1997 USAMO Problems/Problem 2


$\triangle ABC$ is a triangle. Take points $D, E, F$ on the perpendicular bisectors of $BC, CA, AB$ respectively. Show that the lines through $A, B, C$ perpendicular to $EF, FD, DE$ respectively are concurrent.

Solution 1

Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By Carnot's Theorem, The three lines are concurrent if

$FA'^2-EA'^2+EC'^2-DC'^2+DB'^2-FB'^2 = AF^2-AE^2+CE^2-CD^2+BD^2-BF^2 = 0$

But this is clearly true, since D lies on the perpendicular bisector of BC, BD = DC.


Solution 2

We split this into two cases:

Case 1: $D,E,F$ are non-collinear

Observe that since $D,E,F$ lie on perpendicular bisectors, then we get that $DC=DB$, $EC=EA$, and $FA=FB$. This motivates us to construct a circle $\omega_1$ centered at $D$ with radius $DC$, and similarly construct $\omega_2$ and $\omega_3$ respectively for $E$ and $F$.

Now, clearly $\omega_1$ and $\omega_2$ intersect at $C$ and some other point. Now, we know that $DE$ is the line containing the two centers. So, the line perpendicular to $DE$ and passing through $C$ must be the radical axis of $\omega_1$ and $\omega_2$, which is exactly the line that the problem describes! We do this similarly for the others pairs of circles.

Now by the radical lemma, the pairwise radical axes of $\omega_1,\omega_2,\omega_3$ are concurrent, as desired, and they intersect at the radical center.

Case 2: $D,E,F$ are collinear

Now, we are drawing perpendicular lines from $A$, $B$, and $C$ onto the single line $DEF$. Clearly, these lines are parallel and concur at the point of infinity.

See Also

1997 USAMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png