Difference between revisions of "1976 USAMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
− | WLOG, assume that the circle is the unit circle centered at the origin. Then the points <math>A</math> and <math>B</math> have coordinates <math>(-a,b)</math> and <math>(a,b)</math> respectively and <math>X</math> and <math>Y</math> have coordinates <math>(r,s)</math> and <math>(-r,-s)</math>. | + | WLOG, assume that the circle is the unit circle centered at the origin. Then the points <math>A</math> and <math>B</math> have coordinates <math>(-a,b)</math> and <math>(a,b)</math> respectively and <math>X</math> and <math>Y</math> have coordinates <math>(r,s)</math> and <math>(-r,-s)</math>. Note that these coordinates satisfy <math>a^2 + b^2 = 1</math> and <math>r^2 + s^2 = 1</math> since these points are on a unit circle. Now we can find equations for the lines: |
<cmath> \begin{align*} | <cmath> \begin{align*} | ||
AX \longrightarrow y &= \frac{(s-b)x+rb+sa}{r+a}\ | AX \longrightarrow y &= \frac{(s-b)x+rb+sa}{r+a}\ | ||
Line 28: | Line 28: | ||
\end{align*} </cmath> | \end{align*} </cmath> | ||
Solving these simultaneous equations gives coordinates for <math>P</math> in terms of <math>a, b, r,</math> and <math>s</math>: | Solving these simultaneous equations gives coordinates for <math>P</math> in terms of <math>a, b, r,</math> and <math>s</math>: | ||
− | <math>P = (\frac{as}{b},1-ar)</math>. These coordinates can be parametrized as follows: | + | <math>P = \left(\frac{as}{b},\frac{1 - ar}{b}\right)</math>. These coordinates can be parametrized in Cartesian variables as follows: |
<cmath> \begin{align*} | <cmath> \begin{align*} | ||
− | + | x &= \frac{as}{b}\ | |
− | + | y &= \frac{1 - ar}{b}. | |
\end{align*} </cmath> | \end{align*} </cmath> | ||
− | Now | + | Now solve for <math>r</math> and <math>s</math> to get <math>r = \frac{1-by}{a}</math> and <math>s = \frac{bx}{a}</math> . Then since <math>r^2 + s^2 = 1, \left(\frac{bx}{a}\right)^2 + \left(\frac{1-by}{a}\right)^2 = 1</math> which reduces to <math>x^2 + (y-1/b)^2 = \frac{a^2}{b^2}.</math> This equation defines a circle and is the locus of all intersection points <math>P</math>. In order to define this locus more generally, find the slope of this circle function using implicit differentiation: |
− | This | + | <cmath> \begin{align*} |
+ | 2x + 2(y-1/b)y' &= 0\ | ||
+ | (y-1/b)y' &= -x\ | ||
+ | y' &= \frac{-x}{y-1/b}. | ||
+ | \end{align*} </cmath> | ||
+ | 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. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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. | ||
+ | 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}} | ||
− | ==See | + | ==See Also== |
{{USAMO box|year=1976|num-b=1|num-a=3}} | {{USAMO box|year=1976|num-b=1|num-a=3}} | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 22:41, 6 July 2020
Contents
[hide]Problem
If and are fixed points on a given circle and is a variable diameter of the same circle, determine the locus of the point of intersection of lines and . You may assume that is not a diameter.
Solution
WLOG, assume that the circle is the unit circle centered at the origin. Then the points and have coordinates and respectively and and have coordinates and . Note that these coordinates satisfy and since these points are on a unit circle. Now we can find equations for the lines: Solving these simultaneous equations gives coordinates for in terms of and : . These coordinates can be parametrized in Cartesian variables as follows: Now solve for and to get and . Then since which reduces to This equation defines a circle and is the locus of all intersection points . In order to define this locus more generally, find the slope of this circle function using implicit differentiation: Now note that at points and , this slope expression reduces to and respectively, values which are identical to the slopes of lines and . Thus we conclude that the complete locus of intersection points is the circle tangent to lines and at points and respectively.
Solution 2
Notice that (Inscried angle theorem) and that since is a diameter, and thus subtends an arc of . This will hold for all and all , and so by AA similarity, the angle 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 , we know that . We can assume that and intersect the original circle at points , and respectively. This will give us that (Since lies on the perpendicular bisector of , we know that , will be on the same side of .) Now we also know that or . The only case where satisfies the measure of , is when and , implying that and are tangents, and so is the intersection of the tangents from and 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.