# 2005 Austrian Mathematical Olympiad, Problem 3

In an acute-angled triangle ABC two circles C_1 and C_2 are drawn whose diameters are the sides AC and BC. Let E be the foot of the altitude h_b on AC and let F be the foot of the altitude h_a on BC. Let L and N be the intersections of the line BE with the circle C_1 (L on the line BE) and let K and M be the intersections of the line AF with the circle C_2 (K on the line AF).

Show that KLMN is a cyclic quadrilateral.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Abdo & Christine Mouannes of San Diego) Let D be the foot of the altitude from C to AB. Since E is on the circle C_2, BE ┴ AC and because AC is also the diameter of C_1, AC is then the perpendicular bisector of LN. Therefore, CN = CL. Similarly, since BC is the diameter of C_2 and F is on circle C_2, BC is perpendicular bisector of KM and CK = CM.

Now all we need to do is prove CM = CN so that the four points K, L, M, and N will lie on a circle with center at C and radius CM= CN = CK = CL.

Since the two triangles AFC and BEC are similar, we have CF/CE = CA/CB or

CF × CB = CE × CA or CF(CF + BF) = CE(CE + AE) or

```			CF² + CF × BF = CE² + CE × AE 			(*)
```

But BMC is a right triangle at M and F its foot on BC, we have CF × BF = MF² and similarly CE × AE = NE²

Now, rewrite (*) as CF² + MF² = CE² + NE² or CM² = CN² and were done.

Extension of the problem:

Draw circle C_3 with center A and radius AN = AL and circle C_4 with center B and radius BM = BK. Let them intercept each other at J and G with G being inside the circles. We can conclude that the four points D, H, G and C are collinear since CM² = CN² = CG x CJ.