Difference between revisions of "2011 USAJMO Problems/Problem 3"

Latest revision as of 12:37, 29 January 2021

Problem

For a point $P = (a, a^2)$ in the coordinate plane, let $\ell(P)$ denote the line passing through $P$ with slope $2a$ . Consider the set of triangles with vertices of the form $P_1 = (a_1, a_1^2)$ , $P_2 = (a_2, a_2^2)$ , $P_3 = (a_3, a_3^2)$ , such that the intersections of the lines $\ell(P_1)$ , $\ell(P_2)$ , $\ell(P_3)$ form an equilateral triangle $\triangle$ . Find the locus of the center of $\triangle$ as $P_1P_2P_3$ ranges over all such triangles.

Solutions

Solution 1

Note that the lines $l(P_1), l(P_2), l(P_3)$ are $y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,$ respectively. It is easy to deduce that the three points of intersection are $\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).$ The slopes of each side of this equilateral triangle are $2a_1,2a_2,2a_3,$ and we want to find the locus of $\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).$ Define the three complex numbers $w_n = 1+2a_ni$ for $n=1,2,3$ . Then note that the slope - that is, the imaginary part divided by the real part - of all $w_n^3$ is constant, say it is $k$ . Then for $n=1,2,3$ ,

$\begin{align*} \frac{\Im(w_n^3)}{\Re(w_n^3)} &= \frac{\Im((1+2a_ni)^3)}{\Re((1+2a_ni)^3)}\\ &= \frac{\Im(1+6a_ni-12a_n^2-8a_n^3i)}{\Re(1+6a_ni-12a_n^2-8a_n^3i)}\\ &= \frac{6a_n-8a_n^3}{1-12a_n^2}\\ &= k.\\ \end{align*}$

Rearranging, we get that $8a_n^3 -12ka_n^2-6a_n+k=0,$ or $a_n^3-\frac{3ka_n^2}2-\frac{3a_n}4+\frac k8=0.$ Note that this is a cubic, and the roots are $a_1,a_2$ and $a_3$ which are all distinct, and so there are no other roots. Using Vieta's, we get that $a_1+a_2+a_3=\frac{3k}2,$ and $a_1a_2+a_2a_3+a_3a_1=-\frac34.$ Obviously all values of $k$ are possible, and so our answer is the line $\boxed{y=-\frac{1}{4}.}$ $\blacksquare$

~ cocohearts

Solution 2

Note that all the points $P=(a,a^2)$ belong to the parabola $y=x^2$ which we will denote $p$ . This parabola has a focus $F=\left(0,\frac{1}{4}\right)$ and directrix $y=-\frac{1}{4}$ which we will denote $d$ . We will prove that the desired locus is $d$ .

First note that for any point $P$ on $p$ , the line $\ell(P)$ is the tangent line to $p$ at $P$ . This is because $\ell(P)$ contains $P$ and because $[\frac{d}{dx}] x^2=2x$ . If you don't like calculus, you can also verify that $\ell(P)$ has equation $y=2a(x-a)+a^2$ and does not intersect $y=x^2$ at any point besides $P$ . Now for any point $P$ on $p$ let $P'$ be the foot of the perpendicular from $P$ onto $d$ . Then by the definition of parabolas, $PP'=PF$ . Let $q$ be the perpendicular bisector of $\overline{P'F}$ . Since $PP'=PF$ , $q$ passes through $P$ . Suppose $K$ is any other point on $q$ and let $K'$ be the foot of the perpendicular from $K$ to $d$ . Then in right $\Delta KK'P'$ , $KK'$ is a leg and so $KK'<KP'=KF$ . Therefore $K$ cannot be on $p$ . This implies that $q$ is exactly the tangent line to $p$ at $P$ , that is $q=\ell(P)$ . So we have proved Lemma 1: If $P$ is a point on $p$ then $\ell(P)$ is the perpendicular bisector of $\overline{P'F}$ .

