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

Revision as of 18:29, 30 April 2011

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 $\Delta$ . Find the locus of the center of $\Delta$ 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 $y=-\frac{1}{4}x$ 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

@@ Line 2: / Line 2: @@
 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>\Delta</math>. Find the locus of the center of <math>\Delta</math> as <math>P_1P_2P_3</math> ranges over all such triangles.
+==Solution==
+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>.
+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>.
+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>.
+Proof of Lemma 2: Say the reflections of <math>F</math> and <math>H</math> across <math>\overleftrightarrow{YZ}</math> are <math>C'</math> and <math>J</math>, and the reflections of <math>F</math> and <math>H</math> across <math>\overleftrightarrow{XY}</math> are <math>A'</math> and <math>I</math>. Then we angle chase <math>\angle JYZ=\angle HYZ=\angle HXZ=\angle JXZ=m(JZ)/2</math> where <math>m(JZ)</math> is the measure of minor arc <math>JZ</math> on the circumcircle of <math>\Delta XYZ</math>. This implies that <math>J</math> is on the circumcircle of <math>\Delta XYZ</math>, and similarly <math>I</math> is on the circumcircle of <math>\Delta XYZ</math>. Therefore <math>\angle C'HJ=\angle FJH=m(XF)/2</math>, and <math>\angle A'HX=\angle FIX=m(FX)/2</math>. So <math>\angle C'HJ = \angle A'HX</math>. Since <math>J</math>, <math>H</math>, and <math>X</math> are collinear it follows that <math>C'</math>, <math>H</math> and <math>A'</math> are collinear. Similarly, the reflection of <math>F</math> over <math>\overleftrightarrow{XZ}</math> also lies on this line, and so the claim is proved.
+Now suppose <math>A</math>, <math>B</math>, and <math>C</math> are three points of <math>p</math> and let <math>\ell(A)\cap\ell(B)=X</math>, <math>\ell(A)\cap\ell(C)=Y</math>, and <math>\ell(B)\cap\ell(C)=Z</math>. Also let <math>A''</math>, <math>B''</math>, and <math>C''</math> be the midpoints of <math>\overline{A'F}</math>, <math>\overline{B'F}</math>, and <math>\overline{C'F}</math> respectively. Then since <math>\overleftrightarrow{A''B''}\parallel \overline{A'B'}=d</math> and <math>\overleftrightarrow{B''C''}\parallel \overline{B'C'}=d</math>, it follows that <math>A''</math>, <math>B''</math>, and <math>C''</math> are collinear. By Lemma 1, we know that <math>A''</math>, <math>B''</math>, and <math>C''</math> are the feet of the altitudes from <math>F</math> to <math>\overline{XY}</math>, <math>\overline{XZ}</math>, and <math>\overline{YZ}</math>. Therefore by the Simson Line Theorem, <math>F</math> is on the circumcircle of <math>\Delta XYZ</math>. If <math>H</math> is the orthocenter of <math>\Delta XYZ</math>, then by Lemma 2, it follows that <math>H</math> is on <math>\overleftrightarrow{A'C'}=d</math>. It follows that the locus described in the problem is a subset of <math>d</math>.
+Since we claim that the locus described in the problem is <math>d</math>, we still need to show that for any choice of <math>H</math> on <math>d</math> there exists an equilateral triangle with center <math>H</math> such that the lines containing the sides of the triangle are tangent to <math>p</math>. So suppose <math>H</math> is any point on <math>d</math> and let the circle centered at <math>H</math> through <math>F</math> be <math>O</math>. Then suppose <math>A</math> is one of the intersections of <math>d</math> with <math>O</math>. Let <math>\angle HFA=3\theta</math>, and construct the ray through <math>F</math> on the same halfplane of <math>\overleftrightarrow{HF}</math> as <math>A</math> that makes an angle of <math>2\theta</math> with <math>\overleftrightarrow{HF}</math>. Say this ray intersects <math>O</math> in a point <math>B</math> besides <math>F</math>, and let <math>q</math> be the perpendicular bisector of <math>\overline{HB}</math>. Since <math>\angle HFB=2\theta</math> and <math>\angle HFA=3\theta</math>, we have <math>\angle BFA=\theta</math>. By the inscribed angles theorem, it follows that <math>\angle AHB=2\theta</math>. Also since <math>HF</math> and <math>HB</math> are both radii, <math>\Delta HFB</math> is isosceles and <math>\angle HBF=\angle HFB=2\theta</math>. Let <math>P_1'</math> be the reflection of <math>F</math> across <math>q</math>. Then <math>2\theta=\angle FBH=\angle C'HB</math>, and so <math>\angle C'HB=\angle AHB</math>. It follows that <math>P_1'</math> is on <math>\overleftrightarrow{AH}=d</math>, which means <math>q</math> is the perpendicular bisector of <math>\overline{FP_1'}</math>.
+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.
+--Killbilledtoucan

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

Revision as of 18:29, 30 April 2011

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