Difference between revisions of "2014 AMC 12B Problems/Problem 25"
Kevin38017 (talk | contribs) (→Solution) |
Kevin38017 (talk | contribs) m (→Solution) |
||
Line 10: | Line 10: | ||
<cmath>2a(a - b) = 2a^2 - 2</cmath> | <cmath>2a(a - b) = 2a^2 - 2</cmath> | ||
<cmath>ab = 1</cmath> | <cmath>ab = 1</cmath> | ||
− | Notice that either <math>a = 1</math> and <math>b = 1</math> or <math>a = -1</math> and <math>b = -1</math>. For the first case, <math>a = 1</math> only when <math>x = k\pi</math> and <math>k</math> is an integer. <math>b = 1</math> when <math>\frac{2014\pi^2}{k\pi}</math> is an even multiple of <math>\pi</math>, and since <math>2014 = 2*19*53</math>, <math>b =1</math> only when <math>k</math> is an odd divisor of <math>2014</math>. This gives us | + | Notice that either <math>a = 1</math> and <math>b = 1</math> or <math>a = -1</math> and <math>b = -1</math>. For the first case, <math>a = 1</math> only when <math>x = k\pi</math> and <math>k</math> is an integer. <math>b = 1</math> when <math>\frac{2014\pi^2}{k\pi}</math> is an even multiple of <math>\pi</math>, and since <math>2014 = 2*19*53</math>, <math>b =1</math> only when <math>k</math> is an odd divisor of <math>2014</math>. This gives us these possible values for <math>x</math>: |
<cmath>x= \pi, 19\pi, 53\pi, 1007\pi</cmath> | <cmath>x= \pi, 19\pi, 53\pi, 1007\pi</cmath> | ||
For the case where <math>a = -1</math>, <math>\cos{2x} = -1</math>, so <math>x = \frac{m}{2\pi}</math>, where m is odd. <math>\frac{2014\pi^2}{\frac{m}{2\pi}}</math> must also be an odd multiple of \pi in order for <math>b</math> to equal <math>-1</math>, so \frac{4028}{m} must be odd. We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for <math>m</math>, and therefore no cases where <math>a = -1</math> and <math>b = -1</math>. Therefore, the sum of all our possible values for <math>x</math> is | For the case where <math>a = -1</math>, <math>\cos{2x} = -1</math>, so <math>x = \frac{m}{2\pi}</math>, where m is odd. <math>\frac{2014\pi^2}{\frac{m}{2\pi}}</math> must also be an odd multiple of \pi in order for <math>b</math> to equal <math>-1</math>, so \frac{4028}{m} must be odd. We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for <math>m</math>, and therefore no cases where <math>a = -1</math> and <math>b = -1</math>. Therefore, the sum of all our possible values for <math>x</math> is |
Revision as of 18:19, 20 February 2014
Problem
Find the sum of all the positive solutions of
$\textbf{(A)}\ \pi \qquad\textbf{(B)}\ 35\qquad\textbf{(C)}\ 1008\pi \qquad\textbf{(D)}}\ 1080 \pi \qquad\textbf{(E)}\ 1800\pi$ (Error compiling LaTeX. Unknown error_msg)
Solution
Rewrite as
. Now let
, and let
. We have
Notice that either
and
or
and
. For the first case,
only when
and
is an integer.
when
is an even multiple of
, and since
,
only when
is an odd divisor of
. This gives us these possible values for
:
For the case where
,
, so
, where m is odd.
must also be an odd multiple of \pi in order for
to equal
, so \frac{4028}{m} must be odd. We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for
, and therefore no cases where
and
. Therefore, the sum of all our possible values for
is
(Solution by kevin38017)