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Russian Mathematical Olympiad

Geometry Difficulty 7.1 National Olympiad, round 2 Prove it Russia

Points $K$ and $L$ are chosen on a side $AB$ of a convex quadrilateral $ABCD$ , with $K$ lying between $A$ and $L$ . Similarly, points $M$ and $N$ are chosen on a side $CD$ , with $M$ lying between $C$ and $N$ . It appears that $AK = KN = DN$ and $BL = BC = CM$ . Given that $BCNK$ is a cyclic quadrilateral, prove that $ADML$ is also cyclic. (T. Zimanov, P. Kozhevnikov)

На стороне $AB$ выпуклого четырехугольника $ABCD$ взяты точки $K$ и $L$ (точка $K$ лежит между $A$ и $L$ ), а на стороне $CD$ взяты точки $M$ и $N$ (точка $M$ между $C$ и $N$ ). Известно, что $AK = KN = DN$ и $BL = BC = CM$ . Докажите, что если $BCNK$ — вписанный четырехугольник, то и $ADML$ тоже вписан.

Solution

In the case $AB \parallel CD$ , we have $BC = KN$ , so $AK = BL = CM = DN$ . Therefore, the quadrilateral $LMDA$ is obtained from $BCNK$ by a parallel translation by the vector $\vec{BL}$ .

Now suppose $AB$ and $CD$ are not parallel; let $P$ be the intersection point of the lines $AB$ and $CD$ . Since the quadrilateral $BCNK$ is cyclic, triangles $PBC$ and $PNK$ are similar; hence $\frac{PB}{BL} = \frac{PN}{NK} = \frac{PN}{ND}$ . Therefore, $BN \parallel LD$ (see figure). Similarly, $CK \parallel MA$ . From this, we get $\angle ALD = \angle KBN$ and $\angle KCN = \angle AMD$ .

Since the quadrilateral $BCNK$ is cyclic, $\angle KBN = \angle KCN$ . Therefore, $\angle ALD = \angle AMD$ , that is, $ADML$ is also cyclic.

Remark. There are other solutions; for example, from the equalities $\angle AKN = \angle NCB$ and $\angle DNK = \angle KBC$ , it follows that the quadrilaterals $BCML$ and $NKAD$ are similar, so $\angle BLM = \angle MDA$ .

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