Difference between revisions of "2023 AIME I Problems/Problem 9"
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− | ==Problem | + | ==Problem== |
− | Find the number of cubic polynomials <math>p(x) = x^3 + ax^2 + bx + c</math> | + | Find the number of cubic polynomials <math>p(x) = x^3 + ax^2 + bx + c,</math> where <math>a, b,</math> and <math>c</math> are integers in <math>\{-20,-19,-18,\ldots,18,19,20\},</math> such that there is a unique integer <math>m \not= 2</math> with <math>p(m) = p(2).</math> |
− | are integers in <math>\{ -20, -19, -18, \ | ||
− | <math>m \ | ||
− | ==Solution 1 | + | ==Solution 1 (bash) == |
− | |||
− | |||
− | + | Plugging <math>2</math> and <math>m</math> into <math>P(x)</math> and equating them, we get <math>8+4a+2b+c = m^3+am^2+bm+c</math>. Rearranging, we have <cmath>(m^3-8) + (m^2 - 4)a + (m-2)b = 0.</cmath> Note that the value of <math>c</math> won't matter as it can be anything in the provided range, giving a total of <math>41</math> possible choices for <math>c.</math> So what we just need to do is to just find the number of ordered pairs <math>(a, b)</math> that work, and multiply it by <math>41.</math> | |
− | + | We can start by first dividing both sides by <math>m-2.</math> (Note that this is valid since <math>m\neq2:</math> <cmath>m^2 + 2m + 4 + (m+2)a + b = 0.</cmath> We can rearrange this so it is a quadratic in <math>m</math>: <cmath>m^2 + (a+2)m + (4 + 2a + b) = 0.</cmath> Remember that <math>m</math> has to be unique and not equal to <math>2.</math> We can split this into two cases: case <math>1</math> being that <math>m</math> has exactly one solution, and it isn't equal to <math>2</math>; case <math>2</math> being that <math>m</math> has two solutions, one being equal to <math>2,</math> but the other is a unique solution not equal to <math>2.</math> | |
− | |||
− | |||
− | + | <math>\textbf{Case 1:}</math> | |
− | |||
− | |||
− | |||
− | |||
− | + | There is exactly one solution for <math>m,</math> and that solution is not <math>2.</math> This means that the discriminant of the quadratic equation is <math>0,</math> using that, we have <math>(a+2)^2 = 4(4 + 2a + b),</math> rearranging in a neat way, we have <cmath>(a-2)^2 = 4(4 + b)\Longrightarrow a = 2\pm2\sqrt{4+b}.</cmath> Using the fact that <math>4+b</math> must be a perfect square, we can easily see that the values for <math>b</math> can be <math>-4, -3, 0, 5,</math> and <math>12.</math> Also since it's a "<math>\pm</math>" there will usually be <math>2</math> solutions for <math>a</math> for each value of <math>b.</math> The two exceptions for this would be if <math>b = -4</math> and <math>b = 12.</math> For <math>b=-4</math> because it would be a <math>\pm0,</math> which only gives one solution, instead of two. And for <math>b=12</math> because then <math>a = -6</math> and the solution for <math>m</math> would equal to <math>2,</math> and we don't want this. (We can know this by putting the solutions back into the quadratic formula). | |
− | + | So we have <math>5</math> solutions for <math>b,</math> each of which give <math>2</math> values for <math>a,</math> except for <math>2,</math> which only give one. So in total, there are <math>5*2 - 2 = 8</math> ordered pairs of <math>(a,b)</math> in this case. | |
− | + | ||
+ | <math>\textbf{Case 2:}</math> | ||
+ | |||
+ | <math>m</math> has two solutions, but exactly one of them isn't equal to <math>2.</math> This ensures that <math>1</math> of the solutions is equal to <math>2.</math> | ||
+ | |||
+ | Let <math>r</math> be the other value of <math>m</math> that isn't <math>2.</math> By Vieta: | ||
+ | <cmath>\begin{align*} | ||
+ | r+2 &= -a-2\\ | ||
+ | 2r &= 4+2a+b. | ||
+ | \end{align*}</cmath> From the first equation, we subtract both sides by <math>2</math> and double both sides to get <math>2r = -2a - 8</math> which also equals to <math>4+2a+b</math> from the second equation. Equating both, we have <math>4a + b + 12 = 0.</math> We can easily count that there would be <math>11</math> ordered pairs <math>(a,b)</math> that satisfy that. | ||
+ | |||
+ | However, there's an outlier case in which <math>r</math> happens to also equal to <math>2,</math> and we don't want that. We can reverse engineer and find out that <math>r=2</math> when <math>(a,b) = (-6, 12),</math> which we overcounted. So we subtract by one and we conclude that there are <math>10</math> ordered pairs of <math>(a,b)</math> that satisfy this case. | ||
+ | |||
+ | |||
