1997 USAMO Problems/Problem 2
Contents
Problem
is a triangle. Take points on the perpendicular bisectors of respectively. Show that the lines through perpendicular to 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
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: are non-collinear
Observe that since lie on perpendicular bisectors, then we get that , , and . This motivates us to construct a circle centered at with radius , and similarly construct and respectively for and .
Now, clearly and intersect at and some other point. Now, we know that is the line containing the two centers. So, the line perpendicular to and passing through must be the radical axis of and , 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 are concurrent, as desired, and they intersect at the radical center.
Case 2: are collinear
Now, we are drawing perpendicular lines from , , and onto the single line . Clearly, these lines are parallel and concur at the point of infinity.
See Also
1997 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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