Problem
Let be an acute triangle, and let , and denote its -excenter, -excenter, and circumcenter, respectively. Points and are selected on such that and . Similarly, points and are selected on such that and .
Lines and meet at . Prove that and are perpendicular.
Solution
First solution.
Let denote the -excenter and the incenter. Then let denote the foot of the altitude from . Suppose the -excircle is tangent to , , at , , and let , , denote the reflections of across these points. Let denote the circumcenter of .
We begin with the following observation:
Claim: Points , , are collinear. Similarly, points , , are collinear and points , , are collinear.
Proof This basically follows from the "midpoints of altitudes" lemma. To see , , are collinear, recall first that passes through the midpoint of .
Now since , and and are the midpoints of and , it follows from the collinearity of , , that , , are collinear as well.
The other two claims follow in a dual fashion. For example, using the homothety taking the to -excircle, we find that bisects the altitude . (The homothety at will send to the point on the -excircle which is diametrically opposite the foot from to . This diameter of the -excircle is then mapped through homothety through onto the -altitude.) Then since , , are collinear the same argument now gives , , are collinear. The fact that , , are collinear is symmetric.
Observe that .Proceeding similarly on the other sides, we discover and are homothetic. Hence is the center of this homothety (in particular, , , , are collinear). Moreover, lies on the line joining to ,which is the Euler line of , so it passes through the nine-point center of , which is . Consequently, , , are collinear as well.
To finish, we need only prove that . In fact, we claim that is the radical axis of the circumcircles of and . Actually, is the radical center of these two circumcircles and the circle with diameter (which passes through and ). Analogously is the radical center of the circumcircles and the circle with diameter , and the proof is complete.
Second solution (barycentric, outline, Colin Tang). we are going to use barycentric coordinates to show that the line through perpendicular to is concurrent with and .
The displacement vector is proportional to , and so by strong perpendicularity criterion and doing a calculation gives the line
On the other hand, line has equation
and similarly for . Consequently, concurrence of these lines is equivalent to
which is a computation.
The problems on this page are the property of the MAA's American Mathematics Competitions