+ | This all shows that there are a total of <math>8+10 = 18</math> amount of ordered pairs <math>(a,b).</math> Multiplying this by <math>41</math> (the amount of values for <math>c</math>) we get <math>18\cdot41=\boxed{738}</math> as our final answer. | ||
+ | |||
+ | ~s214425 | ||
+ | |||
+ | ==Solution 2 (factor the difference)== | ||
+ | |||
+ | <math>p(x)-p(2)</math> is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers. | ||
+ | |||
+ | There are exactly two distinct roots, so either <math>p(x)=p(2)+(x-2)^2(x-m)</math> or <math>p(x)=p(2)+(x-2)(x-m)^2</math>, with <math>m\neq 2</math>. | ||
+ | |||
+ | In the first case <math>p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)</math>, with <math>|4+4m|\leq 20</math> (which entails <math>|4+m|\leq 20</math>), so <math>m</math> can be <math>-6,-5,-4,-3,-2,-1,0,1, (\textbf{not 2}!), 3,4</math> and <math>-4m+p(2)</math> can be any integer from <math>-20</math> to <math>20</math>, giving <math>410</math> polynomials (since the coefficients are given by linear functions of <math>m</math> and thus are distinct). | ||
+ | |||
+ | In the second case <math>p(x)=x^3-(2+2m)x^2+(4m+m^2)x-2m^2+p(2)</math>, and <math>m</math> can be <math>-6,-5,-4,-3,-2,-1,0,1</math> and <math>-4m+p(2)</math> can be any integer from <math>-20</math> to <math>20</math>, giving <math>328</math> polynomials. | ||
+ | |||
+ | The total is <math>\boxed{738}</math>. | ||
+ | |||
+ | ~EVIN- | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/-Asb_5nTgSg | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
− | |||
==See also== | ==See also== |
Latest revision as of 20:40, 27 December 2023
Contents
Problem
Find the number of cubic polynomials where
and
are integers in
such that there is a unique integer
with
Solution 1 (bash)
Plugging and
into
and equating them, we get
. Rearranging, we have
Note that the value of
won't matter as it can be anything in the provided range, giving a total of
possible choices for
So what we just need to do is to just find the number of ordered pairs
that work, and multiply it by
We can start by first dividing both sides by
(Note that this is valid since
We can rearrange this so it is a quadratic in
:
Remember that
has to be unique and not equal to
We can split this into two cases: case
being that
has exactly one solution, and it isn't equal to
; case
being that
has two solutions, one being equal to
but the other is a unique solution not equal to
There is exactly one solution for and that solution is not
This means that the discriminant of the quadratic equation is
using that, we have
rearranging in a neat way, we have
Using the fact that
must be a perfect square, we can easily see that the values for
can be
and
Also since it's a "
" there will usually be
solutions for
for each value of
The two exceptions for this would be if
and
For
because it would be a
which only gives one solution, instead of two. And for
because then
and the solution for
would equal to
and we don't want this. (We can know this by putting the solutions back into the quadratic formula).
So we have solutions for
each of which give
values for
except for
which only give one. So in total, there are
ordered pairs of
in this case.
has two solutions, but exactly one of them isn't equal to
This ensures that
of the solutions is equal to
Let be the other value of
that isn't
By Vieta:
From the first equation, we subtract both sides by
and double both sides to get
which also equals to
from the second equation. Equating both, we have
We can easily count that there would be
ordered pairs
that satisfy that.
However, there's an outlier case in which happens to also equal to
and we don't want that. We can reverse engineer and find out that
when
which we overcounted. So we subtract by one and we conclude that there are
ordered pairs of
that satisfy this case.
This all shows that there are a total of amount of ordered pairs
Multiplying this by
(the amount of values for
) we get
as our final answer.
~s214425
Solution 2 (factor the difference)
is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers.
There are exactly two distinct roots, so either or
, with
.
In the first case , with
(which entails
), so
can be
and
can be any integer from
to
, giving
polynomials (since the coefficients are given by linear functions of
and thus are distinct).
In the second case , and
can be
and
can be any integer from
to
, giving
polynomials.
The total is .
~EVIN-
Video Solution
~MathProblemSolvingSkills.com
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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