Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1987 OIM Problems/Problem 5"
(solution)
 
(One intermediate revision by the same user not shown)
Line 1: Line 1:
 
== 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
  
 
== Solution ==
 
== Solution ==
−
{{solution}}
+
(a) Consider the polynomial <math>f(x)=x(x-2)(3x-7)-2</math>. When <math>x</math> is negative, the value <math>x(x-2)(3x-7)</math> is clearly negative, so adding this to <math>-2</math> will yield negative <math>f(x)</math>; 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 <math>f(0)=-2,f(1)=2,f(2)=-2,</math> and <math>f(3)=4</math>, which imply three real roots, so clearly <math>r,s,t</math> are all real and are thus positive.
 +
 
 +
(b) Let <math>\alpha=\arctan r,\beta=\arctan s,\gamma=\arctan t</math>. Then <math>\tan\alpha=r,\tan\beta=s,\tan\gamma=t</math>. By multiple applications of the sum of tangents formula:
 +
<cmath>\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}</cmath>
 +
If we expand <math>f(x)</math>, we find that
 +
<cmath>f(x)=3x^3-13x^2+14x-2</cmath>
 +
implying that
 +
<cmath>r+s+t=\frac{13}{3},rs+st+rt=\frac{14}{3},rst=\frac{2}{3}</cmath>
 +
Therefore,
 +
<cmath>\tan(\alpha+\beta+\gamma)=\frac{\frac{13}{3}-\frac{2}{3}}{1-\frac{14}{3}}=\frac{13-2}{3-14}=-1</cmath>
 +
Due to our given range of <math>\arctan x</math>, we know that <math>\alpha+\beta+\gamma=\boxed{\frac{3\pi}{4}}</math>.
 +
 
 +
~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
  
 
== See also ==
 
== See also ==
 
https://www.oma.org.ar/enunciados/ibe2.htm
 
https://www.oma.org.ar/enunciados/ibe2.htm

Latest revision as of 17:13, 19 April 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

(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

See also

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1987_OIM_Problems/Problem_5&oldid=246985"