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&=-\frac{1}{4}.\end{align*}</cmath> Now we show that <math>\frac{a_1+a_2+a_3}{3}</math> can be any real number. Let's say <cmath>\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}=k</cmath> for some real number <math>k.</math> Multiplying both sides by <math>2-\tan^2\theta</math> and rearranging yields a cubic in <math>\tan\theta.</math> Clearly this cubic has at least one real solution. As <math>\tan \theta</math> can take on any real number, all values of <math>k</math> are possible, and our answer is the line <cmath>\boxed{y=-\frac{1}{4}.}</cmath> Of course, as the denominator could equal 0, we must check <math>\tan \theta=\pm \frac{1}{\sqrt{3}}.</math> <cmath>3\tan \theta-\tan^3\theta=k(2-6\tan^2\theta).</cmath> The left side is nonzero, while the right side is zero, so these values of <math>\theta</math> do not contribute to any values of <math>k.</math> So, our answer remains the same. <math>\blacksquare</math> ~ Benq | &=-\frac{1}{4}.\end{align*}</cmath> Now we show that <math>\frac{a_1+a_2+a_3}{3}</math> can be any real number. Let's say <cmath>\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}=k</cmath> for some real number <math>k.</math> Multiplying both sides by <math>2-\tan^2\theta</math> and rearranging yields a cubic in <math>\tan\theta.</math> Clearly this cubic has at least one real solution. As <math>\tan \theta</math> can take on any real number, all values of <math>k</math> are possible, and our answer is the line <cmath>\boxed{y=-\frac{1}{4}.}</cmath> Of course, as the denominator could equal 0, we must check <math>\tan \theta=\pm \frac{1}{\sqrt{3}}.</math> <cmath>3\tan \theta-\tan^3\theta=k(2-6\tan^2\theta).</cmath> The left side is nonzero, while the right side is zero, so these values of <math>\theta</math> do not contribute to any values of <math>k.</math> So, our answer remains the same. <math>\blacksquare</math> ~ Benq | ||
− | ===Solution 4=== | + | ===Solution 4 (Simple, not bashy)=== |
+ | WLOG, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that | ||
== See also == | == See also == | ||
{{USAJMO newbox|year=2011|num-b=2|num-a=4}} | {{USAJMO newbox|year=2011|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:46, 27 January 2021
Contents
Problem
For a point in the coordinate plane, let
denote the line passing through
with slope
. Consider the set of triangles with vertices of the form
,
,
, such that the intersections of the lines
,
,
form an equilateral triangle
. Find the locus of the center of
as
ranges over all such triangles.
Solutions
Solution 1
Note that the lines are
respectively. It is easy to deduce that the three points of intersection are
The slopes of each side of this equilateral triangle are
and we want to find the locus of
Define the three complex numbers
for
. Then note that the slope - that is, the imaginary part divided by the real part - of all
is constant, say it is
. Then for
,
Rearranging, we get thator
Note that this is a cubic, and the roots are
and
which are all distinct, and so there are no other roots. Using Vieta's, we get that
and
Obviously all values of
are possible, and so our answer is the line
~ cocohearts
Solution 2
Note that all the points belong to the parabola
which we will denote
. This parabola has a focus
and directrix
which we will denote
. We will prove that the desired locus is
.
First note that for any point on
, the line
is the tangent line to
at
. This is because
contains
and because
. If you don't like calculus, you can also verify that
has equation
and does not intersect
at any point besides
. Now for any point
on
let
be the foot of the perpendicular from
onto
. Then by the definition of parabolas,
. Let
be the perpendicular bisector of
. Since
,
passes through
. Suppose
is any other point on
and let
be the foot of the perpendicular from
to
. Then in right
,
is a leg and so
. Therefore
cannot be on
. This implies that
is exactly the tangent line to
at
, that is
. So we have proved Lemma 1: If
is a point on
then
is the perpendicular bisector of
.
We need another lemma before we proceed. Lemma 2: If is on the circumcircle of
with orthocenter
, then the reflections of
across
,
, and
are collinear with
.
Proof of Lemma 2: Say the reflections of and
across
are
and
, and the reflections of
and
across
are
and
. Then we angle chase
where
is the measure of minor arc
on the circumcircle of
. This implies that
is on the circumcircle of
, and similarly
is on the circumcircle of
. Therefore
, and
. So
. Since
,
, and
are collinear it follows that
,
and
are collinear. Similarly, the reflection of
over
also lies on this line, and so the claim is proved.
Now suppose ,
, and
are three points of
and let
,
, and
. Also let
,
, and
be the midpoints of
,
, and
respectively. Then since
and
, it follows that
,
, and
are collinear. By Lemma 1, we know that
,
, and
are the feet of the altitudes from
to
,
, and
. Therefore by the Simson Line Theorem,
is on the circumcircle of
. If
is the orthocenter of
, then by Lemma 2, it follows that
is on
. It follows that the locus described in the problem is a subset of
.
Since we claim that the locus described in the problem is , we still need to show that for any choice of
on
there exists an equilateral triangle with center
such that the lines containing the sides of the triangle are tangent to
. So suppose
is any point on
and let the circle centered at
through
be
. Then suppose
is one of the intersections of
with
. Let
, and construct the ray through
on the same halfplane of
as
that makes an angle of
with
. Say this ray intersects
in a point
besides
, and let
be the perpendicular bisector of
. Since
and
, we have
. By the inscribed angles theorem, it follows that
. Also since
and
are both radii,
is isosceles and
. Let
be the reflection of
across
. Then
, and so
. It follows that
is on
, which means
is the perpendicular bisector of
.
Let intersect
in points
and
and let
be the point diametrically opposite to
on
. Also let
intersect
at
. Then
. Therefore
is a
right triangle and so
. So
and by the inscribed angles theorem,
. Since
it follows that
is and equilateral triangle with center
.
By Lemma 2, it follows that the reflections of across
and
, call them
and
, lie on
. Let the intersection of
and the perpendicular to
through
be
, the intersection of
and the perpendicular to
through
be
, and the intersection of
and the perpendicular to
through
be
. Then by the definitions of
,
, and
it follows that
for
and so
,
, and
are on
. By lemma 1,
,
, and
. Therefore the intersections of
,
, and
form an equilateral triangle with center
, which finishes the proof.
--Killbilledtoucan
Solution 3
Note that the lines are
respectively. It is easy to deduce that the three points of intersection are
The slopes of each side of this equilateral triangle are
and we want to find the locus of
We know that
for some
Therefore, we can use the tangent addition formula to deduce
and
Now we show that
can be any real number. Let's say
for some real number
Multiplying both sides by
and rearranging yields a cubic in
Clearly this cubic has at least one real solution. As
can take on any real number, all values of
are possible, and our answer is the line
Of course, as the denominator could equal 0, we must check
The left side is nonzero, while the right side is zero, so these values of
do not contribute to any values of
So, our answer remains the same.
~ Benq
Solution 4 (Simple, not bashy)
WLOG, let . Notice that
See also
2011 USAJMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.