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. (O_m) is circumcircle of MPX. (O_n) is circumcircle of NPX. L_m is a line passing through Q that intersects (O_m) at M₁, M₂. L_n is a line passing through Q that intersects (O_n) 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 L_m at M₄. Other than Q, circumcircle of NXQ intersect line L_n 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:

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Suppose (O') is circumcircle of triangle QM₄N₄.

Let Y and Z be the intersections with MN (L₁) of lines M₁M₂ (L_m) and N₁N₂ (L_n), respectively. Denote the circumcircles of MXQ and NXQ as (O_y), (O_z) respectively.

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

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

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

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

The radical center of (O), (O') and (O_m) is the intersection of the common chord of (O), (O_m) and the common chord of (O') and (O_m). By (1) and (2) it is point Y. So Y is the radical center of (O), (O') and (O_m) 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').

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