We need another lemma before we proceed. Lemma 2: If $F$ is on the circumcircle of $\Delta XYZ$ with orthocenter $H$ , then the reflections of $F$ across $\overleftrightarrow{XY}$ , $\overleftrightarrow{XZ}$ , and $\overleftrightarrow{YZ}$ are collinear with $H$ .

Proof of Lemma 2: Say the reflections of $F$ and $H$ across $\overleftrightarrow{YZ}$ are $C'$ and $J$ , and the reflections of $F$ and $H$ across $\overleftrightarrow{XY}$ are $A'$ and $I$ . Then we angle chase $\angle JYZ=\angle HYZ=\angle HXZ=\angle JXZ=m(JZ)/2$ where $m(JZ)$ is the measure of minor arc $JZ$ on the circumcircle of $\Delta XYZ$ . This implies that $J$ is on the circumcircle of $\Delta XYZ$ , and similarly $I$ is on the circumcircle of $\Delta XYZ$ . Therefore $\angle C'HJ=\angle FJH=m(XF)/2$ , and $\angle A'HX=\angle FIX=m(FX)/2$ . So $\angle C'HJ = \angle A'HX$ . Since $J$ , $H$ , and $X$ are collinear it follows that $C'$ , $H$ and $A'$ are collinear. Similarly, the reflection of $F$ over $\overleftrightarrow{XZ}$ also lies on this line, and so the claim is proved.

Now suppose $A$ , $B$ , and $C$ are three points of $p$ and let $\ell(A)\cap\ell(B)=X$ , $\ell(A)\cap\ell(C)=Y$ , and $\ell(B)\cap\ell(C)=Z$ . Also let $A''$ , $B''$ , and $C''$ be the midpoints of $\overline{A'F}$ , $\overline{B'F}$ , and $\overline{C'F}$ respectively. Then since $\overleftrightarrow{A''B''}\parallel \overline{A'B'}=d$ and $\overleftrightarrow{B''C''}\parallel \overline{B'C'}=d$ , it follows that $A''$ , $B''$ , and $C''$ are collinear. By Lemma 1, we know that $A''$ , $B''$ , and $C''$ are the feet of the altitudes from $F$ to $\overline{XY}$ , $\overline{XZ}$ , and $\overline{YZ}$ . Therefore by the Simson Line Theorem, $F$ is on the circumcircle of $\Delta XYZ$ . If $H$ is the orthocenter of $\Delta XYZ$ , then by Lemma 2, it follows that $H$ is on $\overleftrightarrow{A'C'}=d$ . It follows that the locus described in the problem is a subset of $d$ .

Since we claim that the locus described in the problem is $d$ , we still need to show that for any choice of $H$ on $d$ there exists an equilateral triangle with center $H$ such that the lines containing the sides of the triangle are tangent to $p$ . So suppose $H$ is any point on $d$ and let the circle centered at $H$ through $F$ be $O$ . Then suppose $A$ is one of the intersections of $d$ with $O$ . Let $\angle HFA=3\theta$ , and construct the ray through $F$ on the same halfplane of $\overleftrightarrow{HF}$ as $A$ that makes an angle of $2\theta$ with $\overleftrightarrow{HF}$ . Say this ray intersects $O$ in a point $B$ besides $F$ , and let $q$ be the perpendicular bisector of $\overline{HB}$ . Since $\angle HFB=2\theta$ and $\angle HFA=3\theta$ , we have $\angle BFA=\theta$ . By the inscribed angles theorem, it follows that $\angle AHB=2\theta$ . Also since $HF$ and $HB$ are both radii, $\Delta HFB$ is isosceles and $\angle HBF=\angle HFB=2\theta$ . Let $P_1'$ be the reflection of $F$ across $q$ . Then $2\theta=\angle FBH=\angle C'HB$ , and so $\angle C'HB=\angle AHB$ . It follows that $P_1'$ is on $\overleftrightarrow{AH}=d$ , which means $q$ is the perpendicular bisector of $\overline{FP_1'}$ .

