Difference between revisions of "2004 USAMO Problems/Problem 1"
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==Problem== | ==Problem== | ||
− | Let <math>ABCD</math> be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that | + | (''Titu Andreescu'') Let <math>ABCD</math> be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that |
<cmath>\frac {1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.</cmath> | <cmath>\frac {1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.</cmath> | ||
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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>. | 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>. | ||
− | 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 | + | 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>. |
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. | 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. | ||
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== Resources == | == Resources == | ||
− | {{USAMO newbox|year=2004| | + | {{USAMO newbox|year=2004|before=First problem|num-a=2}} |
− | * <url>viewtopic.php? | + | * <url>viewtopic.php?p=17439&sid=d212b9d95317a1fad7651771b6efa5bb Discussion on AoPS/MathLinks</url> |
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
[[Category:Olympiad Inequality Problems]] | [[Category:Olympiad Inequality Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:21, 17 July 2014
Problem
(Titu Andreescu) Let be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that
When does equality hold?
Solution
By a well-known property of tangential quadrilaterals, the sum of the two pairs of opposite sides are equal; hence Now we factor the desired expression into . Temporarily discarding the case where and , we can divide through by the to get the simplified expression .
Now, draw diagonal . By the law of cosines, . Since each of the interior and exterior angles of the quadrilateral is at least 60 degrees, we have that . Cosine is monotonically decreasing on this interval, so by setting at the extreme values, we see that . Applying the law of cosines analogously to and , we see that ; we hence have and .
We wrap up first by considering the second inequality. Because , . This latter expression is of course greater than or equal to because the inequality can be rearranged to , which is always true. Multiply the first inequality by and we see that it is simply the second inequality with the variables swapped; hence by symmetry it is true as well.
Equality occurs when and , or when is a kite.
Resources
2004 USAMO (Problems • Resources) | ||
Preceded by First problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
- <url>viewtopic.php?p=17439&sid=d212b9d95317a1fad7651771b6efa5bb Discussion on AoPS/MathLinks</url>
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.