55rd Ukrainian National Mathematical Olympiad - Fourth Round

Let lines $AB$ and $CD$ intersect at $E$. We have that $\angle BKC=60^{\circ }$. Let us prove that $BK>\frac{1}{2}BC$. Indeed, one of the angles $\angle KBC$ or $\angle KCB$ is at least $60^{\circ }$. WLOG, this angle is $\angle KBC$ (Fig. 35). Then in $△BMK$ $\angle KBM\ge 60^{\circ }=\angle BKC>\angle BKM$. So $MK>BM$, hence on the segment $BK$ there is point $O^{\prime }$, such that $MK\cdot M{O}^{\prime }=B{M}^2=M{C}^2$. Then circumcircles of $△B{O}^{\prime }K$ and $△C{O}^{\prime }K$ are tangent to $BC$. Hence, $\angle BKM=\angle O^{\prime }BC$, $\angle CKM=\angle O^{\prime }CB$. So $\angle B{O}^{\prime }C=180^{\circ }-\angle O^{\prime }BC-\angle O^{\prime }CB=180^{\circ }-\angle BKM-\angle MKC=180^{\circ }-60^{\circ }=120^{\circ }$



Fig. 35

Then point $O^{\prime }$ is on circumcircle of $△EBC$. Let line $B{O}^{\prime }$ intersect $CE$ at $D^{\prime }$, $C{O}^{\prime }$ intersect $BE$ at $A^{\prime }$. From the fact that $BEC{O}^{\prime }$ is cyclic we conclude that (Fig. 36) $$\angle B{O}^{\prime }E=\angle BCE=60^{\circ }=\angle EBC=\angle E{O}^{\prime }C.$$ Consider $△O^{\prime }EB$ and $△EB{D}^{\prime }$. Two pairs of angles are equal. Hence $\angle BE{O}^{\prime }=\angle E{D}^{\prime }B$. Then $\angle BE{O}^{\prime }=\angle BC{O}^{\prime }=\angle CK{O}^{\prime }=\angle E{D}^{\prime }B$. So point $D^{\prime }$ is on circumcircle of $△K{O}^{\prime }C$. Similarly, point $A^{\prime }$ is on the circle $△BK{O}^{\prime }$. Then $\angle A^{\prime }KB=\angle A^{\prime }{O}^{\prime }B=60^{\circ }=\angle C{O}^{\prime }{D}^{\prime }=\angle CK{D}^{\prime }$. It follows that points $A^{\prime }$, $K$ and $D^{\prime }$ are on the same line.

Point $O$ lies on segment $MK$. If this point is inside $M{O}^{\prime }$, then $A$ lies inside $A^{\prime }B$, $D^{\prime }$ inside $C{D}^{\prime }$. Then $K$ lies strictly outside $△EAD$, but by the problem $K$ is inside $AD$, a contradiction. Similarly, if point $O$ is outside $M{O}^{\prime }$, then $A$, $D$ are outside $E{A}^{\prime }$ and $E{D}^{\prime }$, so $K$ is strictly inside $△EAD$, again a contradiction. Hence $O=O^{\prime }$. Then $A=A^{\prime }$ and $D=D^{\prime }$. But this is already proved that $\angle A^{\prime }KB=\angle CK{D}^{\prime }=60^{\circ }$, which is needed, because $A=A^{\prime }$ and $D=D^{\prime }$.

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Alternative solution.

Let us draw external bisector of $\angle BKC$. Let it intersect lines $BA$ and $CD$ at $A^{\prime }$ and $D^{\prime }$ (Fig. 36). We draw on sides $△BKC$ outside equilateral triangles $BPC$, $BRK$ and $CQK$. Obviously $R$ and $Q$ are on $A^{\prime }{D}^{\prime }$, and $P$ is an intersection of $AB$ and $CD$.

Consider $△BKC$. We saw, that

$$\angle CBR=\angle CBK+60^{\circ }=180^{\circ }-(120^{\circ }-\angle CBK)=\angle KB{A}^{\prime }$$

It implies that $BR$ and $B{A}^{\prime }$ are isogonal, also $KR$ is external bisector. So $A^{\prime }$ and $R$ are isogonal, hence $C{A}^{\prime }$ and $CR$ are isogonal. Similarly we have that segments $B{D}^{\prime }$ and $BQ$ are isogonal. It is easy to see that $P$ - the point of intersection of tangents to circumcircle of $△BKC$, so $\angle PBC=\angle BKC=\angle PCB=60^{\circ }$. For this $KP$ is symmedian of $△BKC$, so $KP$ and $KM$ are isogonal. As it is known $KP$, $BQ$ and $CR$ intersect in the same point, namely at Fermat point, so as lines $KM$, $B{D}^{\prime }$ and $C{A}^{\prime }$ intersect in the same point. Let this be point $O^{\prime }$. Assume that $MO>M{O}^{\prime }$ (Fig. 37). Then points $A^{\prime }$ and $D^{\prime }$ are on sides $BA$ and $CD$, but in this case $AD$ and $A^{\prime }{D}^{\prime }$ don't intersect each other. Similarly if $MO<M{O}^{\prime }$. Hence $MO=M{O}^{\prime }\Rightarrow A=A^{\prime },D=D^{\prime }\Rightarrow \angle AKB=\angle DKC$.



Fig. 37