Let $q$ intersect $O$ in points $Y$ and $Z$ and let $X$ be the point diametrically opposite to $B$ on $O$ . Also let $\overline{HB}$ intersect $q$ at $M$ . Then $HM=HB/2=HZ/2$ . Therefore $\Delta HMZ$ is a $30-60-90$ right triangle and so $\angle ZHB=60^{\circ}$ . So $\angle ZHY=120^{\circ}$ and by the inscribed angles theorem, $\angle ZXY=60^{\circ}$ . Since $ZX=ZY$ it follows that $\Delta ZXY$ is and equilateral triangle with center $H$ .

By Lemma 2, it follows that the reflections of $F$ across $\overleftrightarrow{XY}$ and $\overleftrightarrow{XZ}$ , call them $P_2'$ and $P_3'$ , lie on $d$ . Let the intersection of $\overleftrightarrow{YZ}$ and the perpendicular to $d$ through $P_1'$ be $P_1$ , the intersection of $\overleftrightarrow{XY}$ and the perpendicular to $d$ through $P_2'$ be $P_2$ , and the intersection of $\overleftrightarrow{XZ}$ and the perpendicular to $d$ through $P_3'$ be $P_3$ . Then by the definitions of $P_1'$ , $P_2'$ , and $P_3'$ it follows that $FP_i=P_iP_i'$ for $i=1,2,3$ and so $P_1$ , $P_2$ , and $P_3$ are on $p$ . By lemma 1, $\ell(P_1)=\overleftrightarrow{YZ}$ , $\ell(P_2)=\overleftrightarrow{XY}$ , and $\ell(P_3)=\overleftrightarrow{XZ}$ . Therefore the intersections of $\ell(P_1)$ , $\ell(P_2)$ , and $\ell(P_3)$ form an equilateral triangle with center $H$ , which finishes the proof. --Killbilledtoucan

Solution 3

Note that the lines $l(P_1), l(P_2), l(P_3)$ are $y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,$ respectively. It is easy to deduce that the three points of intersection are $\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).$ The slopes of each side of this equilateral triangle are $2a_1,2a_2,2a_3,$ and we want to find the locus of $\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).$ We know that $2a_1=\tan(\theta), 2a_2=\tan (\theta + 120), 2a_3=\tan (\theta-120)$ for some $\theta.$ Therefore, we can use the tangent addition formula to deduce $\frac{a_1+a_2+a_3}{3}=\frac{\tan(\theta)+\tan (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}$ and $\begin{align*} \frac{a_1a_2+a_2a_3+a_3a_1}{3}&=\frac{\tan\theta (\tan(\theta-120)+\tan(\theta+120))+\tan(\theta-120)\tan(\theta+120)}{12}\\ &=\frac{9\tan^2\theta-3}{12(1-3\tan^2\theta)}\\ &=-\frac{1}{4}.\end{align*}$ Now we show that $\frac{a_1+a_2+a_3}{3}$ can be any real number. Let's say $\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}=k$ for some real number $k.$ Multiplying both sides by $2-\tan^2\theta$ and rearranging yields a cubic in $\tan\theta.$ Clearly this cubic has at least one real solution. As $\tan \theta$ can take on any real number, all values of $k$ are possible, and our answer is the line $\boxed{y=-\frac{1}{4}.}$ Of course, as the denominator could equal 0, we must check $\tan \theta=\pm \frac{1}{\sqrt{3}}.$ $3\tan \theta-\tan^3\theta=k(2-6\tan^2\theta).$ The left side is nonzero, while the right side is zero, so these values of $\theta$ do not contribute to any values of $k.$ So, our answer remains the same. $\blacksquare$ ~ Benq

Work in progress: Solution 4 (Clean algebra)

$[asy] Label f; f.p=fontsize(6); xaxis(-2,2); yaxis(-2,2); real f(real x) { return x^2; } draw(graph(f,-sqrt(2),sqrt(2))); real f(real x) { return (2*sqrt(3)/3)*x-1/3; } draw(graph(f,-5*sqrt(3)/6,2)); real f(real x) { return (-sqrt(3)/9)*x-1/108; } draw(graph(f,-2,2)); real f(real x) { return (-5*sqrt(3)/3)*x-25/12; } draw(graph(f,-49*sqrt(3)/60,-sqrt(3)/60)); [/asy]$

It can be easily shown that the center of $\triangle$ has coordinates $\left(\frac{a_{1}+a_{2}+a_{3}}{3},\frac{a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}}{3}\right)$ .

