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

Revision as of 21:31, 15 April 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.

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 ==Solution==
-{{solution}}
+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,
+<cmath>k=(-r-s-t)^2-3(rs+st+rt)</cmath>
+<cmath>k=r^2+s^2+t^2-rs-st-rt</cmath>
+Now notice that we can multiply both sides by 2, and rearrange terms to get
+<math>2k=(r-s)^2+(s-t)^2+(r-t)^2</math>.
+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>.
+Now, we will show that <math>\sqrt k\le r-t</math>.
+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>.
+Now we already have <math>k=r^2+s^2+t^2-rs-st-rt</math>, so substituting this for k gives
+<cmath>r^2+s^2+t^2-rs-st-rt \le r^2-2rt+t^2</cmath>
+<cmath>s^2-rs-st+rt \le 0</cmath>
+<cmath>s^2-(r+t)s+rt \le 0</cmath>
+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:
+<cmath>s=\frac {r+t\pm \sqrt {(-r-t)^2-4rt} } 2</cmath>
+<cmath>s=\frac {r+t\pm \sqrt {r^2-2rt+t^2} } 2</cmath>
+<cmath>s=\frac {r+t\pm (r-t) } 2</cmath>
+<cmath>s \in \{r, t\}</cmath>
+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, <math>r\ge s\ge t</math>.
+(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.
 ==See Also==

1988 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5
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Difference between revisions of "1988 USAMO Problems/Problem 2"

Revision as of 21:31, 15 April 2014

Problem

Solution

See Also