# Line, Circle, and Fixed Points

The applet below illustrates Problem 3 from the 2007 Irish Mathematical Olympiad:

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.

The applet actually illustrates a slightly different problem. Points R and V are interchangeable in the sense that one uniquely determines the other. As it was more convenient to keep V on a given line than P on a given circle, I followed an equivalent formulation of the problem:

The point P is a fixed point on a circle and Q is a fixed point on a line. The point V (different from Q) is a variable point on the line. The circle through P, Q and V meets the given circle again at R. Show that the line VR passes through a fixed point.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Explanation

We want to establish the following assertion:

The point P is a fixed point on a circle and Q is a fixed point on a line. The point V (different from Q) is a variable point on the line. The circle through P, Q and V meets the given circle again at R. Show that the line VR passes through a fixed point.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

### Proof

Let's pick and fix one position of point V, say V', and consider three circles that circumscribe triangles PQV, PQV', and the given circle C(O). The three circles concur in P.

Three concurrent circles have several engaging properties, one of which plays a central role in the proof.

Given three circles concurrent in P. Let T be the point of intersection of the pair C(O) and C(PQV'). Then any chain of line segments VS (through R), SV' (through T), V'V (through Q) is closed.

(In the applet I actually picked V' at the intersection of SP with the given lines, making T = P and two circles C(O) and C(PQV') tangent - which in itself is not at all important.)

By the construction, V and V' are collinear with Q. By the above statement VR and V'P meet in point S on C(O). Since P and V' are fixed as is the given circle C(O), the point of intersection of PV' with C(O) is also fixed and does not depend on the position of V on line QV'!