Solving 2-degree equations in 3 variables.

Question

We are given 3 equations:

$x^2+\sqrt3 xy + y^2 = 25$

$y^2 + z^2 = 9$

$x^2 +xz+ z^2 = 16$.

$x,y,z$ are positive real numbers.

Then we have to find value of $xy + 2yz + \sqrt3 xz$.

Answer 1

Based on system of equations being sides of a right angled triangle and a point P
inside the triangle such that -

$\angle BPC = 90^0, \angle APC = 150^0, \angle APB = 120^0 $
and, $AP = x, CP = y, BP = z$

We know area of a triangle is $\frac{1}{2} \times$ length of side 1 $\times$ length of side 2 $\times \sin \theta$
where $\theta$ is the angle between side 1 and side 2.

Now sum of area, $\triangle APC + \triangle BPC + \triangle APB = \triangle XYZ$

$\frac{1}{2}(xy\sin150^0 + yz\sin90^0 + xz\sin 120^0) = \frac{1}{2} \times 3 \times 4$

$xy \times \frac{1}{2} + yz + xz \times \frac{\sqrt3}{2} = 12$

$xy + 2yz + \sqrt3 xz = 24$

Answer 2

Note that you have $$\forall x,y,z \in \mathbb{R}^{+}: \left\{\begin{aligned} x^{2}+\sqrt{3} xy + y^2 &=& 25\\ y^{2} + z^{2} &=& 9\\ x^{2} +xz+ z^{2} &=& 16 \end{aligned} \right.$$ if, and only if, $$ \forall x,y,z \in \mathbb{R}^{+}: \left\{\begin{aligned} x^{2}+\sqrt{3} xy + y^2 &=& \color{blue}{5}^{2}\\ y^{2} + z^{2} &=& \color{blue}{3}^{2}\\ x^{2} +xz+ z^{2} &=& \color{blue}{4}^{2} \end{aligned} \right. $$ Now, we can approach this problem as an algebraic geometry problem. Indeed, consider a triangle $\bigtriangleup XYZ$ with side lengths $3,4,5$ and draw a point $P$ inside the triangle such that $XP=x$, $YP=y$, and $ZP=z$. Now, you can considerer he equations in the context of the law of cosines.

Can you continue from here?

Answer 3

Just for the fun !

Using algebra, you can entirely solve the problem. Let $$X=x^2 \qquad Y=y^2 \qquad Z=z^2 \qquad a= xy \qquad b=xz$$So, the equations are now $$X+\sqrt{3} a+Y=25 \tag 1$$ $$Y+Z=9\tag 2$$ $$ X +b+ Z=16\tag 3$$ $$a^2=XY\tag 4$$ $$b^2=XZ\tag 5$$

Using $(1)$,$(2)$ and $(3)$ which are linear equations, we have $$X=-\frac{\sqrt{3} a}{2}-\frac{b}{2}+16\qquad Y=-\frac{\sqrt{3} a}{2}+\frac{b}{2}+9\qquad Z=\frac{\sqrt{3} a}{2}-\frac{b}{2}$$ Replacing, equations $(4)$ and $(5)$ become $$-a^2-50 \sqrt{3} a-b^2+14 b+576=0 \tag 6$$ $$-3 a^2+32 \sqrt{3} a-3 b^2-32 b=0\tag 7$$ Using $(7)$ $$b=\frac{1}{3} \left(\sqrt{-9 a^2+96 \sqrt{3} a+256}-16\right)$$ Plug in $(6)$ to get $$74 \sqrt{-9 a^2+96 \sqrt{3} a+256}-546 \sqrt{3} a+4000=0$$

One squaring step to get

$$a=\frac{96 \left(148+177 \sqrt{3}\right)}{6553}\qquad \implies \qquad b=\frac{384 \left(91 \sqrt{3}-72\right)}{6553}$$ $$x=32 \sqrt{\frac{91-24 \sqrt{3}}{6553}}\qquad y=3 \sqrt{\frac{2185+1152 \sqrt{3}}{6553}} \qquad z=12 \sqrt{\frac{3 \left(91-24 \sqrt{3}\right)}{6553}}$$

Now, you can compute the value of any function $f(x,y,z)$.

Edit

In comments, @WillJagy reported difficulties for the more general case where the equations would be $$X+\alpha\, a+Y=\beta \tag 1$$ $$Y+Z=9\tag 2$$ $$ X +b+ Z=16\tag 3$$ $$a^2=XY\tag 4$$ $$b^2=XZ\tag 5$$

Repeating the same steps, I have $$X=x^2=-\frac{1}{6} \sqrt{-3 a^2 \alpha ^2+6 a \alpha \beta -54 a \alpha -3 \beta ^2+54 \beta +781}-\frac{a \alpha }{2}+\frac{\beta }{2}+\frac{37}{6}$$ $$Y=y^2=+ \frac{1}{6} \sqrt{-3 a^2 \alpha ^2+6 a \alpha \beta -54 a \alpha -3 \beta ^2+54 \beta +781}-\frac{a \alpha }{2}+\frac{\beta }{2}-\frac{37}{6}$$ $$Z=z^2=-\frac{1}{6} \sqrt{-3 a^2 \alpha ^2+6 a \alpha \beta -54 a \alpha -3 \beta ^2+54 \beta +781}+\frac{a \alpha }{2}-\frac{\beta }{2}+\frac{91}{6}$$ The problem now is that the last equation in $a$ is a quartic and no more a quadratic.

To reduce it to a cubic would require $$16 \left(\alpha ^2-3\right)^2 \implies \alpha=\pm \sqrt 3$$ which in fact reduces the last equation to a quadratic the solution of which being $$a=\frac{(\beta +7) \left(\sqrt{3} \left(4 \beta ^2-55 \beta -63\right)+\sqrt{(49-\beta ) (\beta -1)}\right)}{2 \left(12 \beta ^2-54 \beta +403\right)}$$ So, we have the analytical solution for any $1\leq \beta \leq 49$, the variables $(x,y,z)$ being non-negative in this range.

Answer 4

Here is an accurate picture. Draw in some extra lines......

Answer 5

Given a triangle with sides $l_1,l_2,l_3$ we have

$$ \cases{ l_1^2=l_2^2+l_3^2-2l_2l_3\cos\theta_1\\ l_2^2=l_1^2+l_3^2-2l_1l_3\cos\theta_2\\ l_2^2=l_1^2+l_2^2-2l_1 l_2\cos\theta_3 } $$

then making $l_1=x,l_2=y,l_3=z$

$$ \cases{ 2\cos\theta_1=0\\ 2\cos\theta_2=-1\\ 2\cos\theta_3=-\sqrt{3} } $$

It is a rectangle. Etc.