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Problem 1803

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.4 Prove it Baltic Way 2023 Shortlist · Baltic Way · 2023

Let $\omega_1$ and $\omega_2$ be circles with no common points. Points $M$ and $N$ are chosen on the circles $\omega_1$ and $\omega_2$ , respectively, such that the tangent to the circle $\omega_1$ at $M$ and the tangent to the circle $\omega_2$ at $N$ intersect at $P$ and $\triangle PMN$ is an isosceles triangle with apex $P$ . The circles $\omega_1$ and $\omega_2$ meet the segment $MN$ again at $A$ and $B$ , respectively. The line $PA$ meets the circle $\omega_1$ again at $C$ and the line $PB$ meets the circle $\omega_2$ again at $D$ . Prove that $\angle BCN = \angle ADM$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solutions — 2

Solution 1

Since $\triangle MPN$ is an isosceles triangle, we have $\angle PMA = \angle PMN = \angle MNP = \angle BNP$ . By tangent and chord theorem, $\angle MCA = \angle PMA = \angle BNP = \angle BDN$ .
Since $\angle MCP = \angle MNP$ , the quadrilateral $CMPN$ is cyclic. Analogously, from $\angle PDN = \angle PMN$ , we get that $NDMP$ is cyclic. Since $C$ and $D$ both lie on the circumcircle of $\triangle NPM$ , points $P, N, M, C$ and $D$ are concyclic.
From inscribed angles subtending arcs with the same length, we get that $\angle MDP = \angle MCP = \angle MNP = \angle PDN = \angle PMN = \angle PCN$ .
The power of $P$ with respect to $\omega_1$ gives us that $|PM|^2 = |PA| \cdot |PC|$ . The power of $P$ with respect to $\omega_2$ gives us that $|PN|^2 = |PB| \cdot |PD|$ . Since $|PM| = |PN|$ , the powers of $P$ with respect to $\omega_1$ and $\omega_2$ are equal ( $P$ lies on the radical axis). Hence, $|PA| \cdot |PC| = |PB| \cdot |PD|$ , which implies that $ABDC$ is cyclic. From inscribed angles subtending the arc $AB$ , we get that $\angle ACB = \angle ADB$ .
Hence, $\angle BCN = \angle ACN - \angle ACB = \angle MDB - \angle ADB = \angle MDA$ .

Solution 2

Since $\triangle MPN$ is an isosceles triangle, we have $\angle PMA = \angle PMN = \angle MNP = \angle BNP$ . By tangent and chord theorem, $\angle MCA = \angle PMA = \angle BNP = \angle BDN$ .
Since $\angle MCP = \angle MNP$ , the quadrilateral MPNC is cyclic, which means that $P$ lies on the circumcircle of $\triangle MNC$ . Since $\triangle MPN$ is isosceles, the perpendicular bisector of $MN$ passes through $P$ . Since the intersection point of the angle bisector and the perpendicular bisector of the opposite side of the triangle lies on the circumcircle, it follows that $CP$ bisects angle $\angle MCN$ . Hence, $\angle MCP = \angle PCN$ . Analogously, since $\angle PDN = \angle PMN$ , it follows that $NDMP$ is cyclic and the circumcircle of $\triangle MND$ , the perpendicular bisector of $MN$ and the angle bisector of $\angle MDN$ meet at $P$ . Hence, $\angle MDP = \angle PDN = \angle MCP = \angle PCN$ .
Now we continue as in the previous solution.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.