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Source of the problem: Mathematics Education Innovation
Let $k>0$ be some real constant and consider $f_k(x) = x^3 - k , x$ for real $x$. Then one can show that for $k geq sqrt8$ there is a (tilted) square centered at $(0,0)$ with all its corners on the graph of $f_k$. (For $k=sqrt8$ there is exactly one such square, for $k>sqrt8$ there are two.) This minimal $k$ can be derived from a computation with resultants or Gröbner bases on the system of equations $$f_k(x)-y, f_k(y)+x, (3x^2 - k)(3y^2 - k)+1$$
and using $x,y neq 0$ where required. (The third equation expresses that the curves given by the first two equations touch in a corner of the square.)
Now for the actual question: how could $k=sqrt8$ be derived using only high school level mathematics? This is not a very precise question, but resultants and Gröbner bases are definitely out. Equations of “high†degree without apparent structure should be avoided as well.
Playing with the equations above leads to many derived equalities but getting rid of $x$ and $y$ by just playing around is not so simple it seems. (One such equation is $k^2=3x^2 y^2-1$ which suggests to somehow derive $x^2y^2=3$.)
Any method is acceptable by the way, it doesn’t have to be purely algebraic as sketched above. Here’s a nice picture for $k=sqrt8$.
polynomials alternative-proof cubic-equations
asked Jul 23 at 10:51
WimC
23.7k22860
add a comment |Â
up vote
4
down vote
favorite
Source of the problem: Mathematics Education Innovation
Let $k>0$ be some real constant and consider $f_k(x) = x^3 - k , x$ for real $x$. Then one can show that for $k geq sqrt8$ there is a (tilted) square centered at $(0,0)$ with all its corners on the graph of $f_k$. (For $k=sqrt8$ there is exactly one such square, for $k>sqrt8$ there are two.) This minimal $k$ can be derived from a computation with resultants or Gröbner bases on the system of equations $$f_k(x)-y, f_k(y)+x, (3x^2 - k)(3y^2 - k)+1$$
and using $x,y neq 0$ where required. (The third equation expresses that the curves given by the first two equations touch in a corner of the square.)
Now for the actual question: how could $k=sqrt8$ be derived using only high school level mathematics? This is not a very precise question, but resultants and Gröbner bases are definitely out. Equations of “high†degree without apparent structure should be avoided as well.
Playing with the equations above leads to many derived equalities but getting rid of $x$ and $y$ by just playing around is not so simple it seems. (One such equation is $k^2=3x^2 y^2-1$ which suggests to somehow derive $x^2y^2=3$.)
Any method is acceptable by the way, it doesn’t have to be purely algebraic as sketched above. Here’s a nice picture for $k=sqrt8$.
polynomials alternative-proof cubic-equations
asked Jul 23 at 10:51
WimC
23.7k22860
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Source of the problem: Mathematics Education Innovation
Let $k>0$ be some real constant and consider $f_k(x) = x^3 - k , x$ for real $x$. Then one can show that for $k geq sqrt8$ there is a (tilted) square centered at $(0,0)$ with all its corners on the graph of $f_k$. (For $k=sqrt8$ there is exactly one such square, for $k>sqrt8$ there are two.) This minimal $k$ can be derived from a computation with resultants or Gröbner bases on the system of equations $$f_k(x)-y, f_k(y)+x, (3x^2 - k)(3y^2 - k)+1$$
and using $x,y neq 0$ where required. (The third equation expresses that the curves given by the first two equations touch in a corner of the square.)
Now for the actual question: how could $k=sqrt8$ be derived using only high school level mathematics? This is not a very precise question, but resultants and Gröbner bases are definitely out. Equations of “high†degree without apparent structure should be avoided as well.
Playing with the equations above leads to many derived equalities but getting rid of $x$ and $y$ by just playing around is not so simple it seems. (One such equation is $k^2=3x^2 y^2-1$ which suggests to somehow derive $x^2y^2=3$.)
Any method is acceptable by the way, it doesn’t have to be purely algebraic as sketched above. Here’s a nice picture for $k=sqrt8$.
polynomials alternative-proof cubic-equations
asked Jul 23 at 10:51
WimC
23.7k22860
Source of the problem: Mathematics Education Innovation
Let $k>0$ be some real constant and consider $f_k(x) = x^3 - k , x$ for real $x$. Then one can show that for $k geq sqrt8$ there is a (tilted) square centered at $(0,0)$ with all its corners on the graph of $f_k$. (For $k=sqrt8$ there is exactly one such square, for $k>sqrt8$ there are two.) This minimal $k$ can be derived from a computation with resultants or Gröbner bases on the system of equations $$f_k(x)-y, f_k(y)+x, (3x^2 - k)(3y^2 - k)+1$$
and using $x,y neq 0$ where required. (The third equation expresses that the curves given by the first two equations touch in a corner of the square.)
Now for the actual question: how could $k=sqrt8$ be derived using only high school level mathematics? This is not a very precise question, but resultants and Gröbner bases are definitely out. Equations of “high†degree without apparent structure should be avoided as well.
Playing with the equations above leads to many derived equalities but getting rid of $x$ and $y$ by just playing around is not so simple it seems. (One such equation is $k^2=3x^2 y^2-1$ which suggests to somehow derive $x^2y^2=3$.)
Any method is acceptable by the way, it doesn’t have to be purely algebraic as sketched above. Here’s a nice picture for $k=sqrt8$.
polynomials alternative-proof cubic-equations
asked Jul 23 at 10:51
WimC
23.7k22860
edited Jul 23 at 11:10
asked Jul 23 at 10:51
WimC
23.7k22860
asked Jul 23 at 10:51
WimC
23.7k22860
asked Jul 23 at 10:51
WimC
23.7k22860
23.7k22860
add a comment |Â
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2 Answers
2
active
oldest
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up vote
3
down vote
Here is how I would do it.
Consider a pair of perpendicular lines passing through the origin, rendered here as $y=mx$ and $y=-x/m$. Combining each of these linear equations with $y=x^3-kx$ gives in turn the following non-origin intersections, where $L$= length if the "diagonal" between each pair of intersections:
$y=mx rightarrow x_1=pmsqrtk+m,L_1=2sqrt(1+m^2)(k+m)$
$y=-x/m rightarrow x_2=pmsqrtk-(1/m),L_2=frac2sqrt(1+m^2)(k-(1/m))m$
To make a square we then just render the diagonal congruent, thus $L_1=L_2$. After removing common factors, clearing fractions and squaring we ultimately arrive at a polynomial equation:
$m^4+km^3-km+1=0$
This looks ugly, but note that any line which solves this equation for a given $k$ must be accompanied by its perpendicular line. Then the quartic polynomial in the slope $m$ must have pairs of roots with product $-1$ forcing this factorization:
$m^4+km^3-km+1=(m^2+am-1)(m^2+bm-1)=0$
Matching both cubic and linear terms then gives $a+b=k$, while matching quadratic terms yields $ab=2$.
Now to get a single pair of perpendicular lines, corresponding to a single "inscribed" square, we need $a=b$ making the factors identical. Then from the matching equations given above either
$a=b=sqrt2,k=2sqrt2$
or
$a=b=-sqrt2,k=-2sqrt2$.
The second root gives no square in the real plane; instead the vertices of that square have imaginary coordinates. The first root is the one giving a single square in the real plane. The corresponding slopes of the lines for this case, obtained by solving $m^2+(sqrt2)m-1=0$, are $m=(-sqrt2pmsqrt6)/2$.
answered Jul 23 at 13:11
Oscar Lanzi
9,96911632
Nice method, thanks. Getting $k$ as the minimum value of $(m^4+1)/(m-m^3)$ on the interval $(0, 1)$ still seems quite a challenge though.
– WimC
Jul 23 at 13:45
I do not minimize the k function directly. Rather I use the factorization of the quartic to identify the minimum positive k for real factors, which would give the single root. Incidentally, the slopes of the diagonals in the degenerate case are $(-sqrt2±sqrt6)/2$.
– Oscar Lanzi
Jul 23 at 14:50
add a comment |Â
up vote
0
down vote
how could $k=sqrt8$ be derived using only high school level mathematics?
The following is a somewhat speculative argument that "something happens" at $,k=2 sqrt2,$.
