Difference between revisions of "2019 USAJMO Problems/Problem 4"
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− | <math>(*)</math> Let <math>ABC</math> be a triangle with <math>\angle ABC</math> obtuse. The | + | <math>(*)</math> Let <math>ABC</math> be a triangle with <math>\angle ABC</math> obtuse. The <math>A</math>''-excircle'' is a circle in the exterior of <math>\triangle ABC</math> that is tangent to side <math>\overline{BC}</math> of the triangle and tangent to the extensions of the other two sides. Let <math>E</math>, <math>F</math> be the feet of the altitudes from <math>B</math> and <math>C</math> to lines <math>AC</math> and <math>AB</math>, respectively. Can line <math>EF</math> be tangent to the <math>A</math>-excircle? |
==Solution== | ==Solution== |
Revision as of 23:15, 19 April 2019
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?
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.
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 |