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Difference between revisions of "1988 USAMO Problems/Problem 2"

(Created page with "==Problem== The cubic polynomial <math>x^3+ax^2+bx+c</math> has real coefficients and three real roots <math>r\ge s\ge t</math>. Show that <math>k=a^2-3b\ge 0</math> and that <ma...")
 
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==Solution==
 
==Solution==
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{{solution}}
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===Solution 1===
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By Vieta's Formulas, <math>a=-r-s-t</math>, <math>b=rs+st+rt</math>, and <math>c=-rst</math>.
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Now we know <math>k=a^2-3b</math>; in terms of r, s, and t, then,
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<cmath>k=(-r-s-t)^2-3(rs+st+rt)</cmath>
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<cmath>k=r^2+s^2+t^2-rs-st-rt</cmath>
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Now notice that we can multiply both sides by 2, and rearrange terms to get
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<math>2k=(r-s)^2+(s-t)^2+(r-t)^2</math>.
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But since <math>r, s, t\in \mathbb{R}</math>, the three terms of the RHS are all non-negative (as the square of a real number is always non-negative), and therefore their sum is also non-negative -- that is, <math>2k\ge 0 \Rightarrow k\ge 0</math>.
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Now, we will show that <math>\sqrt k\le r-t</math>.
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We can square both sides, and the inequality will hold since they are both non-negative (it is given that <math>r\ge t</math>, therefore <math>r-t\ge 0</math>). This gives <math>k \le r^2-2rt+t^2</math>.
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Now we already have <math>k=r^2+s^2+t^2-rs-st-rt</math>, so substituting this for k gives
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<cmath>r^2+s^2+t^2-rs-st-rt \le r^2-2rt+t^2</cmath>
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<cmath>s^2-rs-st+rt \le 0</cmath>
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<cmath>s^2-(r+t)s+rt \le 0</cmath>
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Note that this is a quadratic. Since its leading coefficient is positive, its value is less than 0 when s is between the two roots. Using the quadratic formula:
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<cmath>s=\frac {r+t\pm \sqrt {(-r-t)^2-4rt} } 2</cmath>
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<cmath>s=\frac {r+t\pm \sqrt {r^2-2rt+t^2} } 2</cmath>
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<cmath>s=\frac {r+t\pm (r-t) } 2</cmath>
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<cmath>s \in \{r, t\}</cmath>
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The quadratic is 0 when s is equal to r or t, and the inequality
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holds when its value is less than or equal to 0 -- that is, <math>r\ge s\ge t</math>.
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(Its value is less than or equal to 0 when s is between the roots, since the
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graph of the quadratic opens upward.)
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In fact, the problem tells us this is true. Q.E.D.
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===Solution 2===
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From Vieta's Formula (which tells us that <math>a = -(r+s+t)</math> and <math>b = rs + st + rt</math>), we have that
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<cmath>k = a^2 - 3b = r^2 + s^2 + t^2 - rs - st - rt = \frac{1}{2} ((r-s)^2 + (s-t)^2 + (r-t)^2),</cmath>
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clearly non-negative. To prove <math>\sqrt{k} \le r - t</math>, it suffices to prove the square of this relation, or <cmath>r^2 + s^2 + t^2 - rs - st - rt \le r^2 - 2rt + t^2.</cmath> This in turn simplifies to <cmath>rs + st - rt - s^2 \ge 0,</cmath> or <cmath>(r - s)(s - t) \ge 0,</cmath> which is clearly true as <math>r \ge s \ge t</math>. This completes the proof.
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==Solution 3==
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By Vieta's Formulas, <math>a = -(r+s+t)</math> and <math>b = rs + st + rt</math>. <math>k = (r+s+t)^2 - 3(rs + st + rt) = r^2+s^2+t^2-rs-st-rt = \frac{(r-s)^2 + (s-t)^2 + (r-t)^2}{2}</math>.
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To show that <math>k \ge 0</math>, simply note that by the trivial inequality, all three squares are greater than <math>0</math> as they are the squares of real numbers.
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To show that <math>\sqrt{k} \le r-t</math>, since both are positive, it is sufficient to show that <math>k \le (r-t)^2</math>. <math>\frac{(r-s)^2 + (s-t)^2 + (r-t)^2}{2} \le (r-t)^2</math> implies that <math>k \le (r-t)^2</math>. <math>\frac{(r-s)^2 + (s-t)^2 - (r-t)^2}{2} \le 0</math>. Let <math>y = r-s</math> and <math>z = s-t</math>. We then have <math>\frac{y^2 + z^2 - (y+z)^2}{2} \le 0 \implies -2yz \le 0</math>, which is clearly true as both <math>y</math> and <math>z</math> are positive.
  
