2019 USAJMO Problems/Problem 4
Let 
be a triangle with 
obtuse. The [i]
-excircle[/i] 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?
Solution
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, ![\[[ABC] = \frac{1}{2} \cdot AC \cdot BE = \frac{AC}{2} \cdot \sqrt{AB^2 - AE^2} = \frac{AC}{2} \cdot \sqrt{AB^2 - AB'^2} = \frac{AC}{2} \cdot \sqrt{AB^2 - AB^2} = 0,\]](http://latex.artofproblemsolving.com/5/1/8/5181e3e69a28830c435bcf89cb3262fdd610b357.png)
contradiction. -alifenix-
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
See also
| 2019 USAJMO (Problems • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAJMO Problems and Solutions | ||
