# 2007 Irish Mathematical Olympiad, Problem 3

The point P is a fixed point on a circle and Q is a fixed point on a line. The point R is a variable point on the circle such that P, Q and R are not collinear. The circle through P, Q and R meets the line again at V. Show that the line VR passes through a fixed point.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to my lovely wife Quynh-Chau) Let C be the circle where the fixed point P is on. Link QR and VR and extend them to intercept C at U and S, respectively.

Let I and IN be the center and the diameter of the circumcircle of triangle PQR, respectively.

Now let ε = ∠SPU. We also have ε = ∠SRU = ∠VRQ = ∠VPQ,

and let θ = ∠PRQ = ∠PVQ = ∠PNQ.

We have ∠SPQ = ∠UPQ + ∠SPU (*)

But O and I are centers of the two circles and P and R are their intersections, we then have

∠IOP = ½ ∠ROP = ∠PUQ, and similarly ∠OIP = ∠PQU. The two triangles OPI and UPQ are then similar and ∠OPI = ∠UPQ.

Equation (*) can now be written as

∠SPQ = ∠OPI + ε = ∠OPI + ∠VPQ = ∠OPI + 180° - ∠PQV - ∠PVQ =

∠OPI + 180° - ∠PQV - ∠PNQ = ∠OPI + 180° - ∠PQV  (90° - ∠NPQ ) =

∠OPI + 180° - ∠PQV  90° + ∠NPQ =

∠OPQ + 90° - ∠PQV

Since both angles ∠OPQ and ∠PQV are constants, ∠SPQ is then constant, and VR passes through a fixed point.

There is another, apparently simpler solution.