 
==See Also==
 
==See Also==
 
{{USAMO box|year=1988|num-b=1|num-a=3}}
 
{{USAMO box|year=1988|num-b=1|num-a=3}}
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{{MAA Notice}}
  
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]

Latest revision as of 17:17, 13 June 2017

Problem

The cubic polynomial $x^3+ax^2+bx+c$ has real coefficients and three real roots $r\ge s\ge t$. Show that $k=a^2-3b\ge 0$ and that $\sqrt k\le r-t$.

Solution

Solution 1

By Vieta's Formulas, $a=-r-s-t$, $b=rs+st+rt$, and $c=-rst$. Now we know $k=a^2-3b$; in terms of r, s, and t, then, \[k=(-r-s-t)^2-3(rs+st+rt)\] \[k=r^2+s^2+t^2-rs-st-rt\] Now notice that we can multiply both sides by 2, and rearrange terms to get $2k=(r-s)^2+(s-t)^2+(r-t)^2$. But since $r, s, t\in \mathbb{R}$, the three terms of the RHS are all non-negative (as the square of a real number is always non-negative), and therefore their sum is also non-negative -- that is, $2k\ge 0 \Rightarrow k\ge 0$.

Now, we will show that $\sqrt k\le r-t$. We can square both sides, and the inequality will hold since they are both non-negative (it is given that $r\ge t$, therefore $r-t\ge 0$). This gives $k \le r^2-2rt+t^2$. Now we already have $k=r^2+s^2+t^2-rs-st-rt$, so substituting this for k gives \[r^2+s^2+t^2-rs-st-rt \le r^2-2rt+t^2\] \[s^2-rs-st+rt \le 0\] \[s^2-(r+t)s+rt \le 0\] Note that this is a quadratic. Since its leading coefficient is positive, its value is less than 0 when s is between the two roots. Using the quadratic formula: \[s=\frac {r+t\pm \sqrt {(-r-t)^2-4rt} } 2\] \[s=\frac {r+t\pm \sqrt {r^2-2rt+t^2} } 2\] \[s=\frac {r+t\pm (r-t) } 2\] \[s \in \{r, t\}\] The quadratic is 0 when s is equal to r or t, and the inequality holds when its value is less than or equal to 0 -- that is, $r\ge s\ge t$. (Its value is less than or equal to 0 when s is between the roots, since the graph of the quadratic opens upward.) In fact, the problem tells us this is true. Q.E.D.

Solution 2

From Vieta's Formula (which tells us that $a = -(r+s+t)$ and $b = rs + st + rt$), we have that \[k = a^2 - 3b = r^2 + s^2 + t^2 - rs - st - rt = \frac{1}{2} ((r-s)^2 + (s-t)^2 + (r-t)^2),\] clearly non-negative. To prove $\sqrt{k} \le r - t$, it suffices to prove the square of this relation, or \[r^2 + s^2 + t^2 - rs - st - rt \le r^2 - 2rt + t^2.\] This in turn simplifies to \[rs + st - rt - s^2 \ge 0,\] or \[(r - s)(s - t) \ge 0,\] which is clearly true as $r \ge s \ge t$. This completes the proof.

Solution 3

By Vieta's Formulas, $a = -(r+s+t)$ and $b = rs + st + rt$. $k = (r+s+t)^2 - 3(rs + st + rt) = r^2+s^2+t^2-rs-st-rt = \frac{(r-s)^2 + (s-t)^2 + (r-t)^2}{2}$.

To show that $k \ge 0$, simply note that by the trivial inequality, all three squares are greater than $0$ as they are the squares of real numbers.

To show that $\sqrt{k} \le r-t$, since both are positive, it is sufficient to show that $k \le (r-t)^2$. $\frac{(r-s)^2 + (s-t)^2 + (r-t)^2}{2} \le (r-t)^2$ implies that $k \le (r-t)^2$. $\frac{(r-s)^2 + (s-t)^2 - (r-t)^2}{2} \le 0$. Let $y = r-s$ and $z = s-t$. We then have $\frac{y^2 + z^2 - (y+z)^2}{2} \le 0 \implies -2yz \le 0$, which is clearly true as both $y$ and $z$ are positive.

See Also

1988 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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