https://artofproblemsolving.com/wiki/index.php?title=1988_IMO_Problems/Problem_1&feed=atom&action=history
1988 IMO Problems/Problem 1 - Revision history
2024-03-29T13:20:05Z
Revision history for this page on the wiki
MediaWiki 1.31.1
https://artofproblemsolving.com/wiki/index.php?title=1988_IMO_Problems/Problem_1&diff=144054&oldid=prev
Hamstpan38825 at 15:32, 30 January 2021
2021-01-30T15:32:35Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 15:32, 30 January 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l41" >Line 41:</td>
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<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">== See Also == {{IMO box|year=1988|before=First Question|num-a=2}}</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">[[Category:Olympiad Geometry Problems]]</ins></div></td></tr>
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Hamstpan38825
https://artofproblemsolving.com/wiki/index.php?title=1988_IMO_Problems/Problem_1&diff=135403&oldid=prev
V v: /* Solution */
2020-10-19T20:11:45Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 20:11, 19 October 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l39" >Line 39:</td>
<td colspan="2" class="diff-lineno">Line 39:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath> MV^2=MU^2=\frac{1}{16}\left(BC^2+AP^2\right).</cmath></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath> MV^2=MU^2=\frac{1}{16}\left(BC^2+AP^2\right).</cmath></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>.</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">LOL JUST TESTING</del></div></td><td colspan="2"> </td></tr>
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V v
https://artofproblemsolving.com/wiki/index.php?title=1988_IMO_Problems/Problem_1&diff=135402&oldid=prev
V v: /* Solution */
2020-10-19T20:11:01Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 20:11, 19 October 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l39" >Line 39:</td>
<td colspan="2" class="diff-lineno">Line 39:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath> MV^2=MU^2=\frac{1}{16}\left(BC^2+AP^2\right).</cmath></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath> MV^2=MU^2=\frac{1}{16}\left(BC^2+AP^2\right).</cmath></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">LOL JUST TESTING</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div></li></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div></li></div></td></tr>
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V v
https://artofproblemsolving.com/wiki/index.php?title=1988_IMO_Problems/Problem_1&diff=38443&oldid=prev
Aopsvd: /* Problem */
2011-05-10T00:33:07Z
<p><span dir="auto"><span class="autocomment">Problem</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 00:33, 10 May 2011</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l2" >Line 2:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>i<del class="diffchange diffchange-inline">.) </del>For which values of <math>\angle OPA</math> is the sum <math>BC^2+CA^2+AB^2</math> extremal?  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline"><ol type="</ins>i<ins class="diffchange diffchange-inline">"></ins></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div> </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline"><li></ins>For which values of <math>\angle OPA</math> is the sum <math>BC^2+CA^2+AB^2</math> extremal?<ins class="diffchange diffchange-inline"></li></ins></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">ii.) </del>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?  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline"><li></ins>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?<ins class="diffchange diffchange-inline"></li></ins></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div> </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline"></ol></ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution==</div></td></tr>
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Aopsvd
https://artofproblemsolving.com/wiki/index.php?title=1988_IMO_Problems/Problem_1&diff=38442&oldid=prev
Aopsvd: /* Solution */
2011-05-10T00:27:06Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 00:27, 10 May 2011</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">'''</del>i<del class="diffchange diffchange-inline">.)'''</del></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline"><ol type="</ins>i<ins class="diffchange diffchange-inline">"></ins></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div> </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline"><li></ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><<del class="diffchange diffchange-inline">math</del>>BC^2+PB^2+PC^2+2AP^2=2BC^2+2AP^2-2PB\cdot PC</<del class="diffchange diffchange-inline">math</del>></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><<ins class="diffchange diffchange-inline">cmath</ins>> BC^2+PB^2+PC^2+2AP^2=2BC^2+2AP^2-2PB\cdot PC </<ins class="diffchange diffchange-inline">cmath</ins>></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>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>. <del class="diffchange diffchange-inline"> </del>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  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>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  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><<del class="diffchange diffchange-inline">math</del>>BC^2=4(R^2-r^2\cos^2 OPA)<del class="diffchange diffchange-inline"></math></del></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><<ins class="diffchange diffchange-inline">cmath</ins>></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div> </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">\begin{align*}</ins></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline"><math></del>AP^2=4r^2\cos^2OPA</<del class="diffchange diffchange-inline">math</del>></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>BC^2<ins class="diffchange diffchange-inline">& </ins>=4(R^2-r^2\cos^2 OPA) <ins class="diffchange diffchange-inline">\\</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>AP^2<ins class="diffchange diffchange-inline">&</ins>=4r^2\cos^2OPA</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">\end{align*}</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div></<ins class="diffchange diffchange-inline">cmath</ins>></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Now it is obvious that <math>BC^2+AP^2=4R^2</math> is constant for all values of <math>\angle OPA</math>.   </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Now it is obvious that <math>BC^2+AP^2=4R^2</math> is constant for all values of <math>\angle OPA</math>.   </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></li></ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"><li></ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">'''ii.)'''</del></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>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 <ins class="diffchange diffchange-inline">isosceles</ins>.  We have</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div> </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><<ins class="diffchange diffchange-inline">cmath</ins>></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>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 <del class="diffchange diffchange-inline">iscoceles</del>.  We have <<del class="diffchange diffchange-inline">math</del>>UT=\frac{1}{2}UV=\frac{1}{4}BC<del class="diffchange diffchange-inline"></math> </del>and <del class="diffchange diffchange-inline"><math></del>MT=\frac{1}{2}PG=\frac{1}{4}AP</<del class="diffchange diffchange-inline">math</del>><del class="diffchange diffchange-inline">. </del>Thus, from the Pythagorean Theorem we have <<del class="diffchange diffchange-inline">math</del>>MV^2=MU^2=\frac{1}{16}\left(BC^2+AP^2\right)</<del class="diffchange diffchange-inline">math</del>><del class="diffchange diffchange-inline">.  </del>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>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">\begin{align*}</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>UT <ins class="diffchange diffchange-inline">&</ins>=\frac{1}{2}UV=\frac{1}{4}BC <ins class="diffchange diffchange-inline">\mbox{ </ins>and <ins class="diffchange diffchange-inline">} \\</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>MT <ins class="diffchange diffchange-inline">&</ins>=\frac{1}{2}PG=\frac{1}{4}AP<ins class="diffchange diffchange-inline">.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">\end{align*}</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div></<ins class="diffchange diffchange-inline">cmath</ins>></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Thus, from the Pythagorean Theorem we have</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><<ins class="diffchange diffchange-inline">cmath</ins>> MV^2=MU^2=\frac{1}{16}\left(BC^2+AP^2\right)<ins class="diffchange diffchange-inline">.</ins></<ins class="diffchange diffchange-inline">cmath</ins>></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>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>.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline"></li></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline"></ol></ins></div></td></tr>
</table>
Aopsvd
https://artofproblemsolving.com/wiki/index.php?title=1988_IMO_Problems/Problem_1&diff=38441&oldid=prev
Aopsvd at 00:00, 10 May 2011
2011-05-10T00:00:17Z
<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 00:00, 10 May 2011</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
<td colspan="2" class="diff-lineno">Line 1:</td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">==Problem==</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l6" >Line 6:</td>
<td colspan="2" class="diff-lineno">Line 7:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">'''</del>Solution<del class="diffchange diffchange-inline">'''</del></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">==</ins>Solution<ins class="diffchange diffchange-inline">==</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>'''i.)'''</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>'''i.)'''</div></td></tr>
<tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l12" >Line 12:</td>
<td colspan="2" class="diff-lineno">Line 13:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><math>BC^2+PB^2+PC^2+2AP^2=2BC^2+2AP^2-2PB\<del class="diffchange diffchange-inline">cdotPC</del></math></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><math>BC^2+PB^2+PC^2+2AP^2=2BC^2+2AP^2-2PB\<ins class="diffchange diffchange-inline">cdot PC</ins></math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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  </div></td></tr>
</table>
Aopsvd
https://artofproblemsolving.com/wiki/index.php?title=1988_IMO_Problems/Problem_1&diff=27486&oldid=prev
Cosinator: page, problem, solution
2008-08-20T19:53:40Z
<p>page, problem, solution</p>
<p><b>New page</b></p><div>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>. <br />
<br />
i.) For which values of <math>\angle OPA</math> is the sum <math>BC^2+CA^2+AB^2</math> extremal? <br />
<br />
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? <br />
<br />
<br />
'''Solution'''<br />
<br />
'''i.)'''<br />
<br />
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<br />
<br />
<math>BC^2+PB^2+PC^2+2AP^2=2BC^2+2AP^2-2PB\cdotPC</math><br />
<br />
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 <br />
<br />
<math>BC^2=4(R^2-r^2\cos^2 OPA)</math><br />
<br />
<math>AP^2=4r^2\cos^2OPA</math><br />
<br />
Now it is obvious that <math>BC^2+AP^2=4R^2</math> is constant for all values of <math>\angle OPA</math>. <br />
<br />
<br />
'''ii.)'''<br />
<br />
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>.</div>
Cosinator