Sharygin Olympiads, the best

Igor Fedorovich Sharygin (13/02/1937 - 12/03/2004, Moscow) - Soviet and Russian mathematician and teacher, specialist in elementary geometry, popularizer of science. He wrote many textbooks on geometry and created a number of beautiful problems. He headed the mathematics section of the Russian Soros Olympiads. After his death, Russia annually hosts the Geometry Olympiad for high school students. It consists of two rounds – correspondence and final. The correspondence round lasts 3 months.

The best problems of these Olympiads will be published. The numbering contains the year of the Olympiad and the serial number of the problem. Solutions are often different from the original ones.

2024, Problem 23

A point $P$ moves along a circle $\Omega.$ Let $A$ and $B$ be fixed points of $\Omega,$ and $C$ be an arbitrary point inside $\Omega.$

The common external tangents to the circumcircles of triangles $\triangle APC$ and $\triangle BCP$ meet at point $Q.$

Prove that all points $Q$ lie on two fixed lines.

Solution

Denote $A' = AC \cap \Omega, B' = BC \cap \Omega, \omega = \odot APC, \omega' = \odot BPC.$ $\theta = \odot ACB', \theta' = \odot BCA'.$

$O$ is the circumcenter of $\triangle APC, O'$ is the circumcenter of $\triangle BPC.$

Let $K$ and $L$ be the midpoints of the arcs $\overset{\Large\frown}{CB'}$ of $\theta.$

Let $K'$ and $L'$ be the midpoints of the arcs $\overset{\Large\frown}{CA'}$ of $\theta'.$

These points not depends from position of point $P.$

Suppose, $P \in \overset{\Large\frown} {B'ABA'} ($ see diagram). $\angle A'BC = 2 \alpha = \angle B'AC \implies \angle D'BC = \angle DAC = \alpha \implies \angle DOC = \angle D'O'C = 2 \alpha.$ $O'D' = O'C, OC = OD \implies \triangle OCD \sim \triangle O'D'C \implies OC||O'D'.$ Let $F= CD \cup OO' \implies \frac {FO}{FO'} = \frac {OC}{O'D'} \implies Q = F.$ $\angle LCB' = \alpha = \angle B'BL' \implies LC || L'B.$ Similarly, $AL || CL' \implies \triangle DLC \sim \triangle CL'D' \implies \frac {LC}{L'D'} = \frac {DC}{CD'} = \frac {OC}{O'D'}.$

Let $F' = LL' \cap DD' \implies \frac {F'C}{F'D'} = \frac {LC}{L'D'} = \frac {OC}{O'D'}= \frac {FC}{FD'} \implies F' = F.$

Therefore $Q \in LL'.$ Similarly, if $P \in \overset{\Large\frown} {B'A'}$ then $Q \in KK'.$

vladimir.shelomovskii@gmail.com, vvsss

2024, Problem 22

A segment $AB$ is given. Let $C$ be an arbitrary point of the perpendicular bisector to $AB, O$ be the point on the circumcircle of $\triangle ABC$ opposite to $C,$ and an ellipse centered at $O$ touche $AB, BC, CA.$

Find the locus of touching points $P$ of the ellipse with the line $BC.$

Solution

Denote $M$ the midpoint $AB, D$ the point on the line $CO, DO = MO, \alpha = \angle CBM, b = OM.$

$\angle CBO = 90^\circ \implies \angle COB = \alpha, MB = b \tan \alpha,$ $CB = \frac {b \sin \alpha}{\cos^2 \alpha}, CO = \frac {b} {\cos^2 \alpha}, CN = b \left (1 + \frac {1} {\cos^2 \alpha} \right ).$ In order to find the ordinate of point $P,$ we perform an affine transformation (compression along axis $AB)$ which will transform the ellipse $MPD$ into a circle with diameter $MD.$ The tangent of the $CP$ maps into the tangent of the $CE, E = \odot CBO \cup \odot MD, PF \perp CO.$ $\angle OEF = \angle ECO \implies OF = OE \sin \angle OEF = OE \sin \angle ECO = b \cos^2 \alpha.$ $CP = \frac {CF}{\sin \alpha} = \frac {b}{\sin \alpha}\left ( \frac {1} {\cos^2 \alpha} - \cos^2 \alpha \right ) = b \sin \alpha \left ( \frac {1}{\cos^2 \alpha } + 1 \right).$ $\frac {CP}{CD} = \sin \alpha , \angle PCD = 90^\circ - \alpha \implies \angle CPD = 90^\circ.$ $BP = CP – CB = \sin \alpha$ Denote $Q = AB \cup DP \implies BQ = \frac {BP}{\cos \alpha} = b \tan \alpha = MB.$

So point $Q$ is the fixed point ( $P$ not depends from angle $\alpha, \angle BPQ = 90^\circ .$ ) Therefore point $P$ lies on the circle with diameter $BQ$ (except points $B$ and $Q.)$

vladimir.shelomovskii@gmail.com, vvsss

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Sharygin Olympiads, the best

2024, Problem 23

2024, Problem 22