Five Concyclic Points
The applet below illustrates a problem proposed by Bui Quang Tuan:
Given two lines L_{1}, L_{2} that meet at point X. M, N are two points on L_{1}. P, Q are two points on L_{2}. (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 M_{1}, M_{2}. L_{n} is a line passing through Q that intersects (O_{n}) at N_{1}, N_{2}. Of course, M_{1}, M_{2}, N_{1}, N_{2} are concyclic on a circle, say, (O). Circle (O) intersects line MN (L_{1}) at M_{3}, N_{3}. Other than Q, circumcircle of MXQ intersects line L_{m} at M_{4}. Other than Q, circumcircle of NXQ intersect line L_{n} at N_{4}. Prove that five points Q, M_{3}, N_{3}, M_{4}, N_{4} are concyclic! |
What if applet does not run? |
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Copyright © 1996-2018 Alexander BogomolnySolution
Here is a proof by Bui Quang Tuan:
What if applet does not run? |
Suppose (O') is circumcircle of triangle QM_{4}N_{4}.
Let Y and Z be the intersections with MN (L_{1}) of lines M_{1}M_{2} (L_{m}) and N_{1}N_{2} (L_{n}), respectively. Denote the circumcircles of MXQ and NXQ as (O_{y}), (O_{z}) respectively.
Y is the intersection of M_{1}M_{2} 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 QM_{4} 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 L_{1}, L_{1} is the common chord of (O), (O'). Therefore M_{3}, N_{3} are the intersections of the two circles (O), (O'). This exactly means that Q, M_{3}, N_{3}, M_{4}, N_{4} are concyclic on (O').
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