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1998 APMO Problem #4

Let ABC be a triangle and D the foot of the altitude from A. Let E and F be on a line passing though D such that AE is perpendicular to BE, AF is perpendicular to CF, and E and F are different from D. Let M and N be the midpoints of the line segments BC and EF, respectively. Prove that AN is perpendicular to NM.

 

Solution #1 by Tim Kirchner

Let w 1 be the circle with diameter AB (shown in green), let w 2 be the circle with diameter AC (shown in blue), and let w 3 be the circle with diameter AM (shown in red).

AE^ BE Þ EÎ w 1 and AF^ CF Þ FÎ w 2

Let l be the line containing D, E, F, and let D,NÎ EFÇ w 3.

(i.e. since l intersects w 3 at two points, one is D, call the other one N)

Since NÎ w 3, we know that AN^ MN.

We are given that N is the midpoint of EF, so it suffices to prove that N=N, that is to say that N is the midpoint of EF.

We now draw in the line l, giving us the following diagram:


Now, Ð ABC and Ð AEF each subtend arc AD of w 1. Thus Ð ABC = Ð AEF.

Similarly, Ð ACB and Ð AFE each subtend arc AD of w 2. Thus Ð ACB = Ð AFE.

Furthermore, Ð AMB and Ð ANE each subtend arc AD of w 3. Thus Ð AMB = Ð ANE.

From this, we know that D ABC ~ D AEF, and D ABM ~ D ABN.

Thus, since M is the midpoint of BC, N is the midpoint of EF.

Therefore, N = N.

\ NÎ w 3 Þ AN^ MN. ÿ