2019 CIME I Problems/Problem 13

Suppose $\text{P}$ is a monic polynomial whose roots $a$ , $b$ , and $c$ are real numbers, at least two of which are positive, that satisfy the relation $a(a-b)=b(b-c)=c(c-a)=1.$ Find the greatest integer less than or equal to $100|P(\sqrt{3})|$ .

Solution

We can rearrange the equations to yield $b=a-\frac{1}{a}$ and equivalent. Adding all three of these equations and simplifying results in $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$ , so $\frac{ab+bc+ca}{abc}=0$ and thus $ab+bc+ca=0$ .

Next, take the three given equations and add them, resulting in $a^2+b^2+c^2-ab-bc-ca=3$ . This is equivalent to $(a+b+c)^2-3(ab-bc-ca)=3$ by direct expansion, so by substituting what we found earlier, we know that $a+b+c=\pm\sqrt{3}$ .

Finally, return to the three equations $b=a-\frac{1}{a}$ and equivalent. Squaring all three results in $b^2=a^2-2+\frac{1}{a^2}$ and equivalent. Adding the equations and simplifying results in $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=6$ , so $a^2b^2+b^2c^2+c^2a^2=6(abc)^2$ . But notice that $a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=\mp2\sqrt{3}abc$ Thus, $\mp2\sqrt{3}abc=6(abc)^2$ . Notice that if any one of $a,b,c$ is $0$ , then the given equations cannot be true; thus we can divide $abc$ from both sides, resulting in $\mp2\sqrt{3}=6abc$ , so $abc=\mp\frac{\sqrt{3}}{3}$ .

This finally means that by Vieta’s Formulas: $P(x)=x^3\pm\sqrt{3}x^2\mp\frac{\sqrt{3}}{3}$ We can easily show using derivatives that the sign of the $x^2$ -coefficient must be negative and vice versa for the constant (due to the existence of two or more positive roots). Thus, $P(x)=x^3-\sqrt{3}x^2+\frac{\sqrt{3}}{3}$ Then, $P(\sqrt{3})=\frac{\sqrt{3}}{3}$ , so the answer is $\frac{100\sqrt{3}}{3}>\boxed{057}$ .

~ eevee9406

2019 CIME I (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2019 CIME I Problems/Problem 13

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