2017 IMO Problems/Problem 4

Revision as of 01:55, 2 September 2022 by Vvsss (talk | contribs) (Solution 2)

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 $\Gamma$.


2017 IMO 4.png

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

Quadrangle $RJSK$ is cyclic $\implies \angle RSJ = \angle RKJ.$

Quadrangle $AJST$ 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 to $\omega$ swap $T$ and $S \implies  \Gamma$ maps into $\Gamma (\Gamma = \Gamma').$

Let $O$ be the center of $\Gamma.$

Inversion with respect to $\omega$ maps $K$ into $K'$. $K$ belong $KT \implies$ circle $K'SR = \omega_1$ is the image of $KT$. Let $Q$ be the center of $\omega_1.$

$K'T$ is the image of $\Omega$ at this inversion, $l = AR$ is tangent line to $\Omega$ at $R,$ so $K'T||AR.$

$K'$ is image K at this inversion $\implies K \in RK' \implies RK'||AT \implies ARK'T$ is parallelogram.

$S$ is the midpoint of $RT \implies S$ is the center of symmetry of $ATK'R \implies$ $\triangle RSK'$ is symmetrical to $\triangle TSA$ with respect to $S \implies$ $\omega_1$ is symmetrical to $\Gamma$ with respect to $S \implies$ $O$ is symmetrycal $Q$ with respect to $S.$

$S$ lies on $\Gamma$ and on $\omega_1 \implies \Gamma$ is tangent to $\omega_1 \implies$ line $KT$ is tangent to $\Gamma.$

vladimir.shelomovskii@gmail.com, vvsss

Solution 2

2017 IMO 4a.png

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 $RJSK$ is cyclic $\implies \angle RSJ = \angle RKJ.$

Quadrangle $AJST$ is cyclic $\implies \angle RSJ = \angle TAJ$ \[\implies AT||RK.\]

(One can use Reim’s theorem – it is shorter way.)

Let $B$ be symmetric to $A$ with respect to $S \implies$ $ATBR$ is parallelogram. \[\angle KST = \angle SRK + \angle SKR = \angle KRA\] $\angle RBT = \angle RAT \implies \angle KST + \angle KBT = 180^\circ$ $\implies SKBT$ is cyclic. \[\angle SBK = \angle STK  = \angle SAT \implies\]

Inscribed angle of $\Gamma (\angle SAT)$ is equal to angle between $KT$ and chord $ST \implies$

$KT$ is tangent to $\Gamma$ by the inverse of tangent-chord theorem.

vladimir.shelomovskii@gmail.com, vvsss