Given six numbers $x,y,z,a,b,c$ which satisfy the following relations. Express $x+y+z$ in terms of $a,b,c$

Question

Given six numbers $x,y,z,a,b,c$ which satisfy the following relations
$y^2+yz+z^2=a^2$
$z^2+zx+x^2=b^2$
$x^2+xy+y^2=c^2$
Express $x+y+z$ in terms of $a,b,c$

My attempt:
$\dfrac{(y^3-z^3)}{(y-z)}=a^2,\dfrac{(z^3-x^3)}{(z-x)}=b^2,\dfrac{(x^3-y^3)}{(x-y)}=c^2$
But this step didn't seem to help.
So I tried adding and expressing $k(a+b+c)^2$ but some other unmanagable elements came along. I could not think of any other way to approach the problem. Please help.

Answer 1

Let $A=a^2,B=b^2,C=c^2$. We have

$$ \begin{array}{lcl} B-A &=& (x-y)(x+y+z) \\ C-A &=& (x-z)(x+y+z) \\ C-B &=& (y-z)(x+y+z) \\ \end{array}\tag{1} $$

Then, if we put $D=(B-A)^2+(C-A)^2+(C-B)^2$, we deduce

$$ \begin{array} D &=& (x+y+z)^2\Bigg((x-y)^2+(x-z)^2+(y-z)^2\Bigg) \\ &=& 2(x+y+z)^2\Bigg(x^2+y^2+z^2-xy-xz-yz\Bigg) \\ &=& 2(x+y+z)^2\Bigg( \big(2(x^2+y^2+z^2)+xy+xz+yz\big)-\big(x^2+y^2+z^2+2(xy+xz+yz)\big) \Bigg) \\ &=& 2(x+y+z)^2\Bigg( \big(A+B+C\big)-\big((x+y+z)^2\big) \Bigg) \\ \end{array}\tag{2} $$

So the sum $w=x+y+z$ satisfies

$$ 2w^4-2(A+B+C)w^2+(B-A)^2+(C-A)^2+(C-B)^2=0 \tag{3} $$

If we simplify by 2 we obtain

$$ w^4-(a^2+b^2+c^2)w^2+(a^4+b^4+c^4-((ab)^2+(ac)^2+(bc)^2)=0 \tag{4} $$

Answer 2

I'm using $p$, $q$, $r$ for $x$, $y$, $z$, because GeoGebra doesn't like the latter for labels. Also, I'm assuming $p$, $q$, $r$ non-negative. (I believe that minor tweaks to the argument would allow for negative values.)

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Arranging edges of length $p$, $q$, $r$ on symmetric rays about point $O$ gives $\triangle ABC$.

The problem's relations for $a^2$, $b^2$, $c^2$ are in fact the Law of Cosines applied to the sub-triangles (whose angles at $O$ have measure $120^\circ$). For instance,

$$a^2 = q^2 + r^2 - 2 q r \cos 120^\circ = q^2 + r^2 - 2 q r \cdot \left(-\frac{1}{2}\right) = q^2 + r^2 + q r$$

A little more trig gives $$\begin{align} |\triangle ABC| &= |\triangle OBC| + |\triangle OCA| + |\triangle OAB| \\[6pt] &= \frac{1}{2}qr\sin 120^\circ + \frac{1}{2}rp\sin 120^\circ + \frac{1}{2}pq\sin 120^\circ \\[6pt] &= \frac{1}{2}\frac{\sqrt{3}}{2}\left(\;pq+qr+rp\;\right) \\[6pt] &= \frac{\sqrt{3}}{4} \left(\;pq+qr+rp\;\right) \end{align}$$ so that $$pq+qr+rp = \frac{4}{\sqrt{3}} |\triangle ABC| = \frac{4\sqrt{3}}{3}|\triangle ABC|$$

Summing-up the three original relations gives $$\begin{align} a^2 + b^2 + c^2 &= 2 p^2 + 2 q^2 + 2 r^2 + p q + q r + r p \\ &= 2 ( p + q + r )^2 - 3\left( pq + qr + rp \right) \\ &= 2 ( p + q + r )^2 - 4\sqrt{3} |\triangle ABC| \end{align}$$ so that $$(p+q+r)^2 = \frac{1}{2}\left(\; a^2 + b^2 + c^2 + 4 \sqrt{3} |\triangle ABC|\;\right)$$ Now, we need only recall Heron's formula for the area of a triangle ...

$$|\triangle ABC|^2 = \frac{1}{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$$

and then we can write

$$(p+q+r)^2 = \frac{1}{2}\left(\; a^2 + b^2 + c^2 + \sqrt{3(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}\;\right)$$