Difference between revisions of "2019 USAJMO Problems/Problem 4"
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Furthermore, as reflection preserves intersection, <math>B'C'</math> is tangent to the reflection of the <math>A</math>-excircle over the <math>A</math>-angle bisector. But it is well-known that the <math>A</math>-excenter lies on the <math>A</math>-angle bisector, so the <math>A</math>-excircle must be preserved under reflection over the <math>A</math>-excircle. Thus <math>B'C'</math> is tangent to the <math>A</math>-excircle.Yet for all lines parallel to <math>EF</math>, there are only two lines tangent to the <math>A</math>-excircle, and only one possibility for <math>EF</math>, so <math>EF = B'C'</math>. | Furthermore, as reflection preserves intersection, <math>B'C'</math> is tangent to the reflection of the <math>A</math>-excircle over the <math>A</math>-angle bisector. But it is well-known that the <math>A</math>-excenter lies on the <math>A</math>-angle bisector, so the <math>A</math>-excircle must be preserved under reflection over the <math>A</math>-excircle. Thus <math>B'C'</math> is tangent to the <math>A</math>-excircle.Yet for all lines parallel to <math>EF</math>, there are only two lines tangent to the <math>A</math>-excircle, and only one possibility for <math>EF</math>, so <math>EF = B'C'</math>. | ||
− | Thus as <math>ABB'</math> is isoceles, <cmath>[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,</cmath> contradiction. | + | Thus as <math>ABB'</math> is isoceles, <cmath>[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,</cmath> contradiction. -alifenix- |
+ | |||
+ | {{MAA Notice}} | ||
+ | |||
+ | ==See also== | ||
+ | {{USAJMO newbox|year=2019|num-b=3|num-a=5}} |
Revision as of 18:08, 19 April 2019
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.
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 |