AIME I 2008 #13 solution?
$begingroup$
This is the problem.
That's the entire given solution.
This is the part I'm confused about.
Essentially, he used the zeroes of the function to express the coefficients in terms of each other and condensed the polynomial. Makes sense, and I got only this far. However, in the line where he says $x(1-x)(1+x)=y(1-y)(1+y)$ (Eq1), I don't understand why he's equating these two. Can someone explain the justification for that?
In the next line(Eq2), given that Eq1 is true, I understand is taken from $3a_1+3a_2+2a_4=0$.
algebra-precalculus
asked Dec 16 '18 at 5:07
Saketh MalyalaSaketh Malyala
7,4231534
$endgroup$
add a comment |
$begingroup$
This is the problem.
That's the entire given solution.
This is the part I'm confused about.
Essentially, he used the zeroes of the function to express the coefficients in terms of each other and condensed the polynomial. Makes sense, and I got only this far. However, in the line where he says $x(1-x)(1+x)=y(1-y)(1+y)$ (Eq1), I don't understand why he's equating these two. Can someone explain the justification for that?
In the next line(Eq2), given that Eq1 is true, I understand is taken from $3a_1+3a_2+2a_4=0$.
algebra-precalculus
asked Dec 16 '18 at 5:07
Saketh MalyalaSaketh Malyala
7,4231534
$endgroup$
add a comment |
$begingroup$
This is the problem.
That's the entire given solution.
This is the part I'm confused about.
Essentially, he used the zeroes of the function to express the coefficients in terms of each other and condensed the polynomial. Makes sense, and I got only this far. However, in the line where he says $x(1-x)(1+x)=y(1-y)(1+y)$ (Eq1), I don't understand why he's equating these two. Can someone explain the justification for that?
In the next line(Eq2), given that Eq1 is true, I understand is taken from $3a_1+3a_2+2a_4=0$.
algebra-precalculus
asked Dec 16 '18 at 5:07
Saketh MalyalaSaketh Malyala
7,4231534
$endgroup$
This is the problem.
That's the entire given solution.
This is the part I'm confused about.
Essentially, he used the zeroes of the function to express the coefficients in terms of each other and condensed the polynomial. Makes sense, and I got only this far. However, in the line where he says $x(1-x)(1+x)=y(1-y)(1+y)$ (Eq1), I don't understand why he's equating these two. Can someone explain the justification for that?
In the next line(Eq2), given that Eq1 is true, I understand is taken from $3a_1+3a_2+2a_4=0$.
algebra-precalculus
algebra-precalculus
asked Dec 16 '18 at 5:07
Saketh MalyalaSaketh Malyala
7,4231534
asked Dec 16 '18 at 5:07
Saketh MalyalaSaketh Malyala
7,4231534
asked Dec 16 '18 at 5:07
Saketh MalyalaSaketh Malyala
7,4231534
asked Dec 16 '18 at 5:07
Saketh MalyalaSaketh Malyala
7,4231534
asked Dec 16 '18 at 5:07
Saketh MalyalaSaketh Malyala
7,4231534
7,4231534
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You want $$x(1-x)(1+x) a_1 + y(1-y)(1+y)a_2 + xy(1-x) a_4 = 0.$$
You have used all the information given in the problem already to write the polynomial $p$ in terms of only three coefficients $a_1, a_2, a_4$ satisfying $$3 a_1 + 3 a_2 + 2 a_4 = 0.$$
With only this information, you can only look for $x$ and $y$ such that the triple $x(1-x)(1+x)$, $y(1-y)(1+y)$, and $xy(1-x)$ have ratio $3:3:2$.
It is possible that a particular polynomial $p$ may have other zeros of the form $(a/c, b/c)$ etc., but the problem is asking for a zero that works for all such $p$. Since you've already used all the information characterizing $p$, the above is all you can do.
answered Dec 16 '18 at 5:22
angryavianangryavian
42.1k23381
$endgroup$
$begingroup$
OHHHHHHHHHHHHH I GET IT. it clicked!!!!!!
