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', | + | 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> 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> | ||
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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<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 the <math>\angle DFE = \angle DLE – \angle FDL = \angle AKC – \angle CBD</math> and | ||
+ | <math>\angle PDE = 180^o – \angle CDK – \angle CDP – \angle LDE</math> | ||
+ | <math>\angle PDE = 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>\ | + | <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> |
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'''vladimir.shelomovskii@gmail.com, vvsss, www.deoma–cmd.ru''' | '''vladimir.shelomovskii@gmail.com, vvsss, www.deoma–cmd.ru''' |
Revision as of 04:14, 23 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
.
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
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
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
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 the
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]