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2004 USAMO Problems/Problem 1: Difference between revisions

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== Resources ==
== Resources ==
{{USAMO newbox|year=2004|before=First problem|num-a=2}}
{{USAMO newbox|year=2004|before=First problem|num-a=2}}
==Problem==
Let <math>ABCD</math> be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that


* <url>viewtopic.php?t=1478389 Discussion on AoPS/MathLinks</url>
<cmath>\frac {1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.</cmath>
 
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 <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 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.
 
Equality occurs when <math>a = b</math> and <math>c = d</math>, or when <math>ABCD</math> is a kite.
 
== Resources ==
{{USAMO newbox|year=2004|before=First problem|num-a=2}}
 
* <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]]

Revision as of 20:47, 7 April 2013

Problem

Let $ABCD$ be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that

\[\frac {1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.\]

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 $a + c = b + d \Rightarrow a - b = d - c \Rightarrow |a - b| = |d - c|$ Now we factor the desired expression into $\frac {|d - c|(c^2 + d^2 + cd)}{3} \le|a - b|(a^2 + b^2 + ab)\le 3|d - c|(c^2 + d^2 + cd)$. Temporarily discarding the case where $a = b$ and $c = d$, we can divide through by the $|a - b| = |d - c|$ to get the simplified expression $(c^2 + d^2 + cd)/3\le a^2 + b^2 + ab\le 3(c^2 + d^2 + cd)$.

Now, draw diagonal $BD$. By the law of cosines, $c^2 + d^2 + 2cd\cos A = BD$. Since each of the interior and exterior angles of the quadrilateral is at least 60 degrees, we have that $A\in [60^{\circ},120^{\circ}]$. Cosine is monotonically decreasing on this interval, so by setting $A$ at the extreme values, we see that $c^2 + d^2 - cd\le BD^2 \le c^2 + d^2 + cd$. Applying the law of cosines analogously to $a$ and $b$, we see that $a^2 + b^2 - ab\le BD^2 \le a^2 + b^2 + ab$; we hence have $c^2 + d^2 - cd\le BD^2 \le a^2 + b^2 + ab$ and $a^2 + b^2 - ab\le BD^2 \le c^2 + d^2 + cd$.

We wrap up first by considering the second inequality. Because $c^2 + d^2 - cd\le BD^2 \le a^2 + b^2 + ab$, $\text{RHS}\ge 3(a^2 + b^2 - ab)$. This latter expression is of course greater than or equal to $a^2 + b^2 + ab$ because the inequality can be rearranged to $2(a - b)^2\ge 0$, which is always true. Multiply the first inequality by $3$ 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 $a = b$ and $c = d$, or when $ABCD$ 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

Problem

Let $ABCD$ be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that

\[\frac {1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.\]

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 $a + c = b + d \Rightarrow a - b = d - c \Rightarrow |a - b| = |d - c|$ Now we factor the desired expression into $\frac {|d - c|(c^2 + d^2 + cd)}{3} \le|a - b|(a^2 + b^2 + ab)\le 3|d - c|(c^2 + d^2 + cd)$. Temporarily discarding the case where $a = b$ and $c = d$, we can divide through by the $|a - b| = |d - c|$ to get the simplified expression $(c^2 + d^2 + cd)/3\le a^2 + b^2 + ab\le 3(c^2 + d^2 + cd)$.

Now, draw diagonal $BD$. By the law of cosines, $c^2 + d^2 + 2cd\cos A = BD$. Since each of the interior and exterior angles of the quadrilateral is at least 60 degrees, we have that $A\in [60^{\circ},120^{\circ}]$. Cosine is monotonically decreasing on this interval, so by setting $A$ at the extreme values, we see that $c^2 + d^2 - cd\le BD^2 \le c^2 + d^2 + cd$. Applying the law of cosines analogously to $a$ and $b$, we see that $a^2 + b^2 - ab\le BD^2 \le a^2 + b^2 + ab$; we hence have $c^2 + d^2 - cd\le BD^2 \le a^2 + b^2 + ab$ and $a^2 + b^2 - ab\le BD^2 \le c^2 + d^2 + cd$.

We wrap up first by considering the second inequality. Because $c^2 + d^2 - cd\le BD^2 \le a^2 + b^2 + ab$, $\text{RHS}\ge 3(a^2 + b^2 - ab)$. This latter expression is of course greater than or equal to $a^2 + b^2 + ab$ because the inequality can be rearranged to $2(a - b)^2\ge 0$, which is always true. Multiply the first inequality by $3$ 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 $a = b$ and $c = d$, or when $ABCD$ 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>
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