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

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.

@@ Line 3: / Line 3: @@
 ==Solution==
+===Solution 1===
 By Vieta's Formulas, <math>a=-r-s-t</math>, <math>b=rs+st+rt</math>, and <math>c=-rst</math>.
 Now we know <math>k=a^2-3b</math>; in terms of r, s, and t, then,
@@ Line 28: / Line 29: @@
 In fact, the problem tells us this is true. Q.E.D.
-==Solution 2==
+===Solution 2===
 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>
-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.
+==Solution 3==
+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>.
+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.
+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==

1988 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

Page

Toolbox

Search