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Difference between revisions of "1983 USAMO Problems/Problem 2"
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==1983 USAMO Problem 1==
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== Problem ==
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Prove that the zeros of
  
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==Solution==
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<cmath>x^5+ax^4+bx^3+cx^2+dx+e=0</cmath>
  
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cannot all be real if <math>2a^2<5b</math>.
  
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'''Lemma:'''
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==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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For all real numbers <math>x_1,x_2,\cdots, x_3</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
  
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<cmath>2(x_1^2+x_2^2+\cdots+x_5^2)\ge</cmath>
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<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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<cmath>x_1x_2+x_1x_3+\cdots+x_4x_5</cmath>
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so <math>5b \le 2a^2</math>, as desired.
  
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We solve this cylicallly by showing
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==Solution 2==
  
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<cmath>\frac{1}{2}x^2+\frac{1}{2}y^2\ge xy</cmath>
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'''Lemma:'''
  
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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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For all real numbers <math>x_1,x_2,\cdots x_5</math>,
  
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<cmath>x^2+y^2\ge 2xy</cmath>
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<cmath>2(x_1^2+x_2^2+\cdots+x_5^2)\ge</cmath>
  
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Dividing by <math>2</math> gives us the desired.
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<cmath>x_1x_2+x_1x_3+\cdots+x_4x_5</cmath>
  
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Making such an inequality for all the variable pairs and summing
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By the trivial inequality,
  
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them, we find the lemma is true.[/hide]
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<cmath>x^2+y^2\ge 2xy \Rightarrow \frac{x^2}{2} + \frac{y^2}{2} \ge xy</cmath>
  
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We start by plugging in our Vieta's: Let our roots be
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Making such an inequality for all the variable pairs and summing them, we find the lemma is true.
  
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<math>x_1,x_2,\cdots,x_5</math>. This means be Vieta's that
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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>
  
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<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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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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If we show that for all real <math>x_1,x_2,\cdots, x_5</math> that <math>2a^2\ge 5b</math>,
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<cmath>2(x_1+x_2+\cdots+x_5)^2\ge 5(x_1x_2+x_1x_3+\cdots+x_4x_5)</cmath>
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then we have a contradiction and all of <math>x_1,x_2,\cdots, x_5</math> cannot
 
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be real. We start by rewriting <math>2a^2\ge 5b</math> as
 
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<cmath>2(x_1+x_2+\cdots+x_5)^2\ge</cmath>
 
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<cmath>5(x_1x_2+x_1x_3+\cdots+x_4x_5)</cmath>
 
  
 
We divide by <math>2</math> and find
 
We divide by <math>2</math> and find
  
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<cmath>(x_1+x_2+\cdots+x_5)^2\ge</cmath>
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<cmath>(x_1+x_2+\cdots+x_5)^2\ge \frac{5}{2}(x_1x_2+x_1x_3+\cdots+x_4x_5)</cmath>
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<cmath>\frac{5}{2}(x_1x_2+x_1x_3+\cdots+x_4x_5)</cmath>
 
  
 
Expanding the LHS, we have
 
Expanding the LHS, we have
  
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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</cmath>
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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>
  
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<cmath>\frac{5}{2}(x_1x_2+x_1x_3+\cdots+x_4x_5)</cmath>
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We subtract the sum in brackets, and then multiply by <math>2</math> to find
  
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Aha! We subtract out the second symmetric sums, and then multiply
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<cmath>2x_1^2+2x_2^2+\cdots+2x_5^2\ge x_1x_2+x_1x_3+\cdots+x_4x_5</cmath>
  
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by <math>2</math> to find
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which is true by our lemma.
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<cmath>2x_1^2+2x_2^2+\cdots+2x_5^2\ge</cmath>
 
  
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<cmath>x_1x_2+x_1x_3+\cdots+x_4x_5</cmath>
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== See Also ==
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{{USAMO box|year=1983|num-b=1|num-a=3}}
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{{MAA Notice}}
  
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which is true by our lemma.
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[[Category:Olympiad Algebra Problems]]
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[[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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