To start with the speculation, the central symmetry of the cubic suggests that an inscribed square might be centered at the origin (though it doesn't prove that it has to). Let $,(a,b),$ be one vertex of such a square, then the other three would be $,(-b,a), (-a,-b), (b,-a),$. The last two follow by symmetry, and the relevant conditions for the first two vertices to lie on the cubic are;
$$
beginalign
begincases
b = a^3 - ka \
a = -b^3 + kb
endcases
endalign
$$
Eliminating (for example) $,a,$ between the equations:
$$
b = aleft(a^2 - kright) = -bleft(b^2-kright)left(b^2left(b^2-kright)^2-kright)
$$
Canceling out the $,b,$ factor, then expanding and collecting:
$$
b^8 - 3 k b^6 + 3 k^2 b^4 - k (k^2 +1) b^2 + k^2 + 1 = 0 \[10px]
iffquad P(c) = c^4 - 3 k c^3 + 3 k^2 c^2 - k (k^2 +1) c + k^2 + 1 = 0 quadquad stylefont-family:inherittextwhere ;; c = b^2
$$
The latter quartic cannot have negative real roots by Descartes' rule of signs, and the nature of the roots can only change at points where there is a double root. Writing the condition as either the discriminant vanishing, or $,deg gcd(P, P') ge 1,$ gives after routine (albeit tedious) calculations:
$$
0 = 4 k^6 - 60 k^4 + 192 k^2 + 256 = 4 (k^2 - 8)^2 (k^2 + 1)
$$
The negative root $,k = -2 sqrt2,$ is uninteresting since $,f(x),$ is an increasing function when $,k lt 0,$. However, the positive root $,k = 2 sqrt2,$ does in fact lead to explicit solutions for $,c,$, then $,b,$ and $,a,$.
answered Jul 24 at 1:40
dxiv
54k64796
1
Sure, that’s how we do such things. But not exactly high school material.
– WimC
Jul 24 at 5:18
@WimC I'd still call it high-school level, with the disclaimer that it's been a long time for me since ;-) The steps I glossed over can be elaborated without much difficulty, though the complete step-by-tiny-step proof would become long'ish to write down in detail. That said, I don't know that you'll find a satisfying short and simple high-school level answer to the question.
– dxiv
Jul 24 at 5:33
I cannot say that I learned about discriminants, other than the quadratic one, in high school.
– Oscar Lanzi
Jul 24 at 20:09
@OscarLanzi Neither can I, which is why I wrote that the polynomial $,gcd,$ can be used, instead, which requires nothing more than euclidean (long) division.
– dxiv
Jul 24 at 23:52
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Here is how I would do it.
Consider a pair of perpendicular lines passing through the origin, rendered here as $y=mx$ and $y=-x/m$. Combining each of these linear equations with $y=x^3-kx$ gives in turn the following non-origin intersections, where $L$= length if the "diagonal" between each pair of intersections:
$y=mx rightarrow x_1=pmsqrtk+m,L_1=2sqrt(1+m^2)(k+m)$
$y=-x/m rightarrow x_2=pmsqrtk-(1/m),L_2=frac2sqrt(1+m^2)(k-(1/m))m$
To make a square we then just render the diagonal congruent, thus $L_1=L_2$. After removing common factors, clearing fractions and squaring we ultimately arrive at a polynomial equation:
$m^4+km^3-km+1=0$
This looks ugly, but note that any line which solves this equation for a given $k$ must be accompanied by its perpendicular line. Then the quartic polynomial in the slope $m$ must have pairs of roots with product $-1$ forcing this factorization:
$m^4+km^3-km+1=(m^2+am-1)(m^2+bm-1)=0$
Matching both cubic and linear terms then gives $a+b=k$, while matching quadratic terms yields $ab=2$.
Now to get a single pair of perpendicular lines, corresponding to a single "inscribed" square, we need $a=b$ making the factors identical. Then from the matching equations given above either
$a=b=sqrt2,k=2sqrt2$
or
$a=b=-sqrt2,k=-2sqrt2$.
The second root gives no square in the real plane; instead the vertices of that square have imaginary coordinates. The first root is the one giving a single square in the real plane. The corresponding slopes of the lines for this case, obtained by solving $m^2+(sqrt2)m-1=0$, are $m=(-sqrt2pmsqrt6)/2$.
answered Jul 23 at 13:11
Oscar Lanzi
9,96911632
Nice method, thanks. Getting $k$ as the minimum value of $(m^4+1)/(m-m^3)$ on the interval $(0, 1)$ still seems quite a challenge though.
