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Bulgarian Mathematical Competitions

Geometry Difficulty 7.1 National Olympiad, round 2 Prove it Bulgaria

Problem:

Let $M$ and $N$ be arbitrary points on the side $AB$ of a triangle $ABC$ such that $M$ lies between $A$ and $N$ . The line through $M$ parallel to $AC$ meets the circumcircle of $\triangle MNC$ at point $P$ , and the line through $M$ parallel to $NC$ meets the circumcircle of $\triangle AMC$ at point $Q$ . Analogously, the line through $N$ parallel to $BC$ meets the circumcircle of $\triangle MNC$ at point $K$ and the line through $N$ parallel to $MC$ meets the circumcircle of $\triangle BNC$ at point $L$ . Prove that:

a) the points $P$ , $Q$ and $C$ are collinear;

b) the points $P$ , $Q$ , $K$ and $L$ are concyclic if and only if $AM = BN$ .

Solution

Solution:

$\begin{aligned} \angle QCP &= \angle QCA + \angle ACN + \angle NCP \\ &= \angle CNA + \angle ACN + \angle NAC \\ &= 180^\circ, \end{aligned}$

i.e. the points $P$ , $C$ and $Q$ are collinear.

b) Set $\angle ACM = \varphi$ , $\angle NCM = \psi$ and denote by $R_1$ and $R_2$ the radii of the circumcircles of $\triangle AMC$ and $\triangle MNC$ , respectively. Then by the Sine theorem we have
$\begin{aligned} QC &= 2R_1 \sin \angle QMC = 2R_1 \sin \psi \\ CP &= 2R_2 \sin \angle PMC = 2R_2 \sin \varphi \\ AM &= 2R_1 \sin \varphi \\ BM &= 2R_2 \sin \psi \end{aligned}$
Therefore $PC \cdot CQ = 4R_1 R_2 \sin \varphi \sin \psi = AM \cdot BM$ .

As in a) we see that the points $L$ , $C$ and $K$ are collinear. Analogously we have $CK \cdot CL = AN \cdot BN$ . Since the circumcircle of $\triangle MNC$ passes through $K$ , $C$ and $P$ , the lines $PQ$ and $KL$ do not coincide. Therefore the points $P$ , $Q$ , $K$ and $L$ are cyclic if and only if
$CP \cdot CQ = CL \cdot CK \Longleftrightarrow AM \cdot BM = AN \cdot BN \Longleftrightarrow AM = BN.$

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