Given the following set of equations, find: $xy + 2yz + 3zx$.

Question

The positive reals x, y, z satisfy the equations $$x^2 + xy + \frac{y^2}{3}=25$$ $$\frac{y^2}{3}+z^2 = 9$$ $$z^2+ zx + x^2 = 16$$ Find $$xy + 2yz + 3zx$$

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My understanding:

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1. What struck me first were the squares $9, 16, 25$. This is the “Egyptian triangle.” It is a hint to the theorem of Pythagoras, to geometry, and geometrical interpretation.

2. Instead of $x, y, z$ only $xy + 2yz + 3zx$ is required. This may be an area, maybe even the area $6$ of the Egyptian triangle. It is also a hint that I should not try to find $x, y, z$.

3. $\frac{y^2}{3}$ occurs twice, so it may be helpful to set $a^2=\frac{y^2}{3}$.

The equations finally become:

$$x^2+ \sqrt{3}xt + t^2 = 25$$

$$t^2+z^2=9$$

$$z^2+ zx + x^2 = 16$$

Also, one more observation, $zx$ must be a perfect square.

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I am not sure if I'm on the right track, but I don't know how to proceed further. Hints and help would be appreciated, thanks.

Answer 1

Adding the second and third equation and subtracting first equation , we get

$2z^2+xz=yx$

Substitute the value for $y$ in the first equation, we get

$$3x^4+7x^2z^2+3x^3z+4z^4+4z^3x=75x^2$$

which easily gets factored into

$$(x^2+z^2+xz)(3x^2+4z^2)=75x^2$$ $$(16)(3x^2+4z^2)=75x^2$$

One possible way to avoid all the algebraic mess is that on solving the above equation we get, $z^2=\frac{27}{64}x^2$ you could take $a^2=\frac{27}{64}$. This simplifies the calculations a lot since $z=ax$ and $y=2a^2x+ax$

On subsitituting in second equation we get $27=4x^2(a^2+a^3+a^4)$

And we have to find $4x^2(a+a^2+a^3)=\frac{27}{a}$.

And we are done. Not the cleanest approach but it is a way if nothing else seems obvious.

Answer 2

Let $f=xy + 2yz + 3zx.$

Second equation is the circle with parametric equations $$y=3\sqrt3\sin t,\quad z=3\cos t$$ Adding the second and third equation and subtracting first equation , we get $$2z^2+xz-xy=0,$$ $$x=\frac{2z^2}{y-z}=\frac{6\cos^2t }{\sqrt3\sin t-\cos t}.$$ Parametric expresions for $x,y,z$ we substitute in $f^2$. Trigonometric simplifications gives $$f^2=-\frac{1458(\cos2t+1)}{\sqrt3\sin2t+\cos2t-2}.$$ From first equation $$\cos2t=\frac{37-32\sqrt3\sin2t}{59}.$$ On subsitituting this in $f^2$ we get $$f^2=1728$$ Then $f=\sqrt{1728}=24\sqrt3$.

For trigonometric simplifications I use CAS (Maxima, Maple, ...)