APMO 98/4 solution

10th APMO 1998

Problem 4

ABC is a triangle. AD is an altitude. X lies on the circle ABD and Y lies on the circle ACD. X, D and Y are collinear. M is the midpoint of XY and M' is the midpoint of BC. Prove that MM' is perpendicular to AM

Solution

Take P, Q so that PADB, AQCD are rectangles. Let N be the midpoint of PQ. Then PD is a diameter of the circumcircle of ABC, so PX is perpendicular to XY. Similarly, QY is perpendicular to XY. N is the midpoint of PQ and M' the midpoint of XY, so NM is parallel to PX and hence perpendicular to XY. NADM' is a rectangle, so ND is a diameter of its circumcircle and M must lie on the circumcircle. But AM' is also a diameter, so ∠AMM' = 90^o.

Thanks to Michael Lipnowski for the above. My original solution is below.

Let P be the circumcenter of ABD and Q the circumcenter of ADC. Let R be the midpoint of AM'. P and Q both lie on the perpendicular bisector of AD, which is parallel to BC and hence also passes through R. We show first that R is the midpoint of PQ.

Let the feet of the perpendiculars from P, Q, R to BC to P', Q', R' respectively. It is sufficient to show that . BP' = BD/2. BR' = BM' + M'R' = (BD + DC)/2 + M'D/2 = (BD + DC)/2 + ( (BD + DC)/2 - DC)/2 = 3BD/4 + DC/4, so P'R' = (BD + DC)/4. Q'C = DC/2, so BQ' = BD + DC/2 and P'Q' = (BD + DC)/2 = 2P'R'.

Now the circumcircle centre P meets XY in X and D, and the circumcircle centre Q meets XY in D and Y. Without loss of generality we may take XD >= DY. Put XD = 4x, DY = 4y. The circle center R through A, M' and D meets XY in a second point, say M''. Let the feet of the perpendiculars from P, Q, R to XY be P'', Q'', R'' respectively. So on XY we have, in order, X, P'', M'', R'', D, Q'', Y. Since R is the midpoint of PQ, R'' is the midpoint of P''Q''. Now P'' is the midpoint of XD and Q'' is the midpoint of DY, so P''Q'' = XY/2 = 2(x+y), so R''Q'' = x+y. But DQ'' = 2y, so R''D = x-y. R'' is the midpoint of M''D, so M''D = 2(x-y) and hence M''Y = M''D + DY = 2(x-y) + 4y = 2(x+y) = XY/2. So M'' is just M the midpoint of XY. Now AM' is a diameter of the circle center R, so AM is perpendicular to MM'.

10th APMO 1998