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Difference between revisions of "1988 IMO Problems/Problem 1"

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==Problem==
 
Consider 2 concentric circles with radii <math>R</math> and <math>r</math> (<math>R>r</math>) with center <math>O</math>.  Fix <math>P</math> on the small circle and consider the variable chord <math>AP</math> of the small circle. Points <math>B</math> and <math>C</math> lie on the large circle; <math>B,P,C</math> are collinear and <math>BC</math> is perpendicular to <math>AP</math>.  
 
Consider 2 concentric circles with radii <math>R</math> and <math>r</math> (<math>R>r</math>) with center <math>O</math>.  Fix <math>P</math> on the small circle and consider the variable chord <math>AP</math> of the small circle. Points <math>B</math> and <math>C</math> lie on the large circle; <math>B,P,C</math> are collinear and <math>BC</math> is perpendicular to <math>AP</math>.  
  
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i.) For which values of <math>\angle OPA</math> is the sum <math>BC^2+CA^2+AB^2</math> extremal?  
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<ol type="i">
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<li>For which values of <math>\angle OPA</math> is the sum <math>BC^2+CA^2+AB^2</math> extremal?</li>
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<li>What are the possible positions of the midpoints <math>U</math> of <math>AB</math> and <math>V</math> of <math>AC</math> as <math>A</math> varies?</li>
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</ol>
  
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ii.) What are the possible positions of the midpoints <math>U</math> of <math>AB</math> and <math>V</math> of <math>AC</math> as <math>A</math> varies?
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==Solution==
  
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<ol type="i">
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<li>
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We claim that the value <math>BC^2+CA^2+AB^2</math> stays constant as <math>\angle OPA</math> varies, and thus achieves its maximum at all value of <math>\angle OPA</math>.  We have from the Pythagorean Theorem that <math>CA^2=AP^2+PC^2</math> and <math>AB^2=AP^2+PB^2</math> and so our expression becomes
  
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'''Solution'''
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<cmath> BC^2+PB^2+PC^2+2AP^2=2BC^2+2AP^2-2PB\cdot PC </cmath>
  
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'''i.)'''
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Since <math>PB\cdot PC</math> is the power of the point <math>P</math>, it stays constant as <math>A</math> varies. Thus, we are left to prove that the value <math>BC^2+AP^2</math> stays constant as <math>\angle OPA</math> varies.  Let <math>G</math> be the midpoint of <math>AP</math> and let <math>H</math> be the midpoint of <math>BC</math>. Since <math>OG</math> is perpendicular to <math>AP</math>, we find that <math>PG=r\cos OPA</math>.  Similarly, we find that <math>OH=r\sin OPC=r\cos OPA</math>.  Thus, by the Pythagorean Theorem, we have
  
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We claim that the value <math>BC^2+CA^2+AB^2</math> stays constant as <math>\angle OPA</math> varies, and thus achieves its maximum at all value of <math>\angle OPA</math>.  We have from the Pythagorean Theorem that <math>CA^2=AP^2+PC^2</math> and <math>AB^2=AP^2+PB^2</math> and so our expression becomes
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<cmath>
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\begin{align*}
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BC^2& =4(R^2-r^2\cos^2 OPA) \\
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AP^2&=4r^2\cos^2OPA
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\end{align*}
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</cmath>
  
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<math>BC^2+PB^2+PC^2+2AP^2=2BC^2+2AP^2-2PB\cdotPC</math>
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Now it is obvious that <math>BC^2+AP^2=4R^2</math> is constant for all values of <math>\angle OPA</math>.   
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</li>
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Since <math>PB\cdot PC</math> is the power of the point <math>P</math>, it stays constant as <math>A</math> varies.  Thus, we are left to prove that the value <math>BC^2+AP^2</math> stays constant as <math>\angle OPA</math> varies.  Let <math>G</math> be the midpoint of <math>AP</math> and let <math>H</math> be the midpoint of <math>BC</math>.  Since <math>OG</math> is perpendicular to <math>AP</math>, we find that <math>PG=r\cos OPA</math>.  Similarly, we find that <math>OH=r\sin OPC=r\cos OPA</math>.  Thus, by the Pythagorean Theorem, we have
 
  
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<math>BC^2=4(R^2-r^2\cos^2 OPA)</math>
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<li>
  
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<math>AP^2=4r^2\cos^2OPA</math>
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We claim that all points <math>U,V</math> lie on a circle centered at the midpoint of <math>OP</math>, <math>M</math> with radius <math>\frac{R}{2}</math>.  Let <math>T</math> be the midpoint of <math>UV</math>.  Since <math>H</math> is the midpoint of <math>BC</math>, it is clear that the projection of <math>T</math> onto <math>BC</math> is the midpoint of <math>H</math> and <math>P</math> (the projection of <math>A</math> onto <math>BC</math>).  Thus, we have that <math>MT</math> is perpendicular to <math>UV</math> and thus the triangle <math>MUV</math> is isosceles.  We have
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<cmath>
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Now it is obvious that <math>BC^2+AP^2=4R^2</math> is constant for all values of <math>\angle OPA</math>.
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\begin{align*}
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UT &=\frac{1}{2}UV=\frac{1}{4}BC \mbox{ and } \\
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MT &=\frac{1}{2}PG=\frac{1}{4}AP.
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\end{align*}
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</cmath>
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Thus, from the Pythagorean Theorem we have
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<cmath> MV^2=MU^2=\frac{1}{16}\left(BC^2+AP^2\right).</cmath>
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Since we have shown already that <math>BC^2+AP^2=4R^2</math> is constant, we have that <math>MV=MU=\frac{R}{2}</math> and the locus of points <math>U,V</math> is indeed a circle of radius <math>\frac{R}{2}</math> with center <math>M</math>.
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</li>
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</ol>
  
  
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'''ii.)'''
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== See Also == {{IMO box|year=1988|before=First Question|num-a=2}}
  
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We claim that all points <math>U,V</math> lie on a circle centered at the midpoint of <math>OP</math>, <math>M</math> with radius <math>\frac{R}{2}</math>.  Let <math>T</math> be the midpoint of <math>UV</math>.  Since <math>H</math> is the midpoint of <math>BC</math>, it is clear that the projection of <math>T</math> onto <math>BC</math> is the midpoint of <math>H</math> and <math>P</math> (the projection of <math>A</math> onto <math>BC</math>).  Thus, we have that <math>MT</math> is perpendicular to <math>UV</math> and thus the triangle <math>MUV</math> is iscoceles.  We have <math>UT=\frac{1}{2}UV=\frac{1}{4}BC</math> and <math>MT=\frac{1}{2}PG=\frac{1}{4}AP</math>. Thus, from the Pythagorean Theorem we have <math>MV^2=MU^2=\frac{1}{16}\left(BC^2+AP^2\right)</math>.  Since we have shown already that <math>BC^2+AP^2=4R^2</math> is constant, we have that <math>MV=MU=\frac{R}{2}</math> and the locus of points <math>U,V</math> is indeed a circle of radius <math>\frac{R}{2}</math> with center <math>M</math>.
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[[Category:Olympiad Geometry Problems]]

Latest revision as of 10:32, 30 January 2021

Problem

Consider 2 concentric circles with radii $R$ and $r$ ($R>r$) with center $O$. Fix $P$ on the small circle and consider the variable chord $AP$ of the small circle. Points $B$ and $C$ lie on the large circle; $B,P,C$ are collinear and $BC$ is perpendicular to $AP$.

  1. For which values of $\angle OPA$ is the sum $BC^2+CA^2+AB^2$ extremal?
  2. What are the possible positions of the midpoints $U$ of $AB$ and $V$ of $AC$ as $A$ varies?

Solution

  1. We claim that the value $BC^2+CA^2+AB^2$ stays constant as $\angle OPA$ varies, and thus achieves its maximum at all value of $\angle OPA$. We have from the Pythagorean Theorem that $CA^2=AP^2+PC^2$ and $AB^2=AP^2+PB^2$ and so our expression becomes \[BC^2+PB^2+PC^2+2AP^2=2BC^2+2AP^2-2PB\cdot PC\] Since $PB\cdot PC$ is the power of the point $P$, it stays constant as $A$ varies. Thus, we are left to prove that the value $BC^2+AP^2$ stays constant as $\angle OPA$ varies. Let $G$ be the midpoint of $AP$ and let $H$ be the midpoint of $BC$. Since $OG$ is perpendicular to $AP$, we find that $PG=r\cos OPA$. Similarly, we find that $OH=r\sin OPC=r\cos OPA$. Thus, by the Pythagorean Theorem, we have \begin{align*} BC^2& =4(R^2-r^2\cos^2 OPA) \\ AP^2&=4r^2\cos^2OPA \end{align*} Now it is obvious that $BC^2+AP^2=4R^2$ is constant for all values of $\angle OPA$.
  2. We claim that all points $U,V$ lie on a circle centered at the midpoint of $OP$, $M$ with radius $\frac{R}{2}$. Let $T$ be the midpoint of $UV$. Since $H$ is the midpoint of $BC$, it is clear that the projection of $T$ onto $BC$ is the midpoint of $H$ and $P$ (the projection of $A$ onto $BC$). Thus, we have that $MT$ is perpendicular to $UV$ and thus the triangle $MUV$ is isosceles. We have \begin{align*} UT &=\frac{1}{2}UV=\frac{1}{4}BC \mbox{ and } \\ MT &=\frac{1}{2}PG=\frac{1}{4}AP. \end{align*} Thus, from the Pythagorean Theorem we have \[MV^2=MU^2=\frac{1}{16}\left(BC^2+AP^2\right).\] Since we have shown already that $BC^2+AP^2=4R^2$ is constant, we have that $MV=MU=\frac{R}{2}$ and the locus of points $U,V$ is indeed a circle of radius $\frac{R}{2}$ with center $M$.


See Also

1988 IMO (Problems) • Resources
Preceded by
First Question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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