$endgroup$
– Saketh Malyala
Dec 16 '18 at 8:58
add a comment |
$begingroup$
The given condition should be true for all polynomials with the then derived relation
- $3a_1+3a_2+2a_4=0$
So, it must be true specifically for
$a_1 = 1, a_2 =-1, a_4 = 0$ and
$a_1 = 2, a_2 =0, a_4 = -3$.
These two sets of values give the two equations.
Comment:
It might be worth noting that $begin{pmatrix}1\-1\0end{pmatrix},begin{pmatrix}2\0\-3end{pmatrix}$ form a basis of the solutions of $3a_1+3a_2+2a_4=0$. This means, for the coefficients of all admissible polynomials you have:
$$begin{pmatrix}a_1\a_2\a_4end{pmatrix} = sbegin{pmatrix}color{blue}{1}\color{blue}{-1}\0end{pmatrix}+tbegin{pmatrix}color{blue}{2}\0\color{blue}{-3}end{pmatrix} mbox{ with } s,t in mathbb{R}$$
So, you get
$$x(1-x)(1+x) a_1 + y(1-y)(1+y)a_2 + xy(1-x) a_4 = s(color{blue}{1}x(1-x)(1+x) color{blue}{-1} y(1-y)(1+y)) + t(color{blue}{2}x(1-x)(1+x) color{blue}{- 3}xy(1-x))$$
From this you can see that any solution to the two specific equations above is a solution for all polynomials satisfying the condition $3a_1+3a_2+2a_4=0$.
answered Dec 16 '18 at 5:30
trancelocationtrancelocation
12.6k1826
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
You want $$x(1-x)(1+x) a_1 + y(1-y)(1+y)a_2 + xy(1-x) a_4 = 0.$$
You have used all the information given in the problem already to write the polynomial $p$ in terms of only three coefficients $a_1, a_2, a_4$ satisfying $$3 a_1 + 3 a_2 + 2 a_4 = 0.$$
With only this information, you can only look for $x$ and $y$ such that the triple $x(1-x)(1+x)$, $y(1-y)(1+y)$, and $xy(1-x)$ have ratio $3:3:2$.
It is possible that a particular polynomial $p$ may have other zeros of the form $(a/c, b/c)$ etc., but the problem is asking for a zero that works for all such $p$. Since you've already used all the information characterizing $p$, the above is all you can do.
answered Dec 16 '18 at 5:22
angryavianangryavian
42.1k23381
$endgroup$
$begingroup$
OHHHHHHHHHHHHH I GET IT. it clicked!!!!!!
$endgroup$
– Saketh Malyala
Dec 16 '18 at 8:58
add a comment |
$begingroup$
You want $$x(1-x)(1+x) a_1 + y(1-y)(1+y)a_2 + xy(1-x) a_4 = 0.$$
You have used all the information given in the problem already to write the polynomial $p$ in terms of only three coefficients $a_1, a_2, a_4$ satisfying $$3 a_1 + 3 a_2 + 2 a_4 = 0.$$
With only this information, you can only look for $x$ and $y$ such that the triple $x(1-x)(1+x)$, $y(1-y)(1+y)$, and $xy(1-x)$ have ratio $3:3:2$.
It is possible that a particular polynomial $p$ may have other zeros of the form $(a/c, b/c)$ etc., but the problem is asking for a zero that works for all such $p$. Since you've already used all the information characterizing $p$, the above is all you can do.
answered Dec 16 '18 at 5:22
angryavianangryavian
42.1k23381
$endgroup$
$begingroup$
OHHHHHHHHHHHHH I GET IT. it clicked!!!!!!
$endgroup$
– Saketh Malyala
Dec 16 '18 at 8:58
add a comment |
$begingroup$
You want $$x(1-x)(1+x) a_1 + y(1-y)(1+y)a_2 + xy(1-x) a_4 = 0.$$
You have used all the information given in the problem already to write the polynomial $p$ in terms of only three coefficients $a_1, a_2, a_4$ satisfying $$3 a_1 + 3 a_2 + 2 a_4 = 0.$$
With only this information, you can only look for $x$ and $y$ such that the triple $x(1-x)(1+x)$, $y(1-y)(1+y)$, and $xy(1-x)$ have ratio $3:3:2$.
