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Difference between revisions of "2017 IMO Problems/Problem 4"

(Created page with "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 lin...")
 
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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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==Solution==
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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>
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Quadrungle <math>RJSK</math> is cyclic <math>\implies \angle RSJ = \angle RKJ.</math>
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Quadrungle <math>AJST</math> is cyclic <math>\implies \angle RSJ = \angle TAJ \implies AT||RK.</math>
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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.</math>
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Inversion with respect <math>\omega</math>  maps <math>K</math> into <math>K'</math>.

Revision as of 10:25, 26 August 2022

Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.

Solution

We construct inversion which maps $KT$ into the circle $\omega_1$ and $\Gamma$ into $\Gamma.$ Than we prove that $\omega_1$ is tangent to $\Gamma.$

Quadrungle $RJSK$ is cyclic $\implies \angle RSJ = \angle RKJ.$ Quadrungle $AJST$ is cyclic $\implies \angle RSJ = \angle TAJ \implies AT||RK.$ We construct circle $\omega$ centered at $R$ which maps $\Gamma$ into $\Gamma.$ Let $C = \omega \cap RT \implies RC^2 = RS \cdot RT.$ Inversion with respect $\omega$ swap $T$ and $S \implies \Gamma$ maps into $\Gamma.$ Inversion with respect $\omega$ maps $K$ into $K'$.

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