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Difference between revisions of "1976 USAMO Problems/Problem 2"

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Now note that at points <math>A</math> and <math>B</math>, this slope expression reduces to <math>y' = \frac{-b}{a}</math> and <math>y' = \frac{b}{a}</math> respectively, values which are identical to the slopes of lines <math>AO</math> and <math>BO</math>. Thus we conclude that the complete locus of intersection points is the circle tangent to lines <math>AO</math> and <math>BO</math> at points <math>A</math> and <math>B</math> respectively.
 
Now note that at points <math>A</math> and <math>B</math>, this slope expression reduces to <math>y' = \frac{-b}{a}</math> and <math>y' = \frac{b}{a}</math> respectively, values which are identical to the slopes of lines <math>AO</math> and <math>BO</math>. Thus we conclude that the complete locus of intersection points is the circle tangent to lines <math>AO</math> and <math>BO</math> at points <math>A</math> and <math>B</math> respectively.
  
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==See also==
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==Solution 2==
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Notice that <math>m\angle AYP=\frac{1}{2}m\overarc{AB}</math> (Inscried angle theorem) and that <math>m\angle XAY=90</math> since <math>XY</math> is a diameter, and thus subtends an arc of <math>180</math>. This will hold for all <math>X</math> and all <math>Y</math>, and so by AA similarity, the angle <math>APY</math> will be constant for all P, thus implying that the points A, B, and all P will be concyclic.
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If we assume that the center of this circle is <math>H</math>, we know that <math>m\angle AHB=2m\angle APB=2(90-\frac{1}{2}\overarc{AB})=180-\overarc{AB}</math>. We can assume that <math>\overline{AH}</math> and <math>\overline{BH}</math> intersect the original circle at points <math>M</math>, and <math>N</math> respectively. This will give us that <math>m\angle AHB=\frac{1}{2}|m\overarc{MN}-m\overarc{AB}|</math> (Since <math>H</math> lies on the perpendicular bisector of <math>AB</math>, we know that <math>M</math>, <math>N</math> will be on the same side of <math>AB</math>.) Now we also know that <math>m\overarc{MN}\leq m\overarc{AB}</math> or <math>m\overarc{MN}\leq 360-m\overarc{AB}</math>. The only case where <math>m\overarc{MN}</math> satisfies the measure of <math>\angle AHB</math>, is when <math>M=A</math> and <math>N=B</math>, implying that <math>AH</math> and <math>BH</math> are tangents, and so <math>H</math> is the intersection of the tangents from <math>A</math> and <math>B</math> to the original circle.
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{{alternate solutions}}
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==See Also==
  
 
{{USAMO box|year=1976|num-b=1|num-a=3}}
 
{{USAMO box|year=1976|num-b=1|num-a=3}}
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{{MAA Notice}}
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Latest revision as of 23:41, 6 July 2020

Problem

If $A$ and $B$ are fixed points on a given circle and $XY$ is a variable diameter of the same circle, determine the locus of the point of intersection of lines $AX$ and $BY$. You may assume that $AB$ is not a diameter.

[asy] size(300); defaultpen(fontsize(8)); real r=10; picture pica, picb; pair A=r*expi(5*pi/6), B=r*expi(pi/6), X=r*expi(pi/3), X1=r*expi(-pi/12), Y=r*expi(4*pi/3), Y1=r*expi(11*pi/12), O=(0,0), P, P1; P = extension(A,X,B,Y);P1 = extension(A,X1,B,Y1); path circ1 = Circle((0,0),r); draw(pica, circ1);draw(pica, B--A--P--Y--X);dot(pica,P^^O); label(pica,"$A$",A,(-1,1));label(pica,"$B$",B,(1,0));label(pica,"$X$",X,(0,1));label(pica,"$Y$",Y,(0,-1));label(pica,"$P$",P,(1,1));label(pica,"$O$",O,(-1,1));label(pica,"(a)",O+(0,-13),(0,0)); draw(picb, circ1);draw(picb, B--A--X1--Y1--B);dot(picb,P1^^O); label(picb,"$A$",A,(-1,1));label(picb,"$B$",B,(1,1));label(picb,"$X$",X1,(1,-1));label(picb,"$Y$",Y1,(-1,0));label(picb,"$P'$",P1,(-1,-1));label(picb,"$O$",O,(-1,-1)); label(picb,"(b)",O+(0,-13),(0,0)); add(pica); add(shift(30*right)*picb); [/asy]

Solution

WLOG, assume that the circle is the unit circle centered at the origin. Then the points $A$ and $B$ have coordinates $(-a,b)$ and $(a,b)$ respectively and $X$ and $Y$ have coordinates $(r,s)$ and $(-r,-s)$. Note that these coordinates satisfy $a^2 + b^2 = 1$ and $r^2 + s^2 = 1$ since these points are on a unit circle. Now we can find equations for the lines: \begin{align*} AX \longrightarrow y &= \frac{(s-b)x+rb+sa}{r+a}\\ BY \longrightarrow y &= \frac{(s+b)x+rb-sa}{r+a}. \end{align*} Solving these simultaneous equations gives coordinates for $P$ in terms of $a, b, r,$ and $s$: $P = \left(\frac{as}{b},\frac{1 - ar}{b}\right)$. These coordinates can be parametrized in Cartesian variables as follows: \begin{align*} x &= \frac{as}{b}\\ y &= \frac{1 - ar}{b}. \end{align*} Now solve for $r$ and $s$ to get $r = \frac{1-by}{a}$ and $s = \frac{bx}{a}$ . Then since $r^2 + s^2 = 1, \left(\frac{bx}{a}\right)^2 + \left(\frac{1-by}{a}\right)^2 = 1$ which reduces to $x^2 + (y-1/b)^2 = \frac{a^2}{b^2}.$ This equation defines a circle and is the locus of all intersection points $P$. In order to define this locus more generally, find the slope of this circle function using implicit differentiation: \begin{align*} 2x + 2(y-1/b)y' &= 0\\ (y-1/b)y' &= -x\\ y' &= \frac{-x}{y-1/b}. \end{align*} Now note that at points $A$ and $B$, this slope expression reduces to $y' = \frac{-b}{a}$ and $y' = \frac{b}{a}$ respectively, values which are identical to the slopes of lines $AO$ and $BO$. Thus we conclude that the complete locus of intersection points is the circle tangent to lines $AO$ and $BO$ at points $A$ and $B$ respectively.

Solution 2

Notice that $m\angle AYP=\frac{1}{2}m\overarc{AB}$ (Inscried angle theorem) and that $m\angle XAY=90$ since $XY$ is a diameter, and thus subtends an arc of $180$. This will hold for all $X$ and all $Y$, and so by AA similarity, the angle $APY$ will be constant for all P, thus implying that the points A, B, and all P will be concyclic. If we assume that the center of this circle is $H$, we know that $m\angle AHB=2m\angle APB=2(90-\frac{1}{2}\overarc{AB})=180-\overarc{AB}$. We can assume that $\overline{AH}$ and $\overline{BH}$ intersect the original circle at points $M$, and $N$ respectively. This will give us that $m\angle AHB=\frac{1}{2}|m\overarc{MN}-m\overarc{AB}|$ (Since $H$ lies on the perpendicular bisector of $AB$, we know that $M$, $N$ will be on the same side of $AB$.) Now we also know that $m\overarc{MN}\leq m\overarc{AB}$ or $m\overarc{MN}\leq 360-m\overarc{AB}$. The only case where $m\overarc{MN}$ satisfies the measure of $\angle AHB$, is when $M=A$ and $N=B$, implying that $AH$ and $BH$ are tangents, and so $H$ is the intersection of the tangents from $A$ and $B$ to the original circle.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1976 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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