Difference between revisions of "1979 USAMO Problems/Problem 3"

(Hint)
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random. Its value is <math>a</math>. Another member is picked at random, independently of the first. Its value is <math>b</math>. Then a third value, <math>c</math>. Show that the probability that <math>a + b +c</math> is divisible by <math>3</math> is at least <math>\frac14</math>.
 
random. Its value is <math>a</math>. Another member is picked at random, independently of the first. Its value is <math>b</math>. Then a third value, <math>c</math>. Show that the probability that <math>a + b +c</math> is divisible by <math>3</math> is at least <math>\frac14</math>.
  
==Hint==
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==First Hint==
 
The given problem is equivalent to proving that <math>4(x^3 + y^3 + z^3 + 6xyz) \ge (x + y + z)^3</math>.
 
The given problem is equivalent to proving that <math>4(x^3 + y^3 + z^3 + 6xyz) \ge (x + y + z)^3</math>.
 +
==Second Hint==
 +
What do you do with ''homogenous'' inequalities? (A function <math>f</math> is ''homogenous'' of degree <math>d</math> if <math>f(kx) = k^df(x)</math> for all x in the domain of <math>f</math>.)
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==Solution==
 
==Solution==
 
{{solution}}
 
{{solution}}

Revision as of 11:07, 19 April 2014

Problem

$a_1, a_2, \ldots, a_n$ is an arbitrary sequence of positive integers. A member of the sequence is picked at random. Its value is $a$. Another member is picked at random, independently of the first. Its value is $b$. Then a third value, $c$. Show that the probability that $a + b +c$ is divisible by $3$ is at least $\frac14$.

First Hint

The given problem is equivalent to proving that $4(x^3 + y^3 + z^3 + 6xyz) \ge (x + y + z)^3$.

Second Hint

What do you do with homogenous inequalities? (A function $f$ is homogenous of degree $d$ if $f(kx) = k^df(x)$ for all x in the domain of $f$.)

Solution

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See Also

1979 USAMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5
All USAMO Problems and Solutions

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