Difference between revisions of "2021 IMO Problems/Problem 3"
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==Solution== | ==Solution== | ||
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[[File:2021 IMO 3i.png|450px|right]] | [[File:2021 IMO 3i.png|450px|right]] | ||
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> | 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> | + | 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{Lemma}</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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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 3.png|450px|right]] | ||
− | [[File:2021 IMO | + | [[File:2021 IMO 3j.png|450px|right]] |
<math>\boldsymbol{Lemma}</math> | <math>\boldsymbol{Lemma}</math> | ||
Revision as of 12:06, 28 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
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, www.deoma–cmd.ru
Video solution
https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems]