https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=AbsolutZer0&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T12:11:54ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2004_USAMO_Problems/Problem_1&diff=335862004 USAMO Problems/Problem 12010-02-18T23:12:19Z<p>AbsolutZer0: /* Solution */</p>
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<div>==Problem==<br />
Let <math>ABCD</math> be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that<br />
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<cmath>\frac {1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.</cmath><br />
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When does equality hold?<br />
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==Solution==<br />
By a well-known property of tangential quadrilaterals, the sum of the two pairs of opposite sides are equal; hence <math>a + c = b + d \Rightarrow a - b = d - c \Rightarrow |a - b| = |d - c|</math> Now we factor the desired expression into <math>\frac {|d - c|(c^2 + d^2 + cd)}{3} \le|a - b|(a^2 + b^2 + ab)\le 3|d - c|(c^2 + d^2 + cd)</math>. Temporarily discarding the case where <math>a = b</math> and <math>c = d</math>, we can divide through by the <math>|a - b| = |d - c|</math> to get the simplified expression <math>(c^2 + d^2 + cd)/3\le a^2 + b^2 + ab\le 3(c^2 + d^2 + cd)</math>.<br />
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Now, draw diagonal <math>BD</math>. By the law of cosines, <math>c^2 + d^2 + 2cd\cos A = BD</math>. Since each of the interior and exterior angles of the quadrilateral is at least 60 degrees, we have that <math>A\in [60^{\circ},120^{\circ}]</math>. Cosine is monotonically decreasing on this interval, so by setting <math>A</math> at the extreme values, we see that <math>c^2 + d^2 - cd\le BD^2 \le c^2 + d^2 + cd</math>. Applying the law of cosines analogously to <math>a</math> and <math>b</math>, we see that <math>a^2 + b^2 - ab\le BD^2 \le a^2 + b^2 + ab</math>; we hence have <math>c^2 + d^2 - cd\le BD^2 \le a^2 + b^2 + ab</math> and <math>a^2 + b^2 - ab\le BD^2 \le c^2 + d^2 + cd</math>. <br />
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We wrap up first by considering the second inequality. Because <math>c^2 + d^2 - cd\le BD^2 \le a^2 + b^2 + ab</math>, <math>\text{RHS}\ge 3(a^2 + b^2 - ab)</math>. This latter expression is of course greater than or equal to <math>a^2 + b^2 + ab</math> because the inequality can be rearranged to <math>2(a - b)^2\ge 0</math>, which is always true. Multiply the first inequality by <math>3</math> and we see that it is simply the second inequality with the variables swapped; hence by symmetry it is true as well.<br />
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Equality occurs when <math>a = b</math> and <math>c = d</math>, or when <math>ABCD</math> is a kite.<br />
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== Resources ==<br />
{{USAMO newbox|year=2004|before=First problem|num-a=2}}<br />
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* <url>viewtopic.php?t=1478389 Discussion on AoPS/MathLinks</url><br />
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[[Category:Olympiad Geometry Problems]]<br />
[[Category:Olympiad Inequality Problems]]</div>AbsolutZer0