1997 USAMO Problems/Problem 2

Problem

$\triangle ABC$ 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.

QED

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.

Solution 3

Notice that $DEF$ and $ABC$ are orthologic, so the perpendiculars concur at the opposite center of orthology.

1997 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6
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1997 USAMO Problems/Problem 2

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also