1987 OIM Problems/Problem 5

Problem

If $r$ , $s$ , and $t$ are all the roots of the equation: $x(x-2)(3x-7)=2$

(a) Prove that $r$ , $s$ , and $t$ are all positive.

(b) Calculate $\arctan r + \arctan s + \arctan t$ .

Note: The range of $\arctan x$ falls between $0$ and $\pi$ , inclusive.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

(a) Consider the polynomial $f(x)=x(x-2)(3x-7)-2$ . When $x$ is negative, the value $x(x-2)(3x-7)$ is clearly negative, so adding this to $-2$ will yield negative $f(x)$ ; therefore, there cannot be a negative root. There also clearly cannot be a root equal to zero, so all that remains is to prove that the roots are all real. This can be achieved by by using the Intermediate Value Theorem; notice that $f(0)=-2,f(1)=2,f(2)=-2,$ and $f(3)=4$ , which imply three real roots, so clearly $r,s,t$ are all real and are thus positive.

(b) Let $\alpha=\arctan r,\beta=\arctan s,\gamma=\arctan t$ . Then $\tan\alpha=r,\tan\beta=s,\tan\gamma=t$ . By multiple applications of the sum of tangents formula: $\tan(\alpha+\beta+\gamma)=\frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\alpha\tan\beta-\tan\beta\tan\gamma-\tan\gamma\tan\alpha}=\frac{r+s+t-rst}{1-rs-st-rt}$ If we expand $f(x)$ , we find that $f(x)=3x^3-13x^2+14x-2$ implying that $r+s+t=\frac{13}{3},rs+st+rt=\frac{14}{3},rst=\frac{2}{3}$ Therefore, $\tan(\alpha+\beta+\gamma)=\frac{\frac{13}{3}-\frac{2}{3}}{1-\frac{14}{3}}=\frac{13-2}{3-14}=-1$ Due to our given range of $\arctan x$ , we know that $\alpha+\beta+\gamma=\boxed{\frac{3\pi}{4}}$ .

~ eevee9406

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1987 OIM Problems/Problem 5

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