https://artofproblemsolving.com/wiki/index.php?title=1989_IMO_Problems/Problem_4&feed=atom&action=history 1989 IMO Problems/Problem 4 - Revision history 2020-09-28T05:24:22Z Revision history for this page on the wiki MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1989_IMO_Problems/Problem_4&diff=27636&oldid=prev Cosinator: New page: Let $ABCD$ be a convex quadrilateral such that the sides $AB,AD,BC$ satisfy $AB=AD+BC$. There exists a point $P$ inside the quadrilateral at a d... 2008-08-30T19:28:28Z <p>New page: Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral such that the sides &lt;math&gt;AB,AD,BC&lt;/math&gt; satisfy &lt;math&gt;AB=AD+BC&lt;/math&gt;. There exists a point &lt;math&gt;P&lt;/math&gt; inside the quadrilateral at a d...</p> <p><b>New page</b></p><div>Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral such that the sides &lt;math&gt;AB,AD,BC&lt;/math&gt; satisfy &lt;math&gt;AB=AD+BC&lt;/math&gt;. There exists a point &lt;math&gt;P&lt;/math&gt; inside the quadrilateral at a distance &lt;math&gt;h&lt;/math&gt; from the line &lt;math&gt;CD&lt;/math&gt; such that &lt;math&gt;AP=h+AD&lt;/math&gt; and &lt;math&gt;BP=h+BC&lt;/math&gt;. Show that:<br /> <br /> &lt;math&gt;\frac{1}{\sqrt{h}}\ge\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}&lt;/math&gt;<br /> <br /> <br /> == Solution ==<br /> Without loss of generality, assume &lt;math&gt;AD\ge BC&lt;/math&gt;.<br /> <br /> Draw the circle with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;AD&lt;/math&gt; as well as the circle with center &lt;math&gt;B&lt;/math&gt; and radius &lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;AB=AD+BC&lt;/math&gt;, the intersection of circle &lt;math&gt;A&lt;/math&gt; with the line &lt;math&gt;AB&lt;/math&gt; is the same as the intersection of circle &lt;math&gt;B&lt;/math&gt; with the line &lt;math&gt;AB&lt;/math&gt;. Thus, the two circles are externally tangent to each other. <br /> <br /> Now draw the circle with center &lt;math&gt;P&lt;/math&gt; and radius &lt;math&gt;h&lt;/math&gt;. Since &lt;math&gt;h&lt;/math&gt; is the shortest distance from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;CD&lt;/math&gt;, we have that the circle &lt;math&gt;P&lt;/math&gt; is tangent to &lt;math&gt;CD&lt;/math&gt;. From the conditions &lt;math&gt;AP=h+AD&lt;/math&gt; and &lt;math&gt;BP=h+BC&lt;/math&gt;, by the same reasoning that we found circles &lt;math&gt;A,B&lt;/math&gt; are tangent, we find that circles &lt;math&gt;P,A&lt;/math&gt; and &lt;math&gt;P,B&lt;/math&gt; are externally tangent also. <br /> <br /> Fix the two externally tangent circles with centers &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; and let the points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; vary about their respective circles (without loss of generality, assume that one can read the letters &lt;math&gt;ABCD&lt;/math&gt; in counter clockwise order). Notice that the value &lt;math&gt;\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}&lt;/math&gt; is fixed. Thus, in order to prove the claim presented in the problem, we must maximize the value of &lt;math&gt;h&lt;/math&gt;. It is clear that all possible circles &lt;math&gt;P&lt;/math&gt; are contained within the region created by circles &lt;math&gt;A,B&lt;/math&gt; as well as their common tangent line. Now it is clear that &lt;math&gt;h&lt;/math&gt; is maximized when circle &lt;math&gt;P&lt;/math&gt; is tangent to the common tangent of circles &lt;math&gt;A,B&lt;/math&gt; as well as to the circles &lt;math&gt;A,B&lt;/math&gt;. We shall prove that the value of &lt;math&gt;h&lt;/math&gt; at this point is &lt;math&gt;\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}&lt;/math&gt;.<br /> <br /> Fix &lt;math&gt;C,D&lt;/math&gt; so that &lt;math&gt;CD&lt;/math&gt; is the common tangent to circles &lt;math&gt;A,B&lt;/math&gt;. Since &lt;math&gt;\angle ADC,\angle BCD&lt;/math&gt; are right, the length &lt;math&gt;CD&lt;/math&gt; is equal to the length of the segment from &lt;math&gt;B&lt;/math&gt; to the projection of &lt;math&gt;B&lt;/math&gt; onto &lt;math&gt;AD&lt;/math&gt; which by the Pythagorean Theorem is &lt;math&gt;\sqrt{(BC+AD)^2-(AD-BC)^2}=2\sqrt{BC\cdot AD}&lt;/math&gt;. Let the projection of &lt;math&gt;P&lt;/math&gt; onto &lt;math&gt;CD&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt; units away from &lt;math&gt;D&lt;/math&gt;. Then, as we found the length of &lt;math&gt;CD&lt;/math&gt;, we find<br /> <br /> &lt;math&gt;x^2+(AD-h)^2=(AD+h)^2\Rightarrow x^2=4AD\cdot h&lt;/math&gt;<br /> <br /> &lt;math&gt;(2\sqrt{BC\cdot AD}-x)^2+(BC-h)^2=(BC+h)^2\Rightarrow (2\sqrt{BC\cdot AD}-x)^2=4BC\cdot h&lt;/math&gt;<br /> <br /> Subtracting the second equation from the first yields<br /> <br /> &lt;math&gt;4x\sqrt{BC\cdot AD}-4BC\cdot AD=4AD\cdot h-4BC\cdot h&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow x=h\left(\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}\right)+\sqrt{BC\cdot AD}&lt;/math&gt;<br /> <br /> Since &lt;math&gt;x^2=2AD\cdot h&lt;/math&gt;, we have<br /> <br /> &lt;math&gt;0=h\left(\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}\right)-2\sqrt{AD\cdot h}+\sqrt{BC\cdot AD}&lt;/math&gt;<br /> <br /> Solving for &lt;math&gt;\sqrt{h}&lt;/math&gt; yields<br /> <br /> &lt;math&gt;\sqrt{h}=\frac{2\sqrt{AD}\pm\sqrt{4AD-4(AD-BC)}}{2\left(\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}\right)}&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow \sqrt{h}=\frac{\sqrt{AD}\pm\sqrt{BC}}{\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}}&lt;/math&gt;<br /> <br /> Since &lt;math&gt;\sqrt{h}\ne \frac{\sqrt{AD}+\sqrt{BC}}{\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}}&lt;/math&gt;, we have<br /> <br /> &lt;math&gt;\sqrt{h}=\frac{\sqrt{AD}-\sqrt{BC}}{\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}}&lt;/math&gt;<br /> <br /> &lt;math&gt;=\frac{\sqrt{AD\cdot BC}}{\sqrt{AD}+\sqrt{BC}}&lt;/math&gt;<br /> <br /> And since this is the maximimal value of &lt;math&gt;\sqrt{h}&lt;/math&gt;, the minimal value of &lt;math&gt;\frac{1}{\sqrt{h}}&lt;/math&gt; is &lt;math&gt;\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}&lt;/math&gt; as desired.</div> Cosinator