Art of Problem Solving

1987 OIM Problems/Problem 5: Difference between revisions

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== Problem ==
== Problem ==
If <math>r</math>, <math>s</math>, and <math>t</math> are all the roots of the equation:
If <math>r</math>, <math>s</math>, and <math>t</math> are all the roots of the equation:
<cmath>x(x-2)3x-7)=2</cmath>
<cmath>x(x-2)(3x-7)=2</cmath>


(a) Prove that <math>r</math>, <math>s</math>, and <math>t</math> are all postive
(a) Prove that <math>r</math>, <math>s</math>, and <math>t</math> are all positive.


(b) Calculate: arctan <math>r</math> + arctan <math>s</math> + arctan <math>t</math>.
(b) Calculate <math>\arctan r + \arctan s + \arctan t</math>.


Note: We define arctan <math>x</math>, as the arc between <math>0</math> and <math>\pi</math> which tangent is <math>x</math>.
Note: The range of <math>\arctan x</math> falls between <math>0</math> and <math>\pi</math>, inclusive.


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

Revision as of 17:04, 22 March 2025

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

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See also

https://www.oma.org.ar/enunciados/ibe2.htm