https://artofproblemsolving.com/wiki/index.php?title=1989_IMO_Problems/Problem_4&feed=atom&action=history1989 IMO Problems/Problem 4 - Revision history2024-03-28T10:42:21ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1989_IMO_Problems/Problem_4&diff=144065&oldid=prevHamstpan38825 at 15:57, 30 January 20212021-01-30T15:57:20Z<p></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 15:57, 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;"></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>And since this is the maximimal value of <math>\sqrt{h}</math>, the minimal value of <math>\frac{1}{\sqrt{h}}</math> is <math>\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}</math> as desired.</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>And since this is the maximimal value of <math>\sqrt{h}</math>, the minimal value of <math>\frac{1}{\sqrt{h}}</math> is <math>\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}</math> as desired.</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=1989|num-b=3|num-a=5}}</ins></div></td></tr>
</table>Hamstpan38825https://artofproblemsolving.com/wiki/index.php?title=1989_IMO_Problems/Problem_4&diff=27636&oldid=prevCosinator: New page: Let <math>ABCD</math> be a convex quadrilateral such that the sides <math>AB,AD,BC</math> satisfy <math>AB=AD+BC</math>. There exists a point <math>P</math> inside the quadrilateral at a d...2008-08-30T19:28:28Z<p>New page: Let <math>ABCD</math> be a convex quadrilateral such that the sides <math>AB,AD,BC</math> satisfy <math>AB=AD+BC</math>. There exists a point <math>P</math> inside the quadrilateral at a d...</p>
<p><b>New page</b></p><div>Let <math>ABCD</math> be a convex quadrilateral such that the sides <math>AB,AD,BC</math> satisfy <math>AB=AD+BC</math>. There exists a point <math>P</math> inside the quadrilateral at a distance <math>h</math> from the line <math>CD</math> such that <math>AP=h+AD</math> and <math>BP=h+BC</math>. Show that:<br />
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<math>\frac{1}{\sqrt{h}}\ge\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}</math><br />
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== Solution ==<br />
Without loss of generality, assume <math>AD\ge BC</math>.<br />
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Draw the circle with center <math>A</math> and radius <math>AD</math> as well as the circle with center <math>B</math> and radius <math>BC</math>. Since <math>AB=AD+BC</math>, the intersection of circle <math>A</math> with the line <math>AB</math> is the same as the intersection of circle <math>B</math> with the line <math>AB</math>. Thus, the two circles are externally tangent to each other. <br />
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Now draw the circle with center <math>P</math> and radius <math>h</math>. Since <math>h</math> is the shortest distance from <math>P</math> to <math>CD</math>, we have that the circle <math>P</math> is tangent to <math>CD</math>. From the conditions <math>AP=h+AD</math> and <math>BP=h+BC</math>, by the same reasoning that we found circles <math>A,B</math> are tangent, we find that circles <math>P,A</math> and <math>P,B</math> are externally tangent also. <br />
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Fix the two externally tangent circles with centers <math>A</math> and <math>B</math> and let the points <math>C</math> and <math>D</math> vary about their respective circles (without loss of generality, assume that one can read the letters <math>ABCD</math> in counter clockwise order). Notice that the value <math>\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}</math> is fixed. Thus, in order to prove the claim presented in the problem, we must maximize the value of <math>h</math>. It is clear that all possible circles <math>P</math> are contained within the region created by circles <math>A,B</math> as well as their common tangent line. Now it is clear that <math>h</math> is maximized when circle <math>P</math> is tangent to the common tangent of circles <math>A,B</math> as well as to the circles <math>A,B</math>. We shall prove that the value of <math>h</math> at this point is <math>\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}</math>.<br />
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Fix <math>C,D</math> so that <math>CD</math> is the common tangent to circles <math>A,B</math>. Since <math>\angle ADC,\angle BCD</math> are right, the length <math>CD</math> is equal to the length of the segment from <math>B</math> to the projection of <math>B</math> onto <math>AD</math> which by the Pythagorean Theorem is <math>\sqrt{(BC+AD)^2-(AD-BC)^2}=2\sqrt{BC\cdot AD}</math>. Let the projection of <math>P</math> onto <math>CD</math> be <math>x</math> units away from <math>D</math>. Then, as we found the length of <math>CD</math>, we find<br />
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<math>x^2+(AD-h)^2=(AD+h)^2\Rightarrow x^2=4AD\cdot h</math><br />
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<math>(2\sqrt{BC\cdot AD}-x)^2+(BC-h)^2=(BC+h)^2\Rightarrow (2\sqrt{BC\cdot AD}-x)^2=4BC\cdot h</math><br />
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Subtracting the second equation from the first yields<br />
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<math>4x\sqrt{BC\cdot AD}-4BC\cdot AD=4AD\cdot h-4BC\cdot h</math><br />
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<math>\Rightarrow x=h\left(\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}\right)+\sqrt{BC\cdot AD}</math><br />
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Since <math>x^2=2AD\cdot h</math>, we have<br />
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<math>0=h\left(\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}\right)-2\sqrt{AD\cdot h}+\sqrt{BC\cdot AD}</math><br />
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Solving for <math>\sqrt{h}</math> yields<br />
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<math>\sqrt{h}=\frac{2\sqrt{AD}\pm\sqrt{4AD-4(AD-BC)}}{2\left(\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}\right)}</math><br />
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<math>\Rightarrow \sqrt{h}=\frac{\sqrt{AD}\pm\sqrt{BC}}{\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}}</math><br />
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Since <math>\sqrt{h}\ne \frac{\sqrt{AD}+\sqrt{BC}}{\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}}</math>, we have<br />
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<math>\sqrt{h}=\frac{\sqrt{AD}-\sqrt{BC}}{\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}}</math><br />
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<math>=\frac{\sqrt{AD\cdot BC}}{\sqrt{AD}+\sqrt{BC}}</math><br />
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And since this is the maximimal value of <math>\sqrt{h}</math>, the minimal value of <math>\frac{1}{\sqrt{h}}</math> is <math>\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}</math> as desired.</div>Cosinator