Difference between revisions of "2021 IMO Problems/Problem 3"
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==Solution== | ==Solution== | ||
− | [[File:2021 IMO | + | [[File:2021 IMO 3j.png|450px|right]] |
− | [[File:2021 IMO | + | [[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> | + | 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> | Consider the circles <math>\omega = ACD</math> centered at <math>O_1, \omega' = A'BD,</math> | ||
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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 | + | 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. | 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> | ||
− | <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>\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 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>\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> | <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> | |
− | |||
− | ==Video solution== | + | |
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | == Video solution == | ||
https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems] | https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems] | ||
+ | |||
+ | == See also == | ||
+ | {{IMO box|year=2021|num-b=2|num-a=4}} | ||
+ | |||
+ | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 09:29, 18 June 2023
Contents
[hide]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
We prove that circles and centered at (the intersection point and have a common chord.
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 (see ).
Consider the circles centered at centered at and
Denote . Then is cyclic), is cyclic, is antiparallel),
is the point of the circle
Let the point be the radical center of the circles It has the same power with respect to these circles. The common chords of the pairs of circles where intersect at this point. has power with respect to since is the radical axis of has power with respect to since containing is the radical axis of and Hence has power with respect to
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, 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 on the segment satisfies Let be the intersection point of the tangent to the circle at the point and the line Let the circle be centered at and has the radius
Then the pairs of points and and are inverse with respect to and and are antiparallel with respect to the sides of an angle
Let the point is symmetric to with respect to bisector Symmetry of points and implies Similarly, we prove that and are antiparallel with respect to angle and the points in triangles and coincide. Hence, and are antiparallel and is cyclic. Note that and so is tangent to the circle
that is, the points and and are inverse with respect to the circle
vladimir.shelomovskii@gmail.com, vvsss
Video solution
https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems]
See also
2021 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |