Let $K$ be the midpoint of the median $AM$ of a triangle $ABC$. Points $X, Y$ lie on $AB, AC$, respectively, such that $\angle KXM =\angle ACB$, $AX>BX$ and similarly $\angle KYM =\angle ABC$, $AY>CY$. Prove that $B, C, X, Y$ are concyclic. Proposed by Mykhailo Shtandenko
Problem
Source: Ukraine MO 2023 11.6
Tags: geometry
05.04.2023 22:50
A really nice problem. Let $\omega$ be a circle through $M$ and tangent to the line parallel $\ell$ to $BC$ through $A$. Redefine $X$ and $Y$ as intersection of $\omega$ with $AB$ and $AC$ respectively. (We can see that $AX > BX$ and $AY > CY$). Because $M$ is the midpoint $BC$, then $-1=(B,C;M,\infty_{BC}) = (X,Y;M,A)$, hence $AXMY$ is harmonic, which means that $\angle KXM = \angle AXY = \angle ACB$, last equality is from tangency in $A$ and parallel lines $\ell$ and $BC$. Same with $\angle KYM = \angle AYX = \angle ABC$, hence $X$ and $Y$ are the same as given in the statement. But now by $\angle AXY = \angle ACB$ we have that $B,C,X,Y$ lie on a circle.
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05.04.2023 22:51
Remark: there is no need that both $AX>BX, AY>CY, X, Y$ must not be the midpoints of the sides, unless $AB=AC$. From the wording I guess you got the idea of solution! Best regards, sunken rock
05.04.2023 23:03
Hmm X is the Second intersection of circle KMN with AB where N is the midpoint of AB. Similarly define Y. From PoP we get AX = AM^2/AB and AY = AM^2/AC, so we have AX*AB = AY*AC hence done.
04.05.2023 09:37
Very cute, let $P=(A+B)/2$ and $Q=(A+C)/2$, note that $P,K,Q$ are collinear and $PQMB$ and $QPMC$ are parallelograms so $\measuredangle KQM=\measuredangle KYM$ and $\measuredangle KPM = \measuredangle KXM$ which means \[AX \cdot AB = 2AX \cdot AP = 2AK \cdot AM = 2 AY\cdot AQ = AY \cdot AC ~ \blacksquare\]
29.09.2023 19:40
Let $D$ and $E$ be the midpoints of $\overline{AB}$ and $\overline{AC}$, so $DKMX$ and $EKMY$ are cyclic. Therefore, $AD\cdot AX=AK\cdot AM=AE\cdot AY \implies AB\cdot AX=AC\cdot AY$; done. $\blacksquare$