Without loss of generality, let $a_{1}>a_{2}>a_{3}$ . Notice that $\ell(P_2)$ is a $60^{\circ}$ clockwise rotation of $\ell(P_1)$ , $\ell(P_3)$ is a $60^{\circ}$ clockwise rotation of $\ell(P_2)$ , and $\ell(P_1)$ is a $60^{\circ}$ clockwise rotation of $\ell(P_3)$ . By definition, $\arctan(2a_{i})$ is the (directed) angle from the x-axis to $\ell(P_{i})$ . Remember that the range of $\arctan(x)$ is $(-90^{\circ},90^{\circ})$ . We have $\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}$

Taking the tangent of both sides of each equation and rearranging, we get $\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}$ We add these equations to get $\sqrt{3}(3+4(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}))=0.$ We solve for $a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}$ to get $a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=-\frac{3}{4}.$ So, the y-coordinate of $\triangle$ is $-\frac{1}{4}$ .

We will prove that the x-coordinate of $\triangle$ can be any real number. If $2a_{1}$ tends to infinity, then $2a_{2}$ tends to $\frac{\sqrt{3}}{2}$ and $2a_{3}$ tends to $-\frac{\sqrt{3}}{2}$ . So, $\frac{a_{1}+a_{2}+a_{3}}{3}$ can be arbitrarily large. Similarly, if we let $2a_{3}$ tend to negative infinity, then $2a_{1}$ tends to $\frac{\sqrt{3}}{2}$ and $2a_{2}$ tends to $-\frac{\sqrt{3}}{2}$ . So, $\frac{a_{1}+a_{2}+a_{3}}{3}$ can be arbitrarily small. Since $\frac{a_{1}+a_{2}+a_{3}}{3}$ is continuous, it can take any real value. So, the locus is the line $y=-\frac{1}{4}$ .

@@ Line 3: / Line 3: @@
 For a point <math>P = (a, a^2)</math> in the coordinate plane, let <math>\ell(P)</math> denote the line passing through <math>P</math> with slope <math>2a</math>. Consider the set of triangles with vertices of the form <math>P_1 = (a_1,  a_1^2)</math>, <math>P_2 = (a_2, a_2^2)</math>, <math>P_3 = (a_3, a_3^2)</math>, such that the intersections of the lines <math>\ell(P_1)</math>, <math>\ell(P_2)</math>, <math>\ell(P_3)</math> form an equilateral triangle <math>\triangle</math>. Find the locus of the center of <math>\triangle</math> as <math>P_1P_2P_3</math> ranges over all such triangles.
-==Solution==
+==Solutions==
+===Solution 1===
+Note that the lines <math>l(P_1), l(P_2), l(P_3)</math> are <cmath>y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,</cmath> respectively. It is easy to deduce that the three points of intersection are <cmath>\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).</cmath> The slopes of each side of this equilateral triangle are <cmath>2a_1,2a_2,2a_3,</cmath> and we want to find the locus of <cmath>\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).</cmath> Define the three complex numbers <math>w_n = 1+2a_ni</math> for <math>n=1,2,3</math>. Then note that the slope - that is, the imaginary part divided by the real part - of all <math>w_n^3</math> is constant, say it is <math>k</math>. Then for <math>n=1,2,3</math>,
+<cmath>\begin{align*}
+\frac{\Im(w_n^3)}{\Re(w_n^3)}
+&= \frac{\Im((1+2a_ni)^3)}{\Re((1+2a_ni)^3)}\\
+&= \frac{\Im(1+6a_ni-12a_n^2-8a_n^3i)}{\Re(1+6a_ni-12a_n^2-8a_n^3i)}\\
+&= \frac{6a_n-8a_n^3}{1-12a_n^2}\\
+&= k.\\
+\end{align*}</cmath>
+Rearranging, we get that<cmath>8a_n^3 -12ka_n^2-6a_n+k=0,</cmath>or<cmath>a_n^3-\frac{3ka_n^2}2-\frac{3a_n}4+\frac k8=0.</cmath>Note that this is a cubic, and the roots are <math>a_1,a_2</math> and <math>a_3</math> which are all distinct, and so there are no other roots. Using Vieta's, we get that
+<cmath>a_1+a_2+a_3=\frac{3k}2,</cmath>and<cmath>a_1a_2+a_2a_3+a_3a_1=-\frac34.</cmath>
+Obviously all values of <math>k</math> are possible, and so our answer is the line <cmath>\boxed{y=-\frac{1}{4}.}</cmath> <math>\blacksquare</math>
+~ cocohearts
+===Solution 2===
 Note that all the points <math>P=(a,a^2)</math> belong to the parabola <math>y=x^2</math> which we will denote <math>p</math>. This parabola has a focus <math>F=\left(0,\frac{1}{4}\right)</math> and directrix <math>y=-\frac{1}{4}</math> which we will denote <math>d</math>. We will prove that the desired locus is <math>d</math>.
-First note that for any point <math>P</math> on <math>p</math>, the line <math>\ell(P)</math> is the tangent line to <math>p</math> at <math>P</math>. This is because <math>\ell(P)</math> contains <math>P</math> and because <math>[\frac{d}{dx}] x^2=2x</math>. If you don't like calculus, you can also verify that <math>\ell(P)</math> has equation <math>y=2a(x-a)+a^2</math> and does not intersect <math>y=x^2</math> at any point besides <math>P</math>. Now for any point <math>P</math> on <math>p</math> let <math>P'</math> be the foot of the perpendicular from <math>P</math> onto <math>d</math>. Then by the definition of parabolas, <math>PP'=PF</math>. Let <math>q</math> be the perpendicular bisector of <math>\oveline{P'F}</math>. Since <math>PP'=PF</math>, <math>q</math> passes through <math>P</math>. Suppose <math>K</math> is any other point on <math>q</math> and let <math>K'</math> be the foot of the perpendicular from <math>K</math> to <math>d</math>. Then in right <math>\Delta KK'P'</math>, <math>KK'</math> is a leg and so <math>KK'<KP'=KF</math>. Therefore <math>K</math> cannot be on <math>p</math>. This implies that <math>q</math> is exactly the tangent line to <math>p</math> at <math>P</math>, that is <math>q=\ell(P)</math>. So we have proved Lemma 1: If <math>P</math> is a point on <math>p</math> then <math>\ell(P)</math> is the perpendicular bisector of <math>\overline{P'F}</math>.
+First note that for any point <math>P</math> on <math>p</math>, the line <math>\ell(P)</math> is the tangent line to <math>p</math> at <math>P</math>. This is because <math>\ell(P)</math> contains <math>P</math> and because <math>[\frac{d}{dx}] x^2=2x</math>. If you don't like calculus, you can also verify that <math>\ell(P)</math> has equation <math>y=2a(x-a)+a^2</math> and does not intersect <math>y=x^2</math> at any point besides <math>P</math>. Now for any point <math>P</math> on <math>p</math> let <math>P'</math> be the foot of the perpendicular from <math>P</math> onto <math>d</math>. Then by the definition of parabolas, <math>PP'=PF</math>. Let <math>q</math> be the perpendicular bisector of <math>\overline{P'F}</math>. Since <math>PP'=PF</math>, <math>q</math> passes through <math>P</math>. Suppose <math>K</math> is any other point on <math>q</math> and let <math>K'</math> be the foot of the perpendicular from <math>K</math> to <math>d</math>. Then in right <math>\Delta KK'P'</math>, <math>KK'</math> is a leg and so <math>KK'<KP'=KF</math>. Therefore <math>K</math> cannot be on <math>p</math>. This implies that <math>q</math> is exactly the tangent line to <math>p</math> at <math>P</math>, that is <math>q=\ell(P)</math>. So we have proved Lemma 1: If <math>P</math> is a point on <math>p</math> then <math>\ell(P)</math> is the perpendicular bisector of <math>\overline{P'F}</math>.
 We need another lemma before we proceed. Lemma 2: If <math>F</math> is on the circumcircle of <math>\Delta XYZ</math> with orthocenter <math>H</math>, then the reflections of <math>F</math> across <math>\overleftrightarrow{XY}</math>, <math>\overleftrightarrow{XZ}</math>, and <math>\overleftrightarrow{YZ}</math> are collinear with <math>H</math>.
@@ Line 19: / Line 38: @@
 Let <math>q</math> intersect <math>O</math> in points <math>Y</math> and <math>Z</math> and let <math>X</math> be the point diametrically opposite to <math>B</math> on <math>O</math>. Also let <math>\overline{HB}</math> intersect <math>q</math> at <math>M</math>. Then <math>HM=HB/2=HZ/2</math>. Therefore <math>\Delta HMZ</math> is a <math>30-60-90</math> right triangle and so <math>\angle ZHB=60^{\circ}</math>. So <math>\angle ZHY=120^{\circ}</math> and by the inscribed angles theorem, <math>\angle ZXY=60^{\circ}</math>. Since <math>ZX=ZY</math> it follows that <math>\Delta ZXY</math> is and equilateral triangle with center <math>H</math>.
-By Lemma 2, it follows that the reflections of <math>F</math> across <math>\overleftrightarrow{XY}</math> and <math>\overleftrightarrow{XZ}</math>, call them <math>P_2'</math> and <math>P_3'</math>, lie on <math>d</math>. Let the intersection of <math>\overleftrightarrow{YZ}</math> and the perpendicular to <math>d</math> through <math>P_1'</math> be <math>P_1</math>, the intersection of <math>\overleftrightarriw{XY}</math> and the perpendicular to <math>d</math> through <math>P_2'</math> be <math>P_2</math>, and the intersection of <math>\overleftrightarrow{XZ}</math> and the perpendicular to <math>d</math> through <math>P_3'</math> be <math>P_3</math>. Then by the definitions of <math>P_1'</math>, <math>P_2'</math>, and <math>P_3'</math> it follows that <math>FP_i=P_iP_i'</math> for <math>i=1,2,3</math> and so <math>P_1</math>, <math>P_2</math>, and <math>P_3</math> are on <math>p</math>. By lemma 1, <math>\ell(P_1)=\overleftrightarrow{YZ}</math>, <math>\ell(P_2)=\overleftrightarrow{XY}</math>, and <math>\ell(P_3)=\overleftrightarrow{XZ}</math>. Therefore the intersections of <math>\ell(P_1)</math>, <math>\ell(P_2)</math>, and <math>\ell(P_3)</math> form an equilateral triangle with center <math>H</math>, which finishes the proof.
+By Lemma 2, it follows that the reflections of <math>F</math> across <math>\overleftrightarrow{XY}</math> and <math>\overleftrightarrow{XZ}</math>, call them <math>P_2'</math> and <math>P_3'</math>, lie on <math>d</math>. Let the intersection of <math>\overleftrightarrow{YZ}</math> and the perpendicular to <math>d</math> through <math>P_1'</math> be <math>P_1</math>, the intersection of <math>\overleftrightarrow{XY}</math> and the perpendicular to <math>d</math> through <math>P_2'</math> be <math>P_2</math>, and the intersection of <math>\overleftrightarrow{XZ}</math> and the perpendicular to <math>d</math> through <math>P_3'</math> be <math>P_3</math>. Then by the definitions of <math>P_1'</math>, <math>P_2'</math>, and <math>P_3'</math> it follows that <math>FP_i=P_iP_i'</math> for <math>i=1,2,3</math> and so <math>P_1</math>, <math>P_2</math>, and <math>P_3</math> are on <math>p</math>. By lemma 1, <math>\ell(P_1)=\overleftrightarrow{YZ}</math>, <math>\ell(P_2)=\overleftrightarrow{XY}</math>, and <math>\ell(P_3)=\overleftrightarrow{XZ}</math>. Therefore the intersections of <math>\ell(P_1)</math>, <math>\ell(P_2)</math>, and <math>\ell(P_3)</math> form an equilateral triangle with center <math>H</math>, which finishes the proof.
 --Killbilledtoucan
+===Solution 3===
+Note that the lines <math>l(P_1), l(P_2), l(P_3)</math> are <cmath>y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,</cmath> respectively. It is easy to deduce that the three points of intersection are <cmath>\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).</cmath> The slopes of each side of this equilateral triangle are <cmath>2a_1,2a_2,2a_3,</cmath> and we want to find the locus of <cmath>\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).</cmath> We know that <cmath>2a_1=\tan(\theta), 2a_2=\tan (\theta + 120), 2a_3=\tan (\theta-120)</cmath> for some <math>\theta.</math> Therefore, we can use the tangent addition formula to deduce <cmath>\frac{a_1+a_2+a_3}{3}=\frac{\tan(\theta)+\tan (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}</cmath> and <cmath>\begin{align*}
+\frac{a_1a_2+a_2a_3+a_3a_1}{3}&=\frac{\tan\theta (\tan(\theta-120)+\tan(\theta+120))+\tan(\theta-120)\tan(\theta+120)}{12}\\
+&=\frac{9\tan^2\theta-3}{12(1-3\tan^2\theta)}\\
+&=-\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
+===Work in progress: Solution 4 (Clean algebra)===
+<asy>
+Label f;
+f.p=fontsize(6);
+xaxis(-2,2);
+yaxis(-2,2);
+real f(real x)
+{
+return x^2;
+}
+draw(graph(f,-sqrt(2),sqrt(2)));
+real f(real x)
+{
+return (2*sqrt(3)/3)*x-1/3;
+}
+draw(graph(f,-5*sqrt(3)/6,2));
+real f(real x)
+{
+return (-sqrt(3)/9)*x-1/108;
+}
+draw(graph(f,-2,2));
+real f(real x)
+{
+return (-5*sqrt(3)/3)*x-25/12;
+}
+draw(graph(f,-49*sqrt(3)/60,-sqrt(3)/60));
+</asy>
+It can be easily shown that the center of <math>\triangle</math> has coordinates <math>\left(\frac{a_{1}+a_{2}+a_{3}}{3},\frac{a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}}{3}\right)</math>.
+Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. Remember that the range of <math>\arctan(x)</math> is <math>(-90^{\circ},90^{\circ})</math>. We have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath>
+Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath> We add these equations to get <cmath>\sqrt{3}(3+4(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}))=0.</cmath> We solve for <math>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}</math> to get <cmath>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=-\frac{3}{4}.</cmath> So, the y-coordinate of <math>\triangle</math> is <math>-\frac{1}{4}</math>.
+We will prove that the x-coordinate of <math>\triangle</math> can be any real number. If <math>2a_{1}</math> tends to infinity, then <math>2a_{2}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{3}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily large. Similarly, if we let <math>2a_{3}</math> tend to negative infinity, then <math>2a_{1}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{2}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily small. Since <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> is continuous, it can take any real value. So, the locus is the line <math>y=-\frac{1}{4}</math>.
+== See also ==
+{{USAJMO newbox|year=2011|num-b=2|num-a=4}}
+{{MAA Notice}}

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