MHB What is the value of $xy+yz+zx$ in a system of equations with real numbers?

[anemone]

Here is this week's POTW:

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If $\{x,y,z\}⊂\Bbb{R}^+$ and

$x^2+xy+y^2=3\\y^2+yz+z^2=1\\x^2+xz+z^2=4$,

find the value of $xy+yz+zx$.

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[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. castor28
2. Opalg
3. kaliprasad

Solution from Opalg:

Spoiler

Given $$x^2+xy+y^2=3, \qquad(1)\\y^2+yz+z^2=1, \qquad(2)\\z^2+zx+x^2=4, \qquad(3)$$ it follows that $$x^3-y^3 = (x-y)(x^2+xy+y^2) = 3(x-y),\\y^3 - z^3 = (y-z)(y^2+yz+z^2) = y-z, \\z^3 - x^3 = (z-x)(z^2+zx+x^2) = 4(z-x).$$ Add those three equations, to get $0 = 3(x-y) + y-z + 4(z-x) = -x - 2y + 3z$, so that $x = 3z-2y$.

Substitute that into (1): $(3z-2y)^2 + y(3z-2y) + y^2 = 3$, from which $3y^2 -9yz + 9z^2 = 3$. Subtract $3\times(2)$ from that, getting $-12yz + 6z^2 = 0$. Therefore $z(z-2y) = 0$. So either $z=0$ or $z=2y$.

If $z=0$ then (2) says that $y = \pm1$ and (3) says that $x = \pm2$. Then (1) is also satisfied provided that $x$ and $y$ have opposite signs. But since the question allows only positive numbers, that solution must be rejected.

If $z = 2y$ then $x = 3z-2y = 4y$. From (2), $7y^2 = 1$. Therefore $(x,y,z) = \frac1{\sqrt7}(4,1,2)$, and finally $xy+yz+zx = \frac{14}7 = 2.$

Alternate solution from kaliprasad:

Spoiler

We are given

$x^2+xy+y^2 = 3\cdots(1)$

$y^2+yz+z^2 = 1\cdots(2)$

$x^2+xy+y^2 = 4\cdots(3)$

Let us consider a $\triangle\, ABO$ with $AB =\sqrt{3}$, $AO = x$ and $BO = y$

Using law of cos we have

$AB^2 = AO^2 + BO^2 - 2 . AO . BO . \cos\, \angle AOB$

or $3 = x^2+y^2 -2 xy \cos\, \angle AOB$

Comparing with $1^{st}$ equation we get

$\cos\, \angle AOB=-\frac{1}{2}$

or $\angle AOB=120^\circ$

Similarly if we take $\triangle\, BOC$ with $BC =1$ $BO = y$ and $CO =z$ and comparing with (2) we get

$\angle BOC=120^\circ$

Similarly if we take $\triangle\,COD$ with $CD =2$ $CO = z$ and $DO =x$ and comparing with (3) we get

$\angle COD=120^\circ$

Now as

$\angle AOB + \angle BOC +\angle COD=360^\circ$

so OA and OD are overlapping and as OA = OD we have A = D

as $AB^2 + BC^2 = CA^2$ we have ABC is a right angled triangle with right angle at B

So area of $\triangle ABC = \frac{AB * BC}{2} = \frac{\sqrt{3}}{2}\cdots(4)$

Another way to calculate area as sum of 3 triangles and each using law of sine

Area of $\triangle AOB = \frac{1}{2} AO * OB\sin\, \angle AOB = \frac{\sqrt{3}}{4}xy$

Area of $\triangle BOC = frac{1}{2} BO* OC\sin\, \angle BOC = \frac{\sqrt{3}}{4}yz$

Area of $\triangle COA = frac{1}{2} CO* OA\sin\, \angle COA = \frac{\sqrt{3}}{4}zx$

Adding the 3 we get

Area of $\triangle ABC = \frac{\sqrt{3}}{4}(xy+yz+zx)\cdots(5)$

From (4) and (5) we get

$xy+yz+zx = 2$