It is possible that a particular polynomial $p$ may have other zeros of the form $(a/c, b/c)$ etc., but the problem is asking for a zero that works for all such $p$. Since you've already used all the information characterizing $p$, the above is all you can do.
answered Dec 16 '18 at 5:22
angryavianangryavian
42.1k23381
$endgroup$
You want $$x(1-x)(1+x) a_1 + y(1-y)(1+y)a_2 + xy(1-x) a_4 = 0.$$
You have used all the information given in the problem already to write the polynomial $p$ in terms of only three coefficients $a_1, a_2, a_4$ satisfying $$3 a_1 + 3 a_2 + 2 a_4 = 0.$$
With only this information, you can only look for $x$ and $y$ such that the triple $x(1-x)(1+x)$, $y(1-y)(1+y)$, and $xy(1-x)$ have ratio $3:3:2$.
It is possible that a particular polynomial $p$ may have other zeros of the form $(a/c, b/c)$ etc., but the problem is asking for a zero that works for all such $p$. Since you've already used all the information characterizing $p$, the above is all you can do.
answered Dec 16 '18 at 5:22
angryavianangryavian
42.1k23381
answered Dec 16 '18 at 5:22
angryavianangryavian
42.1k23381
answered Dec 16 '18 at 5:22
angryavianangryavian
42.1k23381
answered Dec 16 '18 at 5:22
angryavianangryavian
42.1k23381
42.1k23381
$begingroup$
OHHHHHHHHHHHHH I GET IT. it clicked!!!!!!
$endgroup$
– Saketh Malyala
Dec 16 '18 at 8:58
add a comment |
$begingroup$
OHHHHHHHHHHHHH I GET IT. it clicked!!!!!!
$endgroup$
– Saketh Malyala
Dec 16 '18 at 8:58
$begingroup$
OHHHHHHHHHHHHH I GET IT. it clicked!!!!!!
$endgroup$
– Saketh Malyala
Dec 16 '18 at 8:58
$begingroup$
OHHHHHHHHHHHHH I GET IT. it clicked!!!!!!
$endgroup$
– Saketh Malyala
Dec 16 '18 at 8:58
add a comment |
$begingroup$
The given condition should be true for all polynomials with the then derived relation
- $3a_1+3a_2+2a_4=0$
So, it must be true specifically for
$a_1 = 1, a_2 =-1, a_4 = 0$ and
$a_1 = 2, a_2 =0, a_4 = -3$.
These two sets of values give the two equations.
Comment:
It might be worth noting that $begin{pmatrix}1\-1\0end{pmatrix},begin{pmatrix}2\0\-3end{pmatrix}$ form a basis of the solutions of $3a_1+3a_2+2a_4=0$. This means, for the coefficients of all admissible polynomials you have:
$$begin{pmatrix}a_1\a_2\a_4end{pmatrix} = sbegin{pmatrix}color{blue}{1}\color{blue}{-1}\0end{pmatrix}+tbegin{pmatrix}color{blue}{2}\0\color{blue}{-3}end{pmatrix} mbox{ with } s,t in mathbb{R}$$
So, you get
$$x(1-x)(1+x) a_1 + y(1-y)(1+y)a_2 + xy(1-x) a_4 = s(color{blue}{1}x(1-x)(1+x) color{blue}{-1} y(1-y)(1+y)) + t(color{blue}{2}x(1-x)(1+x) color{blue}{- 3}xy(1-x))$$
From this you can see that any solution to the two specific equations above is a solution for all polynomials satisfying the condition $3a_1+3a_2+2a_4=0$.
answered Dec 16 '18 at 5:30
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
The given condition should be true for all polynomials with the then derived relation
- $3a_1+3a_2+2a_4=0$
So, it must be true specifically for
$a_1 = 1, a_2 =-1, a_4 = 0$ and
$a_1 = 2, a_2 =0, a_4 = -3$.
These two sets of values give the two equations.
