Difference between revisions of "2021 IMO Problems/Problem 3"
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Let <math>D</math> be an interior point of the acute triangle <math>ABC</math> with <math>AB > AC</math> so that <math>\angle DAB= \angle CAD</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 FDA= \angle DBC</math>, and the point <math>X</math> on the line <math>AC</math> satisfies <math>CX=BX</math>. Let <math>O_1</math> and <math>O_2</math> be the circumcentres of the triangles <math>ADC</math> and <math>EXD</math> respectively. Prove that the lines <math>BC</math>, <math>EF</math>, and <math>O_1 O_2</math> are concurrent. | Let <math>D</math> be an interior point of the acute triangle <math>ABC</math> with <math>AB > AC</math> so that <math>\angle DAB= \angle CAD</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 FDA= \angle DBC</math>, and the point <math>X</math> on the line <math>AC</math> satisfies <math>CX=BX</math>. Let <math>O_1</math> and <math>O_2</math> be the circumcentres of the triangles <math>ADC</math> and <math>EXD</math> respectively. Prove that the lines <math>BC</math>, <math>EF</math>, and <math>O_1 O_2</math> are concurrent. | ||
− | ==Video solution== | + | ==Solution== |
+ | [[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''' | ||
+ | |||
+ | == 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 |