Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.7 National Olympiad Prove it JBMO

Problem:
Let $A$ and $P$ are the points of intersection of the circles $k_{1}$ and $k_{2}$ with centers $O$ and $K$ , respectively. Let also $B$ and $C$ be the symmetric points of $A$ with respect to $O$ and $K$ , respectively. A line through $A$ intersects the circles $k_{1}$ and $k_{2}$ at the points $D$ and $E$ , respectively. Prove that the centre of the circumcircle of the triangle $D E P$ lies on the circumcircle $O K P$ .

Solution

Solution:
The points $B, P, C$ are collinear, and
$\angle A P C = \angle A P B = 90^\circ$
Let $N$ be the midpoint of $D P$ .
So we have:
$\begin{aligned} & \angle N O P = \angle D A P \\ & = \angle E C P = \angle E C A + \angle A C P \end{aligned}$
Since $O K // B C$ and $O K$ is the bisector of $\angle A K P$ we get

Figure 6
$\angle A C P = O K P$
Also, since $A P \perp O K$ and $M K \perp P E$ we have that
$\angle A P E = \angle M K O$
The points $A, E, C, P$ are co-cyclic, and so $\angle E C A = \angle A P E$ .
Therefore, from (1), (2) and (3) we have that $\angle N O P = \angle M K P$ .
Thus $O, M, K$ and $P$ are co-cyclic.

It is sufficient to prove that the quadrilateral $M O P K$ is circumscrible.
Since $M O$ and $M K$ are perpendicular bisectors of the line segments $P D$ and $P E$ , respectively, we have
Therefore the quadrilateral $M O P K$ is circumscrible.

Figure 7

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