– WimC
Jul 23 at 13:45
I do not minimize the k function directly. Rather I use the factorization of the quartic to identify the minimum positive k for real factors, which would give the single root. Incidentally, the slopes of the diagonals in the degenerate case are $(-sqrt2±sqrt6)/2$.
– Oscar Lanzi
Jul 23 at 14:50
add a comment |Â
up vote
3
down vote
Here is how I would do it.
Consider a pair of perpendicular lines passing through the origin, rendered here as $y=mx$ and $y=-x/m$. Combining each of these linear equations with $y=x^3-kx$ gives in turn the following non-origin intersections, where $L$= length if the "diagonal" between each pair of intersections:
$y=mx rightarrow x_1=pmsqrtk+m,L_1=2sqrt(1+m^2)(k+m)$
$y=-x/m rightarrow x_2=pmsqrtk-(1/m),L_2=frac2sqrt(1+m^2)(k-(1/m))m$
To make a square we then just render the diagonal congruent, thus $L_1=L_2$. After removing common factors, clearing fractions and squaring we ultimately arrive at a polynomial equation:
$m^4+km^3-km+1=0$
This looks ugly, but note that any line which solves this equation for a given $k$ must be accompanied by its perpendicular line. Then the quartic polynomial in the slope $m$ must have pairs of roots with product $-1$ forcing this factorization:
$m^4+km^3-km+1=(m^2+am-1)(m^2+bm-1)=0$
Matching both cubic and linear terms then gives $a+b=k$, while matching quadratic terms yields $ab=2$.
Now to get a single pair of perpendicular lines, corresponding to a single "inscribed" square, we need $a=b$ making the factors identical. Then from the matching equations given above either
$a=b=sqrt2,k=2sqrt2$
or
$a=b=-sqrt2,k=-2sqrt2$.
The second root gives no square in the real plane; instead the vertices of that square have imaginary coordinates. The first root is the one giving a single square in the real plane. The corresponding slopes of the lines for this case, obtained by solving $m^2+(sqrt2)m-1=0$, are $m=(-sqrt2pmsqrt6)/2$.
answered Jul 23 at 13:11
Oscar Lanzi
9,96911632
Nice method, thanks. Getting $k$ as the minimum value of $(m^4+1)/(m-m^3)$ on the interval $(0, 1)$ still seems quite a challenge though.
– WimC
Jul 23 at 13:45
I do not minimize the k function directly. Rather I use the factorization of the quartic to identify the minimum positive k for real factors, which would give the single root. Incidentally, the slopes of the diagonals in the degenerate case are $(-sqrt2±sqrt6)/2$.
– Oscar Lanzi
Jul 23 at 14:50
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Here is how I would do it.
Consider a pair of perpendicular lines passing through the origin, rendered here as $y=mx$ and $y=-x/m$. Combining each of these linear equations with $y=x^3-kx$ gives in turn the following non-origin intersections, where $L$= length if the "diagonal" between each pair of intersections:
$y=mx rightarrow x_1=pmsqrtk+m,L_1=2sqrt(1+m^2)(k+m)$
$y=-x/m rightarrow x_2=pmsqrtk-(1/m),L_2=frac2sqrt(1+m^2)(k-(1/m))m$
To make a square we then just render the diagonal congruent, thus $L_1=L_2$. After removing common factors, clearing fractions and squaring we ultimately arrive at a polynomial equation:
$m^4+km^3-km+1=0$
This looks ugly, but note that any line which solves this equation for a given $k$ must be accompanied by its perpendicular line. Then the quartic polynomial in the slope $m$ must have pairs of roots with product $-1$ forcing this factorization:
$m^4+km^3-km+1=(m^2+am-1)(m^2+bm-1)=0$
Matching both cubic and linear terms then gives $a+b=k$, while matching quadratic terms yields $ab=2$.
Now to get a single pair of perpendicular lines, corresponding to a single "inscribed" square, we need $a=b$ making the factors identical. Then from the matching equations given above either
$a=b=sqrt2,k=2sqrt2$
or
$a=b=-sqrt2,k=-2sqrt2$.
The second root gives no square in the real plane; instead the vertices of that square have imaginary coordinates. The first root is the one giving a single square in the real plane. The corresponding slopes of the lines for this case, obtained by solving $m^2+(sqrt2)m-1=0$, are $m=(-sqrt2pmsqrt6)/2$.
answered Jul 23 at 13:11
Oscar Lanzi
9,96911632
Here is how I would do it.