Comment:
It might be worth noting that $begin{pmatrix}1\-1\0end{pmatrix},begin{pmatrix}2\0\-3end{pmatrix}$ form a basis of the solutions of $3a_1+3a_2+2a_4=0$. This means, for the coefficients of all admissible polynomials you have:
$$begin{pmatrix}a_1\a_2\a_4end{pmatrix} = sbegin{pmatrix}color{blue}{1}\color{blue}{-1}\0end{pmatrix}+tbegin{pmatrix}color{blue}{2}\0\color{blue}{-3}end{pmatrix} mbox{ with } s,t in mathbb{R}$$
So, you get
$$x(1-x)(1+x) a_1 + y(1-y)(1+y)a_2 + xy(1-x) a_4 = s(color{blue}{1}x(1-x)(1+x) color{blue}{-1} y(1-y)(1+y)) + t(color{blue}{2}x(1-x)(1+x) color{blue}{- 3}xy(1-x))$$
From this you can see that any solution to the two specific equations above is a solution for all polynomials satisfying the condition $3a_1+3a_2+2a_4=0$.
answered Dec 16 '18 at 5:30
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
The given condition should be true for all polynomials with the then derived relation
- $3a_1+3a_2+2a_4=0$
So, it must be true specifically for
$a_1 = 1, a_2 =-1, a_4 = 0$ and
$a_1 = 2, a_2 =0, a_4 = -3$.
These two sets of values give the two equations.
Comment:
It might be worth noting that $begin{pmatrix}1\-1\0end{pmatrix},begin{pmatrix}2\0\-3end{pmatrix}$ form a basis of the solutions of $3a_1+3a_2+2a_4=0$. This means, for the coefficients of all admissible polynomials you have:
$$begin{pmatrix}a_1\a_2\a_4end{pmatrix} = sbegin{pmatrix}color{blue}{1}\color{blue}{-1}\0end{pmatrix}+tbegin{pmatrix}color{blue}{2}\0\color{blue}{-3}end{pmatrix} mbox{ with } s,t in mathbb{R}$$
So, you get
$$x(1-x)(1+x) a_1 + y(1-y)(1+y)a_2 + xy(1-x) a_4 = s(color{blue}{1}x(1-x)(1+x) color{blue}{-1} y(1-y)(1+y)) + t(color{blue}{2}x(1-x)(1+x) color{blue}{- 3}xy(1-x))$$
From this you can see that any solution to the two specific equations above is a solution for all polynomials satisfying the condition $3a_1+3a_2+2a_4=0$.
answered Dec 16 '18 at 5:30
trancelocationtrancelocation
12.6k1826
$endgroup$
The given condition should be true for all polynomials with the then derived relation
- $3a_1+3a_2+2a_4=0$
So, it must be true specifically for
$a_1 = 1, a_2 =-1, a_4 = 0$ and
$a_1 = 2, a_2 =0, a_4 = -3$.
These two sets of values give the two equations.
Comment:
It might be worth noting that $begin{pmatrix}1\-1\0end{pmatrix},begin{pmatrix}2\0\-3end{pmatrix}$ form a basis of the solutions of $3a_1+3a_2+2a_4=0$. This means, for the coefficients of all admissible polynomials you have:
$$begin{pmatrix}a_1\a_2\a_4end{pmatrix} = sbegin{pmatrix}color{blue}{1}\color{blue}{-1}\0end{pmatrix}+tbegin{pmatrix}color{blue}{2}\0\color{blue}{-3}end{pmatrix} mbox{ with } s,t in mathbb{R}$$
So, you get
$$x(1-x)(1+x) a_1 + y(1-y)(1+y)a_2 + xy(1-x) a_4 = s(color{blue}{1}x(1-x)(1+x) color{blue}{-1} y(1-y)(1+y)) + t(color{blue}{2}x(1-x)(1+x) color{blue}{- 3}xy(1-x))$$
From this you can see that any solution to the two specific equations above is a solution for all polynomials satisfying the condition $3a_1+3a_2+2a_4=0$.
answered Dec 16 '18 at 5:30
trancelocationtrancelocation
12.6k1826
edited Dec 16 '18 at 5:57
answered Dec 16 '18 at 5:30
trancelocationtrancelocation
12.6k1826
answered Dec 16 '18 at 5:30
trancelocationtrancelocation
12.6k1826
answered Dec 16 '18 at 5:30
trancelocationtrancelocation
12.6k1826
12.6k1826
add a comment |
add a comment |
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