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 "1989 AHSME Problems/Problem 28"

(→‎See also)
(→‎Solution)
Line 9: Line 9:
  
 
Then from the quadratic equation we discover that the product <math>\tan x_1\tan x_2=1</math> which implies that <math>\tan(x_1+x_2)</math> does not exist. The bounds then imply that <math>x_1+x_2=\tfrac\pi{2}</math>. Thus <math>x_1+x_2+\pi+x_1+\pi+x_2=3\pi</math> which is <math>\rm{(D)}</math>.
 
Then from the quadratic equation we discover that the product <math>\tan x_1\tan x_2=1</math> which implies that <math>\tan(x_1+x_2)</math> does not exist. The bounds then imply that <math>x_1+x_2=\tfrac\pi{2}</math>. Thus <math>x_1+x_2+\pi+x_1+\pi+x_2=3\pi</math> which is <math>\rm{(D)}</math>.
 +
 +
<math>t^2-9t+1=0</math>
 +
We treat <math>tan(x_1)</math> and <math>tan(x_2)</math> as the roots of our equation
 +
Because <math>tan(x_1)</math> * <math>tan(x_2)</math> = <math>1</math> by Vieta's formula, x_1 + x_2 = 0.5pi{2}.
 +
Because the principle values of x1 and x2 are acute and our range for x is [<math>0</math>,<math>2pi{2}</math>],
 +
we have four values of x that satisfy the quadratic:
 +
x_1, x_2, x_1+<math>pi{2}</math>, x_2+<math>pi{2}</math>
 +
Summing these, we obtain <math>2(x_1+x_2) + 2pi</math>.
 +
Using the fact that x_1+x_2=<math>0.5pi{2}</math>
 +
<math>2(0.5pi{2}</math>) + <math>2pi{2}</math> = <math>3pi{2}</math>
 +
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1989|num-b=27|num-a=29}}   
 
{{AHSME box|year=1989|num-b=27|num-a=29}}   

Revision as of 06:25, 27 June 2016

Problem

Find the sum of the roots of $\tan^2x-9\tan x+1=0$ that are between $x=0$ and $x=2\pi$ radians.

$\mathrm{(A) \frac{\pi}{2} } \qquad \mathrm{(B) \pi } \qquad \mathrm{(C) \frac{3\pi}{2} } \qquad \mathrm{(D) 3\pi } \qquad \mathrm{(E) 4\pi }$

Solution

The roots of $t^2-9t+1=0$ are positive and distinct, so by considering the graph of $y=\tan x$, the smallest two roots of the original equation $x_1,\ x_2$ are between $0$ and $\tfrac\pi{2}$, and the two other roots are $\pi+x_1,\ \pi+x_2$.

Then from the quadratic equation we discover that the product $\tan x_1\tan x_2=1$ which implies that $\tan(x_1+x_2)$ does not exist. The bounds then imply that $x_1+x_2=\tfrac\pi{2}$. Thus $x_1+x_2+\pi+x_1+\pi+x_2=3\pi$ which is $\rm{(D)}$.

$t^2-9t+1=0$ We treat $tan(x_1)$ and $tan(x_2)$ as the roots of our equation Because $tan(x_1)$ * $tan(x_2)$ = $1$ by Vieta's formula, x_1 + x_2 = 0.5pi{2}. Because the principle values of x1 and x2 are acute and our range for x is [$0$,$2pi{2}$], we have four values of x that satisfy the quadratic: x_1, x_2, x_1+$pi{2}$, x_2+$pi{2}$ Summing these, we obtain $2(x_1+x_2) + 2pi$. Using the fact that x_1+x_2=$0.5pi{2}$ $2(0.5pi{2}$) + $2pi{2}$ = $3pi{2}$

See also

1989 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

tan^2(x) -9tan(x)+1 We treat tan(x1) and tan(x2) as the roots of our equation Because tan(x1) * tan(x2) = 1 by Vieta's formula, x1 + x2 = 0.5pi. Because the principle values of x1 and x2 are acute and our range for x is [0,2pi], we have four values of x that satisfy the quadratic: x1, x2, x1+pi, x2+pi Summing these, we obtain 2(x1+x2) + 2pi. Using the fact that x1+x2=0.5pi 2(0.5pi) + 2pi = 3pi

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AHSME_Problems/Problem_28&oldid=79094"