Consider a pair of perpendicular lines passing through the origin, rendered here as $y=mx$ and $y=-x/m$. Combining each of these linear equations with $y=x^3-kx$ gives in turn the following non-origin intersections, where $L$= length if the "diagonal" between each pair of intersections:
$y=mx rightarrow x_1=pmsqrtk+m,L_1=2sqrt(1+m^2)(k+m)$
$y=-x/m rightarrow x_2=pmsqrtk-(1/m),L_2=frac2sqrt(1+m^2)(k-(1/m))m$
To make a square we then just render the diagonal congruent, thus $L_1=L_2$. After removing common factors, clearing fractions and squaring we ultimately arrive at a polynomial equation:
$m^4+km^3-km+1=0$
This looks ugly, but note that any line which solves this equation for a given $k$ must be accompanied by its perpendicular line. Then the quartic polynomial in the slope $m$ must have pairs of roots with product $-1$ forcing this factorization:
$m^4+km^3-km+1=(m^2+am-1)(m^2+bm-1)=0$
Matching both cubic and linear terms then gives $a+b=k$, while matching quadratic terms yields $ab=2$.
Now to get a single pair of perpendicular lines, corresponding to a single "inscribed" square, we need $a=b$ making the factors identical. Then from the matching equations given above either
$a=b=sqrt2,k=2sqrt2$
or
$a=b=-sqrt2,k=-2sqrt2$.
The second root gives no square in the real plane; instead the vertices of that square have imaginary coordinates. The first root is the one giving a single square in the real plane. The corresponding slopes of the lines for this case, obtained by solving $m^2+(sqrt2)m-1=0$, are $m=(-sqrt2pmsqrt6)/2$.
answered Jul 23 at 13:11
Oscar Lanzi
9,96911632
edited Jul 23 at 15:03
answered Jul 23 at 13:11
Oscar Lanzi
9,96911632
answered Jul 23 at 13:11
Oscar Lanzi
9,96911632
answered Jul 23 at 13:11
Oscar Lanzi
9,96911632
9,96911632
Nice method, thanks. Getting $k$ as the minimum value of $(m^4+1)/(m-m^3)$ on the interval $(0, 1)$ still seems quite a challenge though.
– WimC
Jul 23 at 13:45
I do not minimize the k function directly. Rather I use the factorization of the quartic to identify the minimum positive k for real factors, which would give the single root. Incidentally, the slopes of the diagonals in the degenerate case are $(-sqrt2±sqrt6)/2$.
– Oscar Lanzi
Jul 23 at 14:50
add a comment |Â
Nice method, thanks. Getting $k$ as the minimum value of $(m^4+1)/(m-m^3)$ on the interval $(0, 1)$ still seems quite a challenge though.
– WimC
Jul 23 at 13:45
I do not minimize the k function directly. Rather I use the factorization of the quartic to identify the minimum positive k for real factors, which would give the single root. Incidentally, the slopes of the diagonals in the degenerate case are $(-sqrt2±sqrt6)/2$.
– Oscar Lanzi
Jul 23 at 14:50
Nice method, thanks. Getting $k$ as the minimum value of $(m^4+1)/(m-m^3)$ on the interval $(0, 1)$ still seems quite a challenge though.
– WimC
Jul 23 at 13:45
Nice method, thanks. Getting $k$ as the minimum value of $(m^4+1)/(m-m^3)$ on the interval $(0, 1)$ still seems quite a challenge though.
– WimC
Jul 23 at 13:45
I do not minimize the k function directly. Rather I use the factorization of the quartic to identify the minimum positive k for real factors, which would give the single root. Incidentally, the slopes of the diagonals in the degenerate case are $(-sqrt2±sqrt6)/2$.
– Oscar Lanzi
Jul 23 at 14:50
I do not minimize the k function directly. Rather I use the factorization of the quartic to identify the minimum positive k for real factors, which would give the single root. Incidentally, the slopes of the diagonals in the degenerate case are $(-sqrt2±sqrt6)/2$.
– Oscar Lanzi
Jul 23 at 14:50
add a comment |Â
up vote
0
down vote
how could $k=sqrt8$ be derived using only high school level mathematics?
