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Difference between revisions of "1983 USAMO Problems/Problem 2"

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==1983 USAMO Problem 2==
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== Problem ==
 
Prove that the zeros of
 
Prove that the zeros of
  
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==Solution==
 
==Solution==
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We prove the contrapositive: if the polynomial in question has the five real roots <math>x_1, x_2, x_3, x_4, x_5</math>, then <math>5b \le 2a^2</math>.
  
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Because <math>a = -(x_1 + x_2 + x_3 + x_4 + x_5)</math> and <math>b = x_1x_2 + x_1x_3 + ... + x_4x_5</math> by Vieta's Formulae, we have
 +
 +
<cmath>2b = 2x_1x_2 + 2x_1x_3 + ... + 2x_4x_5 = (x_1 + x_2 + x_3 + x_4 + x_5)^2 - (x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2)</cmath>
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<cmath>=a^2 - \frac{(1+1+1+1+1)(x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2)}{5}</cmath>
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<cmath>\le a^2 - \frac{(x_1 + x_2 + x_3 + x_4 + x_5)^2}{5}</cmath> (by Cauchy-Schwarz)
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<cmath>=\frac{4a^2}{5},</cmath>
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 +
so <math>5b \le 2a^2</math>, as desired.
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 +
==Solution 2==
  
 
'''Lemma:'''
 
'''Lemma:'''
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<cmath>x_1x_2+x_1x_3+\cdots+x_4x_5</cmath>
 
<cmath>x_1x_2+x_1x_3+\cdots+x_4x_5</cmath>
  
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We solve this cyclically by showing
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By the trivial inequality,
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<cmath>\frac{1}{2}x^2+\frac{1}{2}y^2\ge xy</cmath>
 
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By the trivial inequality, <math>(x-y)^2\ge 0</math>, or <math>x^2+y^2-2xy\ge 0</math>.
 
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<cmath>x^2+y^2\ge 2xy</cmath>
 
  
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Dividing by <math>2</math> gives us the desired.
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<cmath>x^2+y^2\ge 2xy \Rightarrow \frac{x^2}{2} + \frac{y^2}{2} \ge xy</cmath>
  
 
Making such an inequality for all the variable pairs and summing them, we find the lemma is true.
 
Making such an inequality for all the variable pairs and summing them, we find the lemma is true.
  
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Now, we start by plugging in our Vieta's: Let our roots be <math>x_1,x_2,\cdots,x_5</math>. This means be Vieta's that <math>a=x_1+x_2+\cdots+x_5, b=x_1x_2+x_1x_3+\cdots+x_4x_5</math>
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Now, let our roots be <math>x_1,x_2,\cdots,x_5</math>. By Vieta's, <math>a=x_1+x_2+\cdots+x_5</math> and <math>b=x_1x_2+x_1x_3+\cdots+x_4x_5</math>
  
 
If we show that for all real <math>x_1,x_2,\cdots, x_5</math> that <math>2a^2\ge 5b</math>, then we have a contradiction and all of <math>x_1,x_2,\cdots, x_5</math> cannot be real. We start by rewriting <math>2a^2\ge 5b</math> as
 
If we show that for all real <math>x_1,x_2,\cdots, x_5</math> that <math>2a^2\ge 5b</math>, then we have a contradiction and all of <math>x_1,x_2,\cdots, x_5</math> cannot be real. We start by rewriting <math>2a^2\ge 5b</math> as
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<cmath>x_1^2+x_2^2+\cdots+x_5^2+2(x_1x_2+x_1x_3+\cdots+x_4x_5)\ge\frac{5}{2}(x_1x_2+x_1x_3+\cdots+x_4x_5)</cmath>
 
<cmath>x_1^2+x_2^2+\cdots+x_5^2+2(x_1x_2+x_1x_3+\cdots+x_4x_5)\ge\frac{5}{2}(x_1x_2+x_1x_3+\cdots+x_4x_5)</cmath>
  
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We subtract out the second symmetric sums, and then multiply by <math>2</math> to find
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We subtract the sum in brackets, and then multiply by <math>2</math> to find
  
 
<cmath>2x_1^2+2x_2^2+\cdots+2x_5^2\ge x_1x_2+x_1x_3+\cdots+x_4x_5</cmath>
 
<cmath>2x_1^2+2x_2^2+\cdots+2x_5^2\ge x_1x_2+x_1x_3+\cdots+x_4x_5</cmath>
  
 
which is true by our lemma.
 
which is true by our lemma.
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 +
== See Also ==
 +
{{USAMO box|year=1983|num-b=1|num-a=3}}
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{{MAA Notice}}
 +
 +
[[Category:Olympiad Algebra Problems]]
 +
[[Category:Olympiad Inequality Problems]]

Latest revision as of 14:40, 19 April 2014

Problem

Prove that the zeros of

\[x^5+ax^4+bx^3+cx^2+dx+e=0\]

cannot all be real if $2a^2<5b$.

Solution

We prove the contrapositive: if the polynomial in question has the five real roots $x_1, x_2, x_3, x_4, x_5$, then $5b \le 2a^2$.

Because $a = -(x_1 + x_2 + x_3 + x_4 + x_5)$ and $b = x_1x_2 + x_1x_3 + ... + x_4x_5$ by Vieta's Formulae, we have

\[2b = 2x_1x_2 + 2x_1x_3 + ... + 2x_4x_5 = (x_1 + x_2 + x_3 + x_4 + x_5)^2 - (x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2)\] \[=a^2 - \frac{(1+1+1+1+1)(x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2)}{5}\] \[\le a^2 - \frac{(x_1 + x_2 + x_3 + x_4 + x_5)^2}{5}\] (by Cauchy-Schwarz) \[=\frac{4a^2}{5},\]

so $5b \le 2a^2$, as desired.

Solution 2

Lemma:

For all real numbers $x_1,x_2,\cdots x_5$,

\[2(x_1^2+x_2^2+\cdots+x_5^2)\ge\]

\[x_1x_2+x_1x_3+\cdots+x_4x_5\]

By the trivial inequality,

\[x^2+y^2\ge 2xy \Rightarrow \frac{x^2}{2} + \frac{y^2}{2} \ge xy\]

Making such an inequality for all the variable pairs and summing them, we find the lemma is true.

Now, let our roots be $x_1,x_2,\cdots,x_5$. By Vieta's, $a=x_1+x_2+\cdots+x_5$ and $b=x_1x_2+x_1x_3+\cdots+x_4x_5$

If we show that for all real $x_1,x_2,\cdots, x_5$ that $2a^2\ge 5b$, then we have a contradiction and all of $x_1,x_2,\cdots, x_5$ cannot be real. We start by rewriting $2a^2\ge 5b$ as

\[2(x_1+x_2+\cdots+x_5)^2\ge 5(x_1x_2+x_1x_3+\cdots+x_4x_5)\]

We divide by $2$ and find

\[(x_1+x_2+\cdots+x_5)^2\ge \frac{5}{2}(x_1x_2+x_1x_3+\cdots+x_4x_5)\]

Expanding the LHS, we have

\[x_1^2+x_2^2+\cdots+x_5^2+2(x_1x_2+x_1x_3+\cdots+x_4x_5)\ge\frac{5}{2}(x_1x_2+x_1x_3+\cdots+x_4x_5)\]

We subtract the sum in brackets, and then multiply by $2$ to find

\[2x_1^2+2x_2^2+\cdots+2x_5^2\ge x_1x_2+x_1x_3+\cdots+x_4x_5\]

which is true by our lemma.

See Also

1983 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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