2019 USAJMO Problems/Problem 4
Let be a triangle with obtuse. The -excircle is a circle in the exterior of that is tangent to side of the triangle and tangent to the extensions of the other two sides. Let , be the feet of the altitudes from and to lines and , respectively. Can line be tangent to the -excircle?
Instead of trying to find a synthetic way to describe being tangent to the -excircle (very hard), we instead consider the foot of the perpendicular from the -excircle to , hoping to force something via the length of the perpendicular. It would be nice if there were an easier way to describe , something more closely related to the -excircle; as we are considering perpendicularity, if we could generate a line parallel to , that would be good.
So we recall that it is well known that triangle is similar to . This motivates reflecting over the angle bisector at to obtain , which is parallel to for obvious reasons.
Furthermore, as reflection preserves intersection, is tangent to the reflection of the -excircle over the -angle bisector. But it is well-known that the -excenter lies on the -angle bisector, so the -excircle must be preserved under reflection over the -excircle. Thus is tangent to the -excircle.Yet for all lines parallel to , there are only two lines tangent to the -excircle, and only one possibility for , so .
Thus as is isoceles, contradiction. -alifenix-
The answer is no.
Suppose otherwise. Consider the reflection over the bisector of . This swaps rays and ; suppose and are sent to and . Note that the -excircle is fixed, so line must also be tangent to the -excircle.
Since is cyclic, we obtain , so . However, as is a chord in the circle with diameter , .
If then too, so then lies inside and cannot be tangent to the excircle.
The remaining case is when . In this case, is also a diameter, so is a rectangle. In particular . However, by the existence of the orthocenter, the lines and must intersect, contradiction.
The answer is .
Suppose for the sake of contradiction that it is possible for to be tangent to the -excircle. Call the tangency point , and let denote the contact points of with the -excircle, respectively. Let denote the semiperimeter of . By equal tangents, we have It is also well known that , so It is well known (by an easy angle chase) that , so we must have the ratio of similitude is . In particular, This results in which is absurd since is a right triangle. We reached a contradiction, so we are done. ~ Mathscienceclass
We claim that the answer is no. We proceed with contradiction. Suppose that is indeed tangent to the a-excenter. Define the point of tangency to be . Let to be the intersection of the a-excircle with the extension of and to be the intersection of the a-excircle with the extension of . Define to be the a-excenter. It is a well-known fact that lies on the angle bisector of . For convenience, let , . Notice that by Power of a Point: Therefore, adding, we see that: It would be rather nice if we could re-write . Indeed, we can: and similarly for : We now seek to re-write . Using the law of cosines: Therefore, Putting this all together, we see that: Now, we seek to write in terms of the side lengths of the triangle. Notice that: where is the semi-perimeter. We get that: Using the fact that , we have that: Therefore, Returning to our original problem: It is a well-known fact that , so: which implies that: which is a contradiction. Hence, our original assumption that is tangent to the a-excircle is incorrect. ~AopsUser101
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