The following is a somewhat speculative argument that "something happens" at $,k=2 sqrt2,$.
To start with the speculation, the central symmetry of the cubic suggests that an inscribed square might be centered at the origin (though it doesn't prove that it has to). Let $,(a,b),$ be one vertex of such a square, then the other three would be $,(-b,a), (-a,-b), (b,-a),$. The last two follow by symmetry, and the relevant conditions for the first two vertices to lie on the cubic are;
$$
beginalign
begincases
b = a^3 - ka \
a = -b^3 + kb
endcases
endalign
$$
Eliminating (for example) $,a,$ between the equations:
$$
b = aleft(a^2 - kright) = -bleft(b^2-kright)left(b^2left(b^2-kright)^2-kright)
$$
Canceling out the $,b,$ factor, then expanding and collecting:
$$
b^8 - 3 k b^6 + 3 k^2 b^4 - k (k^2 +1) b^2 + k^2 + 1 = 0 \[10px]
iffquad P(c) = c^4 - 3 k c^3 + 3 k^2 c^2 - k (k^2 +1) c + k^2 + 1 = 0 quadquad stylefont-family:inherittextwhere ;; c = b^2
$$
The latter quartic cannot have negative real roots by Descartes' rule of signs, and the nature of the roots can only change at points where there is a double root. Writing the condition as either the discriminant vanishing, or $,deg gcd(P, P') ge 1,$ gives after routine (albeit tedious) calculations:
$$
0 = 4 k^6 - 60 k^4 + 192 k^2 + 256 = 4 (k^2 - 8)^2 (k^2 + 1)
$$
The negative root $,k = -2 sqrt2,$ is uninteresting since $,f(x),$ is an increasing function when $,k lt 0,$. However, the positive root $,k = 2 sqrt2,$ does in fact lead to explicit solutions for $,c,$, then $,b,$ and $,a,$.
answered Jul 24 at 1:40
dxiv
54k64796
1
Sure, that’s how we do such things. But not exactly high school material.
– WimC
Jul 24 at 5:18
@WimC I'd still call it high-school level, with the disclaimer that it's been a long time for me since ;-) The steps I glossed over can be elaborated without much difficulty, though the complete step-by-tiny-step proof would become long'ish to write down in detail. That said, I don't know that you'll find a satisfying short and simple high-school level answer to the question.
– dxiv
Jul 24 at 5:33
I cannot say that I learned about discriminants, other than the quadratic one, in high school.
– Oscar Lanzi
Jul 24 at 20:09
@OscarLanzi Neither can I, which is why I wrote that the polynomial $,gcd,$ can be used, instead, which requires nothing more than euclidean (long) division.
– dxiv
Jul 24 at 23:52
add a comment |Â
up vote
0
down vote
how could $k=sqrt8$ be derived using only high school level mathematics?
The following is a somewhat speculative argument that "something happens" at $,k=2 sqrt2,$.
To start with the speculation, the central symmetry of the cubic suggests that an inscribed square might be centered at the origin (though it doesn't prove that it has to). Let $,(a,b),$ be one vertex of such a square, then the other three would be $,(-b,a), (-a,-b), (b,-a),$. The last two follow by symmetry, and the relevant conditions for the first two vertices to lie on the cubic are;
$$
beginalign
begincases
b = a^3 - ka \
a = -b^3 + kb
endcases
endalign
$$
Eliminating (for example) $,a,$ between the equations:
$$
b = aleft(a^2 - kright) = -bleft(b^2-kright)left(b^2left(b^2-kright)^2-kright)
$$
Canceling out the $,b,$ factor, then expanding and collecting:
$$
b^8 - 3 k b^6 + 3 k^2 b^4 - k (k^2 +1) b^2 + k^2 + 1 = 0 \[10px]
iffquad P(c) = c^4 - 3 k c^3 + 3 k^2 c^2 - k (k^2 +1) c + k^2 + 1 = 0 quadquad stylefont-family:inherittextwhere ;; c = b^2
$$
The latter quartic cannot have negative real roots by Descartes' rule of signs, and the nature of the roots can only change at points where there is a double root. Writing the condition as either the discriminant vanishing, or $,deg gcd(P, P') ge 1,$ gives after routine (albeit tedious) calculations:
$$
0 = 4 k^6 - 60 k^4 + 192 k^2 + 256 = 4 (k^2 - 8)^2 (k^2 + 1)
$$
The negative root $,k = -2 sqrt2,$ is uninteresting since $,f(x),$ is an increasing function when $,k lt 0,$. However, the positive root $,k = 2 sqrt2,$ does in fact lead to explicit solutions for $,c,$, then $,b,$ and $,a,$.
answered Jul 24 at 1:40
dxiv
54k64796
1
Sure, that’s how we do such things. But not exactly high school material.
