Difference between revisions of "2017 AMC 12A Problems/Problem 25"
(→Solution 4) |
Isabelchen (talk | contribs) |
||
Line 81: | Line 81: | ||
~Zeric | ~Zeric | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | Let <math>a = \sqrt{2}i</math>, <math>b = \frac{1}{\sqrt{8}}(1+i)</math>, <math>c = \frac{1}{\sqrt{8}}(-1+i)</math>. <math>\bar{a} = -\sqrt{2}i</math>, <math>\bar{b} = \frac{1}{\sqrt{8}}(1-i)</math>, <math>\bar{c} = \frac{1}{\sqrt{8}}(-1-i)</math> | ||
+ | |||
+ | The magnitude of <math>a</math> and <math>\bar{a}</math> is <math>2</math>, while the magnitude of <math>b</math>, <math>\bar{b}</math>, <math>c</math>, and <math>\bar{c}</math> is <math>\frac14</math>. For the product's magnitude to be <math>1</math>, we need <math>8</math> elements from <math>a</math> and <math>\bar{a}</math>, and <math>4</math> elements from <math>b</math>, <math>\bar{b}</math>, <math>c</math>, and <math>\bar{c}</math>. | ||
+ | |||
+ | The product of <math>8</math> elements that are <math>a</math> and <math>\bar{a}</math> will be either <math>16</math> or <math>-16</math>. The product of <math>4</math> elements that are <math>b</math>, <math>\bar{b}</math>, <math>c</math>, and <math>\bar{c}</math> will either be <math>\frac{1}{16}</math> or <math>-\frac{1}{16}</math>. | ||
+ | |||
+ | <cmath>\text{Combinations of a and } \bar{a} \text{ that multiplies to 16} = \left\{ a^8, \bar{a}^8, a^2\bar{a}^6, a^6\bar{a}^2, a^4\bar{a}^4 \right\}</cmath> | ||
+ | |||
+ | <cmath>\text{Combinations of a and } \bar{a} \text{ that multiplies to -16} = \left\{ a\bar{a}^7, a^7\bar{a}, a^5\bar{a}^3, a^3\bar{a}^5 \right\}</cmath> | ||
+ | |||
+ | <cmath>\text{Combinations of b, } \bar{b} \text{, c, and } \bar{c} \text{ that multiplies to } \frac{1}{16} = \left\{ b^2\bar{b}^2, c^2\bar{c}^2, b^2c^2, \bar{b}^2\bar{c}^2, b\bar{b}c\bar{c}, b^3\bar{c}, b\bar{c}^3, \bar{b}^3c, \bar{b}c^3 \right\}</cmath> | ||
+ | |||
+ | <cmath>\text{Combinations of b, } \bar{b} \text{, c, and } \bar{c} \text{ that multiplies to } -\frac{1}{16} = \left\{ b^2\bar{b}c, b\bar{b}^2\bar{c}, bc^2\bar{c}, \bar{b}c\bar{c}^2, b^4, c^4, \bar{b}^4, \bar{c}^4, b^2\bar{c}^2, \bar{b}^2c^2 \right\}</cmath> | ||
+ | |||
==Video Solution by Richard Rusczyk== | ==Video Solution by Richard Rusczyk== |
Revision as of 12:22, 15 December 2023
Contents
Problem
The vertices of a centrally symmetric hexagon in the complex plane are given by
For each
,
, an element
is chosen from
at random, independently of the other choices. Let
be the product of the
numbers selected. What is the probability that
?
Solution 1
It is possible to solve this problem using elementary counting methods. This solution proceeds by a cleaner generating function.
We note that both lie on the imaginary axis and each of the
have length
and angle of odd multiples of
, i.e.
. When we draw these 6 complex numbers out on the complex plane, we get a crystal-looking thing. Note that the total number of ways to choose 12 complex numbers is
. Now we count the number of good combinations.
We first consider the lengths. When we multiply 12 complex numbers together, their magnitudes multiply. Suppose we have of the numbers
; then we must have
. Having
will take care of the length of the product; now we need to deal with the angle.
We require . Letting
be
, we see that the angles we have available are
, where we must choose exactly 8 angles from the set
and exactly 4 from the set
. If we found a good combination where we had
of each angle
, then the amount this would contribute to our count would be
. We want to add these all up. We proceed by generating functions.
Consider The expansion will be of the form
. Note that if we reduced the powers of
mod
and fished out the coefficient of
and plugged in
(and then multiplied by
) then we would be done. Since plugging in
doesn't affect the
's, we do that right away. The expression then becomes
where the last equality is true because we are taking the powers of
mod
. Let
denote the coefficient of
in
. Note
. We use the roots of unity filter, which states
where
. In our case
, so we only need to find the average of the
's.
We plug in
and take the average to find the sum of all coefficients of
. Plugging in
makes all of the above zero except for
and
. Averaging, we get
. Now the answer is simply
Solution 2
By changing to
, we can give a bijection between cases where
and cases where
, so we'll just find the probability that
and divide by
in the end. Multiplying the hexagon's vertices by
doesn't change
, and switching any
with
doesn't change the property
, so the probability that
remains the same if we only select our
's at random from
Since
and
, we must choose
exactly
times to make
. To ensure
is real, we must either choose
times,
times, or both
and
times. This gives us a total of
good sequences
, and hence the final result is
Solution 3
We use generating functions and a roots of unity filter.
Notice that all values in are eighth roots of unity multiplied by a constant. Let
be a primitive eighth root of unity (
). The numbers in
are then
. To have
, we must have that
, so eight of the
must belong to
and the other four must belong to
So, we write the generating function
to describe the product. Note that this assumes that the
that belong to
come first, so we will need to multiply by
at the end. We now apply a roots of unity filter to find the sum of the coefficients of the exponents that are
, or equivalently the coefficients of the powers that are multiples of
of the following function:
Let
. We are looking for
.
, and all of the rest are equal to
. So, we get an answer of
. But wait! We need to multiply by
. So, the answer is
Solution 4
We can write the points in polar form as
Note that when multiplying complex numbers, the
's multiply and the
's add, and since
we need
complex numbers with
and
with
By binomial distribution, the probability of this occurring is
For the
part, note that
must be congruent to
and by using simple symmetry, the probability of the aforementioned occurring is
This is since
is even for all
and the number of ordered quadruples
such that
for all
and
is the same for all
again by using symmetry. Thus, our probability is
-fidgetboss_4000
Solution 5 (Simple)
The absolute value of the first two complex numbers is while the absolute value of the latter four is
. For the absolute value of the product to be
, we need
elements with absolute value
and
elements with absolute value of
.
We pick elements from
and
elements from
. We also need to choose which
of
will be chosen from
which gives us
cases. However, suppose we have chosen
elements and we need to choose one more element from
. The product of these
elements can have any of these values:
. For either of these values, there is just one value of
such that
so we must divide by
which gives
cases.
Because there are ways to pick any
elements, our probability is simply
~Zeric
Solution 6
Let ,
,
.
,
,
The magnitude of and
is
, while the magnitude of
,
,
, and
is
. For the product's magnitude to be
, we need
elements from
and
, and
elements from
,
,
, and
.
The product of elements that are
and
will be either
or
. The product of
elements that are
,
,
, and
will either be
or
.
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=CfO4zIeYrSY
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.