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 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> | + | 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>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. | ||
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The common chords of the pairs of circles <math>A'B, AC, DT</math> intersect at this point. | The common chords of the pairs of circles <math>A'B, AC, DT</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 the 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_1</math> since <math>XE</math> containing <math>Y</math> is the radical axis of <math>\Omega</math> and <math>\Omega_1.</math> | + | <math>Y</math> has the power <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 power <math>\nu</math> with respect to <math>\omega, \omega', \Omega.</math> | + | Hence <math>Y</math> has the same power <math>\nu</math> with respect to <math>\omega, \omega', \Omega.</math> |
'''vladimir.shelomovskii@gmail.com, vvsss, www.deoma–cmd.ru''' | '''vladimir.shelomovskii@gmail.com, vvsss, www.deoma–cmd.ru''' |
Revision as of 17:46, 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 power
with respect to these circles.
The common chords of the pairs of circles
intersect at this point.
has the power
with respect to
since
is the radical axis of
has the power
with respect to
since
containing
is the radical axis of
and
Hence has the same power
with respect to
vladimir.shelomovskii@gmail.com, vvsss, www.deoma–cmd.ru
Video solution
https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems]