https://artofproblemsolving.com/wiki/index.php?title=1979_IMO_Problems/Problem_4&feed=atom&action=history1979 IMO Problems/Problem 4 - Revision history2024-03-28T14:20:50ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1979_IMO_Problems/Problem_4&diff=143993&oldid=prevHamstpan38825 at 03:01, 30 January 20212021-01-30T03:01:54Z<p></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 03:01, 30 January 2021</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>Hence, the point <math>R</math> for which the ratio <math>\frac{QP+PR}{QR}</math> is maximum is the point of intersection of the ray <math>PT</math> with the sphere with center <math>P</math> and radius <math>PQ</math> (or, respectively, it can be any arbitrary point on the intersection of the plane <math>p</math> with the sphere with center <math>P</math> and radius <math>PQ</math> if <math>PQ\perp p</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>Hence, the point <math>R</math> for which the ratio <math>\frac{QP+PR}{QR}</math> is maximum is the point of intersection of the ray <math>PT</math> with the sphere with center <math>P</math> and radius <math>PQ</math> (or, respectively, it can be any arbitrary point on the intersection of the plane <math>p</math> with the sphere with center <math>P</math> and radius <math>PQ</math> if <math>PQ\perp p</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>== See Also == {{IMO box|year=1979|num-b=<del class="diffchange diffchange-inline">1</del>|num-a=<del class="diffchange diffchange-inline">3</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">This solution was posted and copyrighted by darij grinberg. The original thread for this problem can be found here: [https://aops.com/community/p392254]</ins></div></td></tr>
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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> </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>== See Also == {{IMO box|year=1979|num-b=<ins class="diffchange diffchange-inline">3</ins>|num-a=<ins class="diffchange diffchange-inline">5</ins>}}</div></td></tr>
</table>Hamstpan38825https://artofproblemsolving.com/wiki/index.php?title=1979_IMO_Problems/Problem_4&diff=143992&oldid=prevHamstpan38825: Created page with "==Problem== We consider a point <math>P</math> in a plane <math>p</math> and a point <math>Q \not\in p</math>. Determine all the points <math>R</math> from <math>p</math> for..."2021-01-30T03:01:06Z<p>Created page with "==Problem== We consider a point <math>P</math> in a plane <math>p</math> and a point <math>Q \not\in p</math>. Determine all the points <math>R</math> from <math>p</math> for..."</p>
<p><b>New page</b></p><div>==Problem==<br />
We consider a point <math>P</math> in a plane <math>p</math> and a point <math>Q \not\in p</math>. Determine all the points <math>R</math> from <math>p</math> for which<cmath> \frac{QP+PR}{QR} </cmath>is maximum.<br />
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==Solution==<br />
Let <math>T</math> be the orthogonal projection of the point <math>Q</math> on the plane <math>p</math>. Then, the line <math>PT</math> is the orthogonal projection of the line <math>PQ</math> on the plane <math>p</math>, and thus forms the least angle with the line <math>PQ</math> among all lines through the point <math>P</math> which lie in the plane <math>p</math>; hence, <math>\measuredangle RPQ\geq\measuredangle TPQ</math>, and equality holds if and only if the point <math>R</math> lies on the ray <math>PT</math> (the only exception is when <math>PQ\perp p</math>; in this case, <math>P = T</math>, so the ray <math>PT</math> is undefined, and equality holds for all points <math>R</math> in the plane <math>p</math>, since we always have <math>\angle RPQ = \angle TPQ = 90^{\circ}</math>).<br />
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From <math>\measuredangle RPQ\geq\measuredangle TPQ</math>, it follows that <math>\frac{\measuredangle RPQ}{2}\geq\frac{\measuredangle TPQ}{2}</math>; also, since the angles <math>\angle RPQ</math> and <math>\angle TPQ</math> are <math>180^{\circ}</math>, their half-angles <math>\frac{\measuredangle RPQ}{2}</math> and <math>\frac{\measuredangle TPQ}{2}</math> are < 90°, so that from <math>\frac{\measuredangle RPQ}{2}\geq\frac{\measuredangle TPQ}{2}</math> we can conclude that <math>\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}</math>. Equality holds, as in the above, if and only if the point <math>R</math> lies on the ray <math>PT</math> (and, respectively, for all points <math>R</math> in the plane <math>p</math> if <math>PQ\perp p</math>).<br />
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On the other hand, we obviously have <math>\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1</math> with equality if and only if <math>\frac{\measuredangle QRP-\measuredangle PQR}{2}=0^{\circ}</math>, i. e. if and only if < QRP = < PQR, i. e. if and only if triangle PQR is isosceles with base <math>QR</math>, i. e. if and only if <math>PR = PQ</math>, i. e. if and only if the point <math>R</math> lies on the sphere with center <math>P</math> and radius <math>PQ</math>.<br />
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Now, applying the Mollweide theorem in triangle <math>QPR</math>, we get<br />
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<math>\frac{QP+PR}{QR}=\frac{\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}}{\sin\frac{\measuredangle RPQ}{2}}</math><br />
<math>\leq\frac{1}{\sin\frac{\measuredangle RPQ}{2}}</math> (since <math>\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1</math>)<br />
<math>\leq\frac{1}{\sin\frac{\measuredangle TPQ}{2}}</math> (since <math>\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}</math>),<br />
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and equality holds here if and only if equality holds in both of the inequalities <math>\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}</math> and <math>\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1</math> that we have used, i. e. if and only if the point <math>R</math> lies both on the ray <math>PT</math> (this condition should be ignored if <math>PQ\perp p</math>) and on the sphere with center <math>P</math> and radius <math>PQ</math>.<br />
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Hence, the point <math>R</math> for which the ratio <math>\frac{QP+PR}{QR}</math> is maximum is the point of intersection of the ray <math>PT</math> with the sphere with center <math>P</math> and radius <math>PQ</math> (or, respectively, it can be any arbitrary point on the intersection of the plane <math>p</math> with the sphere with center <math>P</math> and radius <math>PQ</math> if <math>PQ\perp p</math>).<br />
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== See Also == {{IMO box|year=1979|num-b=1|num-a=3}}</div>Hamstpan38825