## Five Concyclic Points

The applet below illustrates a problem proposed by Bui Quang Tuan:

 Given two lines L1, L2 that meet at point X. M, N are two points on L1. P, Q are two points on L2. (Om) is circumcircle of MPX. (On) is circumcircle of NPX. Lm is a line passing through Q that intersects (Om) at M1, M2. Ln is a line passing through Q that intersects (On) at N1, N2. Of course, M1, M2, N1, N2 are concyclic on a circle, say, (O). Circle (O) intersects line MN (L1) at M3, N3. Other than Q, circumcircle of MXQ intersects line Lm at M4. Other than Q, circumcircle of NXQ intersect line Ln at N4. Prove that five points Q, M3, N3, M4, N4 are concyclic!

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Solution ### Solution

Here is a proof by Bui Quang Tuan:

### 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.

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Suppose (O') is circumcircle of triangle QM4N4.

Let Y and Z be the intersections with MN (L1) of lines M1M2 (Lm) and N1N2 (Ln), respectively. Denote the circumcircles of MXQ and NXQ as (Oy), (Oz) respectively.

Y is the intersection of M1M2 and MX. Therefore Y is radical center of (O), (Om) and (Oy). This means that

 (1) Y is on common chord of (O), (Om).

Y is the intersection of QM4 with MX. Therefore Y is radical center of (O'), (Om) and (Oy) implying that

 (2) Y is on common chord of (O'), (Om).

The radical center of (O), (O') and (Om) is the intersection of the common chord of (O), (Om) and the common chord of (O') and (Om). By (1) and (2) it is point Y. So Y is the radical center of (O), (O') and (Om) meaning that the common chord of (O), (O') passes through Y.

Similarly, we can show that common chord of (O), (O') passes through Z. Since both Y and Z are on L1, L1 is the common chord of (O), (O'). Therefore M3, N3 are the intersections of the two circles (O), (O'). This exactly means that Q, M3, N3, M4, N4 are concyclic on (O'). 