– WimC
Jul 24 at 5:18
@WimC I'd still call it high-school level, with the disclaimer that it's been a long time for me since ;-) The steps I glossed over can be elaborated without much difficulty, though the complete step-by-tiny-step proof would become long'ish to write down in detail. That said, I don't know that you'll find a satisfying short and simple high-school level answer to the question.
– dxiv
Jul 24 at 5:33
I cannot say that I learned about discriminants, other than the quadratic one, in high school.
– Oscar Lanzi
Jul 24 at 20:09
@OscarLanzi Neither can I, which is why I wrote that the polynomial $,gcd,$ can be used, instead, which requires nothing more than euclidean (long) division.
– dxiv
Jul 24 at 23:52
add a comment |Â
up vote
0
down vote
up vote
0
down vote
how could $k=sqrt8$ be derived using only high school level mathematics?
The following is a somewhat speculative argument that "something happens" at $,k=2 sqrt2,$.
To start with the speculation, the central symmetry of the cubic suggests that an inscribed square might be centered at the origin (though it doesn't prove that it has to). Let $,(a,b),$ be one vertex of such a square, then the other three would be $,(-b,a), (-a,-b), (b,-a),$. The last two follow by symmetry, and the relevant conditions for the first two vertices to lie on the cubic are;
$$
beginalign
begincases
b = a^3 - ka \
a = -b^3 + kb
endcases
endalign
$$
Eliminating (for example) $,a,$ between the equations:
$$
b = aleft(a^2 - kright) = -bleft(b^2-kright)left(b^2left(b^2-kright)^2-kright)
$$
Canceling out the $,b,$ factor, then expanding and collecting:
$$
b^8 - 3 k b^6 + 3 k^2 b^4 - k (k^2 +1) b^2 + k^2 + 1 = 0 \[10px]
iffquad P(c) = c^4 - 3 k c^3 + 3 k^2 c^2 - k (k^2 +1) c + k^2 + 1 = 0 quadquad stylefont-family:inherittextwhere ;; c = b^2
$$
The latter quartic cannot have negative real roots by Descartes' rule of signs, and the nature of the roots can only change at points where there is a double root. Writing the condition as either the discriminant vanishing, or $,deg gcd(P, P') ge 1,$ gives after routine (albeit tedious) calculations:
$$
0 = 4 k^6 - 60 k^4 + 192 k^2 + 256 = 4 (k^2 - 8)^2 (k^2 + 1)
$$
The negative root $,k = -2 sqrt2,$ is uninteresting since $,f(x),$ is an increasing function when $,k lt 0,$. However, the positive root $,k = 2 sqrt2,$ does in fact lead to explicit solutions for $,c,$, then $,b,$ and $,a,$.
answered Jul 24 at 1:40
dxiv
54k64796
how could $k=sqrt8$ be derived using only high school level mathematics?
The following is a somewhat speculative argument that "something happens" at $,k=2 sqrt2,$.
