Difference between revisions of "1989 AHSME Problems/Problem 28"
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== Solution == | == Solution == | ||
− | The roots of <math>t^2-9t+1=0</math> are positive and distinct, so by considering the graph of <math>y=\tan x</math>, the smallest two roots of the original equation <math>x_1,\ x_2</math> are between <math>0</math> and <math>\tfrac\pi{2}</math>, and the two other roots are <math>\pi+x_1, | + | The roots of <math>t^2-9t+1=0</math> are positive and distinct, so by considering the graph of <math>y=\tan x</math>, the smallest two roots of the original equation <math>x_1,\ x_2</math> are between <math>0</math> and <math>\tfrac\pi{2}</math>, and the two other roots are <math>\pi+x_1, \pi+x_2</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>. | + | 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>. |
− | == | + | ==Solution 2== |
<math>t^2-9t+1=0</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 | + | We treat <math>\tan(x_1)</math> and <math>\tan(x_2)</math> as the roots of our equation. |
− | Because <math>tan(x_1) | + | Because <math>\tan(x_1) \times \tan(x_2) = 1</math> by Vieta's formula, <math>x_1 + x_2 = 0.5\pi</math>. |
− | Because the | + | Because the principal values of <math>x_1</math> and <math>x_2</math> are acute and our range for <math>x</math> is <math>[0,2\pi]</math>, |
− | we have four values of x that satisfy the quadratic: | + | we have four values of <math>x</math> that satisfy the quadratic: |
− | <math>x_1, x_2, x_1+\pi, x_2+\pi</math> | + | <math>x_1, x_2, x_1+\pi, x_2+\pi.</math> |
Summing these, we obtain <math>2(x_1+x_2) + 2\pi</math>. | Summing these, we obtain <math>2(x_1+x_2) + 2\pi</math>. | ||
− | Using the fact that <math>x_1+x_2=0.5\pi</math> | + | Using the fact that <math>x_1+x_2=0.5\pi</math>, |
− | <math>2(0.5\pi) + 2\pi = 3\pi</math> | + | we get <math>2(0.5\pi) + 2\pi = 3\pi.</math> |
== See also == | == See also == | ||
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[[Category: Intermediate Algebra Problems]] | [[Category: Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
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Latest revision as of 21:14, 2 March 2019
Contents
Problem
Find the sum of the roots of that are between
and
radians.
Solution
The roots of are positive and distinct, so by considering the graph of
, the smallest two roots of the original equation
are between
and
, and the two other roots are
.
Then, from the quadratic equation, we discover that the product , which implies that
does not exist. The bounds then imply that
. Thus
which is
.
Solution 2
:
We treat
and
as the roots of our equation.
Because
by Vieta's formula,
.
Because the principal values of
and
are acute and our range for
is
,
we have four values of
that satisfy the quadratic:
Summing these, we obtain
.
Using the fact that
,
we get
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.