Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 USAJMO Problems/Problem 4"

(Created page with "<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<...")
 
Line 1: Line 1:
 
<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 19:06, 19 April 2019

$(*)$ Let $ABC$ be a triangle with $\angle ABC$ obtuse. The [i]$A$-excircle[/i] is a circle in the exterior of $\triangle ABC$ that is tangent to side $\overline{BC}$ of the triangle and tangent to the extensions of the other two sides. Let $E$, $F$ be the feet of the altitudes from $B$ and $C$ to lines $AC$ and $AB$, respectively. Can line $EF$ be tangent to the $A$-excircle?

Solution

Instead of trying to find a synthetic way to describe $EF$ being tangent to the $A$-excircle (very hard), we instead consider the foot of the perpendicular from the $A$-excircle to $EF$, hoping to force something via the length of the perpendicular. It would be nice if there were an easier way to describe $EF$, something more closely related to the $A$-excircle; as we are considering perpendicularity, if we could generate a line parallel to $EF$, that would be good.

So we recall that it is well known that triangle $AEF$ is similar to $ABC$. This motivates reflecting $BC$ over the angle bisector at $A$ to obtain $B'C'$, which is parallel to $EF$ for obvious reasons.

Furthermore, as reflection preserves intersection, $B'C'$ is tangent to the reflection of the $A$-excircle over the $A$-angle bisector. But it is well-known that the $A$-excenter lies on the $A$-angle bisector, so the $A$-excircle must be preserved under reflection over the $A$-excircle. Thus $B'C'$ is tangent to the $A$-excircle.Yet for all lines parallel to $EF$, there are only two lines tangent to the $A$-excircle, and only one possibility for $EF$, so $EF = B'C'$.

Thus as $ABB'$ is isoceles, \[[ABC] = \frac{1}{2} \cdot AC \cdot BE = \frac{AC}{2} \cdot \sqrt{AB^2 - AE^2} = \frac{AC}{2} \cdot \sqrt{AB^2 - AB'^2} = \frac{AC}{2} \cdot \sqrt{AB^2 - AB^2} = 0,\] contradiction. $\square$ -alifenix-

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_USAJMO_Problems/Problem_4&oldid=105364"