To start with the speculation, the central symmetry of the cubic suggests that an inscribed square might be centered at the origin (though it doesn't prove that it has to). Let $,(a,b),$ be one vertex of such a square, then the other three would be $,(-b,a), (-a,-b), (b,-a),$. The last two follow by symmetry, and the relevant conditions for the first two vertices to lie on the cubic are;
$$
beginalign
begincases
b = a^3 - ka \
a = -b^3 + kb
endcases
endalign
$$
Eliminating (for example) $,a,$ between the equations:
$$
b = aleft(a^2 - kright) = -bleft(b^2-kright)left(b^2left(b^2-kright)^2-kright)
$$
Canceling out the $,b,$ factor, then expanding and collecting:
$$
b^8 - 3 k b^6 + 3 k^2 b^4 - k (k^2 +1) b^2 + k^2 + 1 = 0 \[10px]
iffquad P(c) = c^4 - 3 k c^3 + 3 k^2 c^2 - k (k^2 +1) c + k^2 + 1 = 0 quadquad stylefont-family:inherittextwhere ;; c = b^2
$$
The latter quartic cannot have negative real roots by Descartes' rule of signs, and the nature of the roots can only change at points where there is a double root. Writing the condition as either the discriminant vanishing, or $,deg gcd(P, P') ge 1,$ gives after routine (albeit tedious) calculations:
$$
0 = 4 k^6 - 60 k^4 + 192 k^2 + 256 = 4 (k^2 - 8)^2 (k^2 + 1)
$$
The negative root $,k = -2 sqrt2,$ is uninteresting since $,f(x),$ is an increasing function when $,k lt 0,$. However, the positive root $,k = 2 sqrt2,$ does in fact lead to explicit solutions for $,c,$, then $,b,$ and $,a,$.
answered Jul 24 at 1:40
dxiv
54k64796
answered Jul 24 at 1:40
dxiv
54k64796
answered Jul 24 at 1:40
dxiv
54k64796
answered Jul 24 at 1:40
dxiv
54k64796
54k64796
1
Sure, that’s how we do such things. But not exactly high school material.
– WimC
Jul 24 at 5:18
@WimC I'd still call it high-school level, with the disclaimer that it's been a long time for me since ;-) The steps I glossed over can be elaborated without much difficulty, though the complete step-by-tiny-step proof would become long'ish to write down in detail. That said, I don't know that you'll find a satisfying short and simple high-school level answer to the question.
– dxiv
Jul 24 at 5:33
I cannot say that I learned about discriminants, other than the quadratic one, in high school.
– Oscar Lanzi
Jul 24 at 20:09
@OscarLanzi Neither can I, which is why I wrote that the polynomial $,gcd,$ can be used, instead, which requires nothing more than euclidean (long) division.
– dxiv
Jul 24 at 23:52
add a comment |Â
1
Sure, that’s how we do such things. But not exactly high school material.
– WimC
Jul 24 at 5:18
@WimC I'd still call it high-school level, with the disclaimer that it's been a long time for me since ;-) The steps I glossed over can be elaborated without much difficulty, though the complete step-by-tiny-step proof would become long'ish to write down in detail. That said, I don't know that you'll find a satisfying short and simple high-school level answer to the question.
– dxiv
Jul 24 at 5:33
I cannot say that I learned about discriminants, other than the quadratic one, in high school.
– Oscar Lanzi
Jul 24 at 20:09
@OscarLanzi Neither can I, which is why I wrote that the polynomial $,gcd,$ can be used, instead, which requires nothing more than euclidean (long) division.
– dxiv
Jul 24 at 23:52
1
1
Sure, that’s how we do such things. But not exactly high school material.
– WimC
Jul 24 at 5:18
Sure, that’s how we do such things. But not exactly high school material.
– WimC
Jul 24 at 5:18
@WimC I'd still call it high-school level, with the disclaimer that it's been a long time for me since ;-) The steps I glossed over can be elaborated without much difficulty, though the complete step-by-tiny-step proof would become long'ish to write down in detail. That said, I don't know that you'll find a satisfying short and simple high-school level answer to the question.
– dxiv
Jul 24 at 5:33
@WimC I'd still call it high-school level, with the disclaimer that it's been a long time for me since ;-) The steps I glossed over can be elaborated without much difficulty, though the complete step-by-tiny-step proof would become long'ish to write down in detail. That said, I don't know that you'll find a satisfying short and simple high-school level answer to the question.
– dxiv
Jul 24 at 5:33
I cannot say that I learned about discriminants, other than the quadratic one, in high school.
– Oscar Lanzi
Jul 24 at 20:09
I cannot say that I learned about discriminants, other than the quadratic one, in high school.
– Oscar Lanzi
Jul 24 at 20:09
@OscarLanzi Neither can I, which is why I wrote that the polynomial $,gcd,$ can be used, instead, which requires nothing more than euclidean (long) division.
– dxiv
Jul 24 at 23:52
@OscarLanzi Neither can I, which is why I wrote that the polynomial $,gcd,$ can be used, instead, which requires nothing more than euclidean (long) division.
– dxiv
Jul 24 at 23:52
add a comment |Â
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