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*,*N’Î
EFÇ
w
*_{3}.

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

Since *N’Î
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
*AN’E* each subtend arc *AD* of w
_{3}. Thus Ð
*AMB* = Ð
*AN’E.*

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