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The system is:

$$x - 2y + z = 2\alpha \\ 3(xy + xz + yz) = 3\alpha - 4$$

I have to find for which values of $\alpha$, the system has an unique solution.

I've tried to simplify both expressions, set $x,y,z$ to some special values, and even apply some well-known inequalities to check for validity.

Any hint will be greatly appreciated!

Gerry Myerson
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Alexandru Dinu
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  • Could you perhaps give a reference of where you found the system? The second equation is not linear, which I find peculiar. – Mussé Redi May 27 '14 at 12:27
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    The first equation's a plane, the second, some kind of quadric surface, you're looking for the value of $a$ that makes the plane tangent to the surface. You know about tangent planes to surfaces? – Gerry Myerson May 27 '14 at 12:29
  • I've only tried algebraic approaches, but, as I said, any hint will be appreciated. – Alexandru Dinu May 27 '14 at 12:31

3 Answers3

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Here is a "by hand" method to make some progress.

You have from the first equation $x+z=2(y+\alpha)$

The second equation can be rewritten as $(x+z)y+xz=\alpha-\frac 43=2y(y+\alpha)+xz$

Both equations are symmetric in $x,z$ so if there is a single solution we must have $x=z$

Hence $x=z=y+\alpha$ and $x^2=\alpha-\frac 43-2y(y+\alpha)=(y+\alpha)^2$

Now there is a quadratic in $y$ namely $$9y^2+12\alpha y+3\alpha^2-3\alpha+4=0$$ or $$(3y+2\alpha)^2=\alpha^2+3\alpha-4=(\alpha+4)(\alpha-1)$$This has a single root for $y$ only if $\alpha=-4$ or $\alpha =1$

This identifies possible values for $\alpha$ on the assumption that the solution is unique. However it remains to be checked whether the solutions obtained are unique for these values of $\alpha$. The method above can be used to obtain a quadratic for $x$ and $z$ with coefficients which are functions of $y$. The discriminant of this quadratic - a function of $y$ can be used to check for uniqueness.

Mark Bennet
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  • How are both equations,$(x+z)=y+xz \text{ and } 2y(y+\alpha)+xz$, symmetric in $x,z$? What do you mean by symmetric? And why must we then have $x=z$? – Mussé Redi May 27 '14 at 16:08
  • @MusséRedi If you exchange $x$ and $z$ you get the same equation back again. So if you have a solution $(x,z)=(a,b)$ then $(x,z)=(b,a)$ is also a solution. Since you want there to be only one solution you need to have $a=b$, and this can be enforced by setting $x=z$. But you then have to check that there are no solutions with $x\neq z$ for any value of $\alpha$ you find by this method. – Mark Bennet May 27 '14 at 16:12
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I assume that the variables $x,y,z$ and the parameter $\alpha$ are supposed to be real. To begin with, one could paramtetrise the plane $x-2y+z=2\alpha$ e.g. as follows: $$ \begin{align} x &= s + 2t, \\ y &= t - \alpha, \\ z &= -s. \end{align} $$ Substituting the above expressions for $x$, $y$ and $z$ in the equation of the quadric and simplifying, one obtains the equation $$ -s^2 - 2st + 2t^2 - 2\alpha t = \alpha - \frac{4}{3}.$$ This can be rewritten into the equivalent form $$ -\left(s^2 + 2st + t^2\right) + 3\left(t^2 - \frac{2\alpha}{3}t + \frac{\alpha^2}{9}\right) = \frac{\alpha^2}{3} + \alpha - \frac{4}{3},$$ or $$ -\left(s+t\right)^2 + 3\left(t-\frac{\alpha}{3}\right)^2 = \frac{1}{3}(\alpha+4)(\alpha-1).$$ If $\alpha = -4$ or $\alpha = 1$, the above equation describes a pair of concurrent lines in the real plane with coordinates $(s,t)$. In all other cases, the equation represents a hyperbola in that plane. Therefore, the original system of equations has infinitely many solutions for all values of $\alpha$ (unless I made a mistake somewhere).

ivanpenev
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  • How did you find your parametrization of the plane $x-2y+z=2\alpha$? – Mussé Redi May 27 '14 at 16:18
  • In order to give an affine parametrisation of a plane one needs the coordinates of a point lying on the plane and of two linearly independent vectors parallel to that plane. To obtain a point, one gives values to all but one of the variables, and then uses the implicit equation of the plane to solve for the last one. In this case, I set $x = z = 0$ and determine $y = -\alpha$, so the point is $p=(0,-\alpha,0)$. To obtain two linearly independent vectors $v$ and $w$ parallel to the plane, one considers the associated homogeneous equation $x - 2y + z = 0$. (to be continued). – ivanpenev May 27 '14 at 16:51
  • (continued from last comment) One selects a variable for which one can solve the homogeneous equation of the plane in terms of the remaining variables. In the present case, this can be any variable. I chose $x$. Then, one gives two sets of values to the remaining variables $y$ and $z$, so that the two $2$-component vectors thus obtained be linearly independent. I chose $y = 0$, $z=-1$ for the first vector and $y=1$, $z=0$ for the second one. After solving for $x$, one obtains $v = (1,0,-1)$ and $w = (2,1,0)$. Finally, the parametric equation of the plane in vector form is $r = p + sv + tw$. – ivanpenev May 27 '14 at 17:00
  • Here $s$ and $t$ are arbitrary real parameters. The above vector equation is equivalent to the three scalar equations given in the answer above. – ivanpenev May 27 '14 at 17:03
  • If one hasn't seen this procedure from linear algebra before, it might seem unnecessarily complicated. Of course, one could have obtained an equivalent result by solving the equation $x-2y+z=2\alpha$ for $x$ in terms of $y$ and $z$ and plugging the resulting expression for $x$ in the equation of the quadric to obtain an equation in two variables, $y$ and $z$. This is of course just a variant of the same approach as above, only much simpler to explain. – ivanpenev May 27 '14 at 17:43
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Here is an alternative, shorter, but calculus based solution. The system has a unique solution only when the plane and the surface are tangent. When this occurs the two surfaces have the same normal direction. A normal to the plane is $(1,-2,1)$. The normal to $f=xy+xz+yz$ is given by the gradient, $(\frac{f}{dx} ,\frac{f}{dy} ,\frac{f}{dz})=(y+z, x+z, x+y )$ so we have the system of equations $$y+z=\beta$$ $$x+z=-2\beta$$ $$x+y=\beta.$$ Solving we have $$x=-\beta$$ $$y=2\beta$$ $$z=-\beta.$$

If we substitute these into the original equations we get,

$$-6\beta=2\alpha$$ $$-9\beta^2=3\alpha -4$$

which gives the equation for $\beta$, $$9\beta^2-9\beta-4=0$$ which factors $$(3\beta+1)(3\beta-4)=0$$ so $\beta=\frac{4}{3}, -\frac{1}{3}$ and $$\alpha=-3\beta=-4, 1$$

Which (thankfully) is the same obtained by the other methods. We also have the unique solutions in these cases, $$(-\frac{4}{3}, \frac{8}{3}, -\frac{4}{3})$$ and $$(\frac{1}{3}, -\frac{2}{3}, \frac{1}{3})$$

Rene Schipperus
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  • Consider, for example, that the plane tangent to the quadric $z = xy$ at the point $(0,0,0)$, i.e. the plane $z=0$, intersects the quadric in a pair of intersecting lines, namely the $x$- and $y$-axes. – ivanpenev May 27 '14 at 17:10