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 [i]<math>A</math>-excircle[/i] 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? | <math>(*)</math> Let <math>ABC</math> be a triangle with <math>\angle ABC</math> obtuse. The [i]<math>A</math>-excircle[/i] 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== | ||
+ | |||
+ | Instead of trying to find a synthetic way to describe <math>EF</math> being tangent to the <math>A</math>-excircle (very hard), we instead consider the foot of the perpendicular from the <math>A</math>-excircle to <math>EF</math>, hoping to force something via the length of the perpendicular. It would be nice if there were an easier way to describe <math>EF</math>, something more closely related to the <math>A</math>-excircle; as we are considering perpendicularity, if we could generate a line parallel to <math>EF</math>, that would be good. | ||
+ | |||
+ | So we recall that it is well known that triangle <math>AEF</math> is similar to <math>ABC</math>. This motivates reflecting <math>BC</math> over the angle bisector at <math>A</math> to obtain <math>B'C'</math>, which is parallel to <math>EF</math> for obvious reasons. | ||
+ | |||
+ | 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. <math>\square</math> -alifenix- |
Revision as of 18:06, 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-