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| ==Problem==
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| Let <math>ABCD</math> be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that
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| <cmath>\frac {1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.</cmath>
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| When does equality hold?
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| ==Solution==
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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>.
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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>.
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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.
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| Equality occurs when <math>a = b</math> and <math>c = d</math>, or when <math>ABCD</math> is a kite.
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| == Resources ==
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| {{USAMO newbox|year=2004|before=First problem|num-a=2}}
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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 | | Let <math>ABCD</math> be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that |
Revision as of 20:47, 7 April 2013
Problem
Let 
be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that
![\[\frac {1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.\]](//latex.artofproblemsolving.com/d/7/5/d7576f585c60ebcc099d1c90d4c8eaf0158e292d.png)
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 ![$A\in [60^{\circ},120^{\circ}]$](//latex.artofproblemsolving.com/9/9/9/999a7ecbf631e0637f709565ea2f5b091c65dd93.png)
. 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
- <url>viewtopic.php?p=17439&sid=d212b9d95317a1fad7651771b6efa5bb Discussion on AoPS/MathLinks</url>
be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that
![\[\frac {1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.\]](http://latex.artofproblemsolving.com/d/7/5/d7576f585c60ebcc099d1c90d4c8eaf0158e292d.png)
Now we factor the desired expression into 
. Temporarily discarding the case where 
and 
, we can divide through by the 
to get the simplified expression 
.
. By the law of cosines, 
. Since each of the interior and exterior angles of the quadrilateral is at least 60 degrees, we have that ![$A\in [60^{\circ},120^{\circ}]$](http://latex.artofproblemsolving.com/9/9/9/999a7ecbf631e0637f709565ea2f5b091c65dd93.png)
. 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 
.
. 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.
