = \omega \cap \omega',</math> intersect at this point. <math>Y</math> has power <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 power <math>\nu</math> with respect to <math>\Omega</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 power <math>\nu</math> with respect to <math>\omega, \omega', \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>DY \cdot YT = DY \cdot YT' \implies</math> the points <math>T</math> and <math>T'</math> coincide. 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. [[File:2021 IMO 3.png|450px|right]] [[File:2021 IMO 3j.png|450px|right]] <math>\boldsymbol{Claim}</math> Let <math>AK</math> be bisector of the triangle <math>ABC</math>, point <math>D</math> lies on <math>AK.</math> The point <math>E</math> on the segment <math>AC</math> satisfies <math>\angle ADE= \angle BCD</math>. The point <math>F</math> on the segment <math>AB</math> satisfies <math>\angle ADF= \angle CBD.</math> Let <math>P</math> be the intersection point of the tangent to the circle <math>BDC</math> at the point <math>D</math> and the line <math>BC.</math> Let the circle <math>\Omega_0</math> be centered at <math>P</math> and has the radius <math>PD.</math> Then the pairs of points <math>F</math> and <math>E, B</math> and <math>C</math> are inverse with respect to <math>\Omega_0</math> and <math>EF</math> and <math>BC</math> are antiparallel with respect to the sides of an angle <math>A.</math> <math>\boldsymbol{Proof}</math> Let the point <math>E'</math> is symmetric to <math>E</math> with respect to bisector <math>AK, E'L || BC.</math> Symmetry of points <math>E</math> and <math>E'</math> implies <math>\angle AEL = \angle AE'L.</math> <cmath>\angle DCK = \angle E'DL, \angle DKC = \angle E'LD \implies</cmath> <cmath> \triangle DCK \sim \triangle E'DL \implies \frac {E'L}{KD}= \frac {DL}{KC}.</cmath> <cmath>\triangle ALE' \sim \triangle AKB \implies \frac {E'L}{BK}= \frac {AL}{AK}\implies</cmath> <cmath> \frac {AL}{DL} = \frac {AK \cdot DK}{BK \cdot KC}.</cmath> Similarly, we prove that <math>FL</math> and <math>BC</math> are antiparallel with respect to angle <math>A,</math> and the points <math>L</math> in triangles <math>\triangle EDL</math> and <math>\triangle FDL</math> coincide. Hence, <math>FE</math> and <math>BC</math> are antiparallel and <math>BCEF</math> is cyclic. Note that <math>\angle DFE = \angle DLE – \angle FDL = \angle AKC – \angle CBD</math> and <math>\angle PDE = 180^o – \angle CDK – \angle CDP – \angle LDE = 180^o – (180^o – \angle AKC – \angle BCD) – \angle CBD – \angle BCD</math> <math>\angle PDE = \angle AKC – \angle CBD = \angle DFE,</math> so <math>PD</math> is tangent to the circle <math>DEF.</math> <math>PD^2 = PC \cdot PB = PE \cdot PF,</math> that is, the points <math>B</math> and <math>C, E</math> and <math>F</math> are inverse with respect to the circle <math>\Omega_0.</math> '''vladimir.shelomovskii@gmail.com, vvsss'''
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