Difference between revisions of "2019 USAJMO Problems/Problem 4"
(Added a new solution) |
(→Solution 3) |
||
Line 28: | Line 28: | ||
The answer is <math>\boxed{\text{no}}</math>. | The answer is <math>\boxed{\text{no}}</math>. | ||
− | Suppose for the sake of contradiction that it is possible for <math>EF</math> to be tangent to the <math>A</math>-excircle. Call the tangency point <math>T</math>, and let <math>S_1, S_2</math> denote the contact points of <math>AB, AC</math> with the <math>A</math>-excircle, respectively. Let <math>s</math> denote the semiperimeter of <math>ABC</math>. By equal tangents, we have <cmath>ET = ES_2, FT = FS_1 \implies EF = ES_2+FS_2</cmath>It is also well known that <math>AS_1 = AS_2 = \frac{s}{2}</math>, so <cmath>EF = ES_2+FS_2 = (AS_2-AE)+(AS_1-AF) = s-AE-AF \implies s=AE+AF+EF</cmath>It is well known (by an easy angle chase) that <math>\triangle AEF \sim \triangle ABC</math>, so we must have the ratio of similitude is <math>2</math>. In particular, <cmath>AB=2 \cdot AE, AC=2 \cdot AF</cmath>This results in <cmath>\angle ABE = 30^{\circ}, \angle CBF = 30^{\circ} \implies \angle EBC = 120^{\circ}</cmath>which is absurd since <math>\triangle BEC</math> is a right triangle. We reached a contradiction, so we are done. | + | Suppose for the sake of contradiction that it is possible for <math>EF</math> to be tangent to the <math>A</math>-excircle. Call the tangency point <math>T</math>, and let <math>S_1, S_2</math> denote the contact points of <math>AB, AC</math> with the <math>A</math>-excircle, respectively. Let <math>s</math> denote the semiperimeter of <math>ABC</math>. By equal tangents, we have <cmath>ET = ES_2, FT = FS_1 \implies EF = ES_2+FS_2</cmath>It is also well known that <math>AS_1 = AS_2 = \frac{s}{2}</math>, so <cmath>EF = ES_2+FS_2 = (AS_2-AE)+(AS_1-AF) = s-AE-AF \implies s=AE+AF+EF</cmath>It is well known (by an easy angle chase) that <math>\triangle AEF \sim \triangle ABC</math>, so we must have the ratio of similitude is <math>2</math>. In particular, <cmath>AB=2 \cdot AE, AC=2 \cdot AF</cmath>This results in <cmath>\angle ABE = 30^{\circ}, \angle CBF = 30^{\circ} \implies \angle EBC = 120^{\circ}</cmath>which is absurd since <math>\triangle BEC</math> is a right triangle. We reached a contradiction, so we are done. <math>\blacksquare</math> ~ Mathscienceclass |
==See also== | ==See also== | ||
{{MAA Notice}} | {{MAA Notice}} | ||
{{USAJMO newbox|year=2019|num-b=3|num-a=5}} | {{USAJMO newbox|year=2019|num-b=3|num-a=5}} |
Revision as of 17:03, 24 May 2020
Contents
[hide]Problem
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-
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.
Solution 3
The answer is .
Suppose for the sake of contradiction that it is possible for to be tangent to the
-excircle. Call the tangency point
, and let
denote the contact points of
with the
-excircle, respectively. Let
denote the semiperimeter of
. By equal tangents, we have
It is also well known that
, so
It is well known (by an easy angle chase) that
, so we must have the ratio of similitude is
. In particular,
This results in
which is absurd since
is a right triangle. We reached a contradiction, so we are done.
~ Mathscienceclass
See also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
2019 USAJMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |