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

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==Solution==
 
==Solution==
  
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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}x</math> which we will denote <math>d</math>. We will prove that the desired locus is <math>d</math>.
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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>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>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>.

Revision as of 07:31, 4 March 2012

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.

Solution

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 $d/dx x^2=2x$. 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 $\oveline{P'F}$ (Error compiling LaTeX. Unknown error_msg). 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 $\overleftrightarriw{XY}$ (Error compiling LaTeX. Unknown error_msg) 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

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