Difference between revisions of "2017 IMO Problems/Problem 4"
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Quadrungle <math>AJST</math> is cyclic <math>\implies \angle RSJ = \angle TAJ \implies AT||RK.</math> | Quadrungle <math>AJST</math> is cyclic <math>\implies \angle RSJ = \angle TAJ \implies AT||RK.</math> | ||
− | We construct circle <math>\omega</math> centered at <math>R</math> which maps <math>\Gamma</math> into <math>\Gamma.</math> Let <math>C = \omega \cap RT \implies RC^2 = RS \cdot RT.</math> Inversion with respect <math>\omega</math> swap <math>T</math> and <math>S \implies \Gamma</math> maps into <math>\Gamma (\Gamma = \Gamma').</math> Let <math>O</math> be the center of <math>\Gamma.</math> | + | We construct circle <math>\omega</math> centered at <math>R</math> which maps <math>\Gamma</math> into <math>\Gamma.</math> |
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+ | Let <math>C = \omega \cap RT \implies RC^2 = RS \cdot RT.</math> Inversion with respect <math>\omega</math> swap <math>T</math> and <math>S \implies \Gamma</math> maps into <math>\Gamma (\Gamma = \Gamma').</math> | ||
+ | |||
+ | Let <math>O</math> be the center of <math>\Gamma.</math> | ||
Inversion with respect <math>\omega</math> maps <math>K</math> into <math>K'</math>. | Inversion with respect <math>\omega</math> maps <math>K</math> into <math>K'</math>. | ||
− | <math>K</math> belong <math>KT \implies</math> circle <math>K'SR</math> is the image of <math>KT</math>. Let <math>Q</math> be the center of | + | <math>K</math> belong <math>KT \implies</math> circle <math>K'SR = \omega_1</math> is the image of <math>KT</math>. Let <math>Q</math> be the center of <math>\omega_1.</math> |
<math>K'T</math> is the image of <math>\Omega</math> at this inversion, <math>l = AR</math> is tangent line to <math>\Omega</math> at <math>R,</math> so <math>K'T||AR.</math> | <math>K'T</math> is the image of <math>\Omega</math> at this inversion, <math>l = AR</math> is tangent line to <math>\Omega</math> at <math>R,</math> so <math>K'T||AR.</math> | ||
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<math>S</math> is the midpoint of <math>RT \implies S</math> is the center of symmetry of <math>ATK'R \implies</math> | <math>S</math> is the midpoint of <math>RT \implies S</math> is the center of symmetry of <math>ATK'R \implies</math> | ||
<math>\triangle RSK'</math> is symmetrical to <math>\triangle TSA</math> with respect to <math>S \implies</math> | <math>\triangle RSK'</math> is symmetrical to <math>\triangle TSA</math> with respect to <math>S \implies</math> | ||
− | + | <math>\omega_1</math> is symmetrical to <math>\Gamma</math> with respect to <math>S \implies</math> | |
<math>O</math> is symmetrycal <math>Q</math> with respect to <math>S, S</math> lies on <math>\Gamma</math> and on <math>|omega_1 \implies</math> | <math>O</math> is symmetrycal <math>Q</math> with respect to <math>S, S</math> lies on <math>\Gamma</math> and on <math>|omega_1 \implies</math> | ||
− | <math>\Gamma</math> is tangent <math>\omega_1 \implies S\Gamma</math> is tangent <math> | + | <math>\Gamma</math> is tangent <math>\omega_1 \implies S\Gamma</math> is tangent <math>TK.</math> |
Revision as of 11:09, 26 August 2022
Let and
be different points on a circle
such that
is not a diameter. Let
be the tangent line to
at
. Point
is such that
is the midpoint of the line segment
. Point
is chosen on the shorter arc
of
so that the circumcircle
of triangle
intersects
at two distinct points. Let
be the common point of
and
that is closer to
. Line
meets
again at
. Prove that the line
is tangent to
.
Solution
We construct inversion which maps into the circle
and
into
Than we prove that
is tangent to
Quadrungle is cyclic
Quadrungle
is cyclic
We construct circle centered at
which maps
into
Let Inversion with respect
swap
and
maps into
Let be the center of
Inversion with respect maps
into
.
belong
circle
is the image of
. Let
be the center of
is the image of
at this inversion,
is tangent line to
at
so
is image K at this inversion
is parallelogramm.
is the midpoint of
is the center of symmetry of
is symmetrical to
with respect to
is symmetrical to
with respect to
is symmetrycal
with respect to
lies on
and on
is tangent
is tangent