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Difference between revisions of "1988 USAMO Problems/Problem 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
 
From Vieta's Formula (which tells us that <math>a = -(r+s+t)</math> and <math>b = rs + st + rt</math>), we have that
 
<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>
 
<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.
+
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.
  
 
==See Also==
 
==See Also==

Revision as of 16:55, 5 October 2014

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

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.

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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