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,
contradiction. -alifenix-
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Solution 2
The answer is no. [asy] pair A, B, C, E, F, Ep, Fp, I, J, K, L; //A = dir(120); //B = 2*foot(origin, A, A+dir(250)) - A; //C = 2*foot(origin, A, A+dir(290)) - A; A = dir(155); B = dir(230); C = dir(310); E = foot(B, A, C); F = foot(C, A, B); I = incenter(A, B, C); J = 2*circumcenter(B, I, C) - I; K = foot(J, A, B); L = foot(J, A, C); Ep = reflect(A, I) * E; Fp = reflect(A, I) * F;
draw(B--K^^C--L, rgb(1, 0.7, 0.2)); //draw(unitcircle, dotted); draw(A--B^^A--C, rgb(0.7, 0.3, 0)); draw(B--C, heavycyan); draw(E--F, heavycyan); draw(Ep--Fp, heavycyan + dashed); filldraw(circle(J, abs(J-foot(J, B, C))), opacity(0.5) + pink, red); draw(B--E^^C--F, heavymagenta); draw(E--Ep^^F--Fp, dotted);
dot("", A, dir(110));
dot("
", B, dir(190));
dot("
", C, dir(50));
dot("
", E, dir(50));
dot("
", F, dir(190));
dot("
", Ep, dir(190));
dot("
", Fp, dir(50));
[/asy]
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.
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 |