Difference between revisions of "2004 USAMO Problems/Problem 1"
m (fix) |
(→Solution) |
||
(5 intermediate revisions by 4 users not shown) | |||
Line 1: | Line 1: | ||
==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> | ||
Line 6: | Line 6: | ||
When does equality hold? | When does equality hold? | ||
− | ==Solution== | + | ==Solution 1== |
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | 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> | ||
+ | |||
+ | ==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>. | ||
− | 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. | ||
Line 18: | Line 24: | ||
{{USAMO newbox|year=2004|before=First problem|num-a=2}} | {{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 16:28, 1 September 2021
Contents
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 1
It suffices to show that because the other side of the inequality follows by symmetry. By Pitot's Theorem, we have
. WLOG, let
. Then we have that
, so we must show
. If
, we are done. Therefore, we will assume for the rest of the proof that
,
.
Then we would like to show that
By the Law of Cosines, we have that
where
(this is because
). Therefore,
Then we wish to show that
and we are done.
Solution 2
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.