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Difference between revisions of "2004 USAMO Problems/Problem 1"

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==Problem==
 
==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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(''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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When does equality hold?
 
When does equality hold?
  
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==Solution==
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==Solution 1==
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It suffices to show that <cmath>|BC^3-CD^3| \leq 3|AB^3-AD^3|,</cmath> because the other side of the inequality follows by symmetry. By Pitot's Theorem, we have <math>AB-AD=BC-CD</math>. WLOG, let <math>AB \geq AD</math>. Then we have that <math>BC \geq CD</math>, so we must show <math>BC^3-CD^3 \leq 3(AB^3-AD^3)</math>. If <math>AB=AD</math>, we are done. Therefore, we will assume for the rest of the proof that <math>AB>AD</math>, <math>BC>CD</math>.
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Then we would like to show that <cmath>BC^2+BC\cdot CD+CD^2 \leq 3(AB^2+AB\cdot AD+AD^2).</cmath>
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By the Law of Cosines, we have that <cmath>BD^2=BC^2+k_1(BC\cdot CD)+CD^2=AB^2+k_2(AB\cdot AD)+AD^2,</cmath> where <math>-1 \leq k_1, k_2 \leq 1</math> (this is because <math>60^{\circ} \leq \angle BAD,\angle BCD \leq 120^{\circ}</math>). Therefore, <cmath>3(BC^2-BC\cdot CD+CD^2) \leq 3(BC^2+k_1(BC\cdot CD)+CD^2)=3(AB^2+k_2(AB\cdot AD)+AD^2)\leq 3(AB^2+AB\cdot AD+AD^2).</cmath> Then we wish to show that <cmath>3(BC^2-BC\cdot CD+CD^2)-(BC^2+BC\cdot CD+CD^2) \geq 0 \rightarrow 2(BC-CD)^2 \geq 0,</cmath> and we are done. <math>\blacksquare</math>
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==Solution 2==
 
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>.
  
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[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Inequality Problems]]
 
[[Category:Olympiad Inequality Problems]]
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{{MAA Notice}}

Latest revision as of 15:28, 1 September 2021

Problem

(Titu Andreescu) 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 1

It suffices to show that \[|BC^3-CD^3| \leq 3|AB^3-AD^3|,\] because the other side of the inequality follows by symmetry. By Pitot's Theorem, we have $AB-AD=BC-CD$. WLOG, let $AB \geq AD$. Then we have that $BC \geq CD$, so we must show $BC^3-CD^3 \leq 3(AB^3-AD^3)$. If $AB=AD$, we are done. Therefore, we will assume for the rest of the proof that $AB>AD$, $BC>CD$.

Then we would like to show that \[BC^2+BC\cdot CD+CD^2 \leq 3(AB^2+AB\cdot AD+AD^2).\] By the Law of Cosines, we have that \[BD^2=BC^2+k_1(BC\cdot CD)+CD^2=AB^2+k_2(AB\cdot AD)+AD^2,\] where $-1 \leq k_1, k_2 \leq 1$ (this is because $60^{\circ} \leq \angle BAD,\angle BCD \leq 120^{\circ}$). Therefore, \[3(BC^2-BC\cdot CD+CD^2) \leq 3(BC^2+k_1(BC\cdot CD)+CD^2)=3(AB^2+k_2(AB\cdot AD)+AD^2)\leq 3(AB^2+AB\cdot AD+AD^2).\] Then we wish to show that \[3(BC^2-BC\cdot CD+CD^2)-(BC^2+BC\cdot CD+CD^2) \geq 0 \rightarrow 2(BC-CD)^2 \geq 0,\] and we are done. $\blacksquare$

Solution 2

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)
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Problem 2
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All USAMO Problems and Solutions
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