, 2015

Geometry Difficulty 6.6 National olympiad Prove it Romania

Let $ABCD$ be a cyclic quadrangle, let $\gamma$ be its circumcircle, and let $M$ be the midpoint of the arc $AB$ of $\gamma$ , not containing the vertices $C$ and $D$ . The line through $M$ and the point where the diagonals $AC$ and $BD$ cross one another, crosses $\gamma$ again at $N$ . Let $P$ and $Q$ be points on the side $CD$ such that $\angle AQD = \angle DAP$ and $\angle BPC = \angle CBQ$ . Show that the circles $NPQ$ and $\gamma$ are tangent to one another.
Flavian Georgescu

Solution

Since the lines $AC$ , $BD$ and $MN$ are concurrent, $(AM/MB)(BC/CN)(ND/DA) = 1$ , by Ceva's theorem in trigonometric form along with the sine law in the triangle $ABN$ ; and since $MA = MB$ , it follows that $CN/DN = BC/AD$ .

Next, the triangles $ADP$ and $QDA$ are similar, so $AD^2 = DP \cdot DQ$ . Similarly, $BC^2 = CP \cdot CQ$ , so, by the preceding, $(CN/DN)^2 = (CP/DP)(CQ/DQ)$ .

By a well-known theorem of Steiner, the lines $NP$ and $NQ$ are isogonal with respect to the lines $NC$ and $ND$ ; that is, the angles $CNP$ and $DNQ$ are congruent.

Finally, if $X$ is a point on the tangent $t$ at $N$ to $\gamma$ , other than $N$ , then a standard angle chase shows the angle $QNX$ congruent to one of the angles $CPN$ , $NPQ$ , depending on which side of $N$ the point $X$ lies on $t$ , so $t$ is also tangent at $N$ to the circle $NPQ$ . The conclusion follows.

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