Difference between revisions of "2017 IMO Problems/Problem 4"
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+ | ==Problem== | ||
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Let <math>R</math> and <math>S</math> be different points on a circle <math>\Omega</math> such that <math>RS</math> is not a diameter. Let <math>\ell</math> be the tangent line to <math>\Omega</math> at <math>R</math>. Point <math>T</math> is such that <math>S</math> is the midpoint of the line segment <math>RT</math>. Point <math>J</math> is chosen on the shorter arc <math>RS</math> of <math>\Omega</math> so that the circumcircle <math>\Gamma</math> of triangle <math>JST</math> intersects <math>\ell</math> at two distinct points. Let <math>A</math> be the common point of <math>\Gamma</math> and <math>\ell</math> that is closer to <math>R</math>. Line <math>AJ</math> meets <math>\Omega</math> again at <math>K</math>. Prove that the line <math>KT</math> is tangent to <math>\Gamma</math>. | Let <math>R</math> and <math>S</math> be different points on a circle <math>\Omega</math> such that <math>RS</math> is not a diameter. Let <math>\ell</math> be the tangent line to <math>\Omega</math> at <math>R</math>. Point <math>T</math> is such that <math>S</math> is the midpoint of the line segment <math>RT</math>. Point <math>J</math> is chosen on the shorter arc <math>RS</math> of <math>\Omega</math> so that the circumcircle <math>\Gamma</math> of triangle <math>JST</math> intersects <math>\ell</math> at two distinct points. Let <math>A</math> be the common point of <math>\Gamma</math> and <math>\ell</math> that is closer to <math>R</math>. Line <math>AJ</math> meets <math>\Omega</math> again at <math>K</math>. Prove that the line <math>KT</math> is tangent to <math>\Gamma</math>. | ||
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We construct inversion which maps <math>KT</math> into the circle <math>\omega_1</math> and <math>\Gamma</math> into <math>\Gamma.</math> Than we prove that <math>\omega_1</math> is tangent to <math>\Gamma.</math> | We construct inversion which maps <math>KT</math> into the circle <math>\omega_1</math> and <math>\Gamma</math> into <math>\Gamma.</math> Than we prove that <math>\omega_1</math> is tangent to <math>\Gamma.</math> | ||
− | + | Quadrangle <math>RJSK</math> is cyclic <math>\implies \angle RSJ = \angle RKJ.</math> | |
− | + | ||
+ | Quadrangle <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> | 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>C = \omega \cap RT \implies RC^2 = RS \cdot RT.</math> Inversion with respect to <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> | 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 to <math>\omega</math> maps <math>K</math> into <math>K'</math>. |
<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</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> | ||
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==Solution 2== | ==Solution 2== | ||
[[File:2017 IMO 4a.png|500px|right]] | [[File:2017 IMO 4a.png|500px|right]] | ||
− | We use the | + | We use the tangent-chord theorem: the angle formed between a chord and a tangent line to a circle is equal to the inscribed angle on the other side of the chord. |
+ | |||
+ | Quadrangle <math>RJSK</math> is cyclic <math>\implies \angle RSJ = \angle RKJ.</math> | ||
− | + | Quadrangle <math>AJST</math> is cyclic <math>\implies \angle RSJ = \angle TAJ</math> <cmath>\implies AT||RK.</cmath> | |
− | + | (One can use Reim’s theorem – it is shorter way.) | |
Let <math>B</math> be symmetric to <math>A</math> with respect to <math>S \implies</math> | Let <math>B</math> be symmetric to <math>A</math> with respect to <math>S \implies</math> | ||
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<cmath>\angle SBK = \angle STK = \angle SAT \implies </cmath> | <cmath>\angle SBK = \angle STK = \angle SAT \implies </cmath> | ||
− | Inscribed angle of <math>\Gamma (\angle SAT)</math> is equal to angle between <math>KT</math> and chord <math>ST \implies KT</math> is tangent to <math>\Gamma | + | Inscribed angle of <math>\Gamma (\angle SAT)</math> is equal to angle between <math>KT</math> and chord <math>ST \implies</math> |
+ | |||
+ | <math>KT</math> is tangent to <math>\Gamma</math> by the inverse of tangent-chord theorem. | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2017|num-b=3|num-a=5}} |
Latest revision as of 01:41, 19 November 2023
Contents
Problem
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
Quadrangle is cyclic
Quadrangle is cyclic
We construct circle centered at
which maps
into
Let Inversion with respect to
swap
and
maps into
Let be the center of
Inversion with respect to 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 parallelogram.
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 to
line
is tangent to
vladimir.shelomovskii@gmail.com, vvsss
Solution 2
We use the tangent-chord theorem: the angle formed between a chord and a tangent line to a circle is equal to the inscribed angle on the other side of the chord.
Quadrangle is cyclic
Quadrangle is cyclic
(One can use Reim’s theorem – it is shorter way.)
Let be symmetric to
with respect to
is parallelogram.
is cyclic.
Inscribed angle of is equal to angle between
and chord
is tangent to
by the inverse of tangent-chord theorem.
vladimir.shelomovskii@gmail.com, vvsss
See Also
2017 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |