Difference between revisions of "2016 AMC 12B Problems/Problem 14"
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==Solution 5 (Quadratic formula)== | ==Solution 5 (Quadratic formula)== | ||
− | As in the previous solutions, <math>S=\frac{1}{r-r^2} \implies Sr^2-Sr+1=0 \implies \text{discriminant}=S^2-4S \ge 0</math> since we know that such an <math>r</math> exists. We then have <math>S \ge 4,</math> and this is achievable when <math>r=.5,</math> so the answer is <math>4.</math> | + | As in the previous solutions, <math>S=\frac{1}{r-r^2} \implies Sr^2-Sr+1=0 \implies \text{discriminant}=S^2-4S \ge 0</math> since we know that such an <math>r</math> exists. We then have <math>S \ge 4</math> for positive <math>S,</math> and this is achievable when <math>r=.5,</math> so the answer is <math>4.</math> |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}} | {{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:57, 5 July 2022
Contents
Problem
The sum of an infinite geometric series is a positive number , and the second term in the series is
. What is the smallest possible value of
Solution
The second term in a geometric series is , where
is the common ratio for the series and
is the first term of the series. So we know that
and we wish to find the minimum value of the infinite sum of the series. We know that:
and substituting in
, we get that
. From here, you can either use calculus or AM-GM.
Let , then
. Since
and
are undefined
. This means that we only need to find where the derivative equals
, meaning
. So
, meaning that
For 2 positive real numbers and
,
. Let
and
. Then:
. This implies that
. or
. Rearranging :
. Thus, the smallest value is
.
Solution 1
The sum of the geometric sequence is where
is the first term and
is the common ratio. We know the second term,
is equal to
Thus
This means,
In order to minimize
we maximize the denominator. By AM-GM,
Equality occurs at
This gives the minimum value of
as
Solution 2
A geometric sequence always looks like
and they say that the second term . You should know that the sum of an infinite geometric series (denoted by
here) is
. We now have a system of equations which allows us to find
in one variable.
We seek the smallest positive value of . We proceed by graphing in the
plane (if you want to be rigorous, only stop graphing when you know all the rest you didn't graph is just the approaching of asymptotes and infinities) and find the answer is
We seek the smallest positive value of . We proceed by graphing in the
plane (if you want to be rigorous, only stop graphing when you know all the rest you didn't graph is just the approaching of asymptotes and infinities) and find the answer is
We seek the smallest positive value of .
and
at
and
.
and
is negative (implying a relative maximum occurs at
) and
is positive (implying a relative minimum occurs at
). At
,
. Since this a competition math problem with an answer, we know this relative minimum must also be the absolute minimum among the "positive parts" of
and that our answer is indeed
However, to be sure of this outside of this cop-out, one can analyze the end behavior of
, how
behaves at its asymptotes, and the locations of its maxima and minima relative to the asymptotes to be sure that 4 is the absolute minimum among the "positive parts" of
.
We seek the smallest positive value of . We could use calculus like we did in the solution immediately above this one, but that's a lot of work and we don't have a ton of time. To minimize the positive value of this fraction, we must maximize the denominator. The denominator is a quadratic that opens down, so its maximum occurs at its vertex. The vertex of this quadratic occurs at
and
.
Solution 3
Let
be the common ratio. If the second term is
, the first must be
. By the infinite geometric series formula, the sum must be
This equals
. To find the minimum value of S, we must find the maximum value of the denominator,
, which is
, completing the square. Thus, the minimum value of
is
.
Solution 4 (no AM-GM or Calculus)
Our sequence is . Since we know this is a converging series, our ratio is in
. Because the 2nd term in the sequence is a 1, the ratio must be
, so we can write
as
. With some manipulation we get
. Since S has to be a "positive number," we come to think
is
(makes S positive & we know a sequence/series of a ratio
is definitely convergent). So our sequence is
,
.
-thedodecagon
Solution 5 (Quadratic formula)
As in the previous solutions, since we know that such an
exists. We then have
for positive
and this is achievable when
so the answer is
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 | |
All AMC 12 Problems and Solutions |
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