Algebra Challenge: Proving $p,\,q,\,r,\,s,\,t$ Satisfy Equation

[anemone]

Let $p,\,q,\,r,\,s,\,t \in \mathbb {R_+}$ satisfying

$p^2+pq+q^2=s^2\\ q^2+qr+r^2=t^2\\r^2+rp+p^2=s^2-st+t^2$

Prove that

$\dfrac{s^2-st+t^2}{s^2t^2}=\dfrac{r^2}{q^2t^2}+\dfrac{p^2}{q^2s^2}-\dfrac{pr}{q^2ts}$

 

[jvicens]

Dear forum members,

I would like to provide a proof for the given statement.

First, we can rewrite the given equations as:
$p^2+pq+q^2=s^2\\ q^2+qr+r^2=t^2\\(p^2+q^2+pq)-(st)=t^2-st+s^2$

Next, we can rearrange the third equation to get:
$(p^2+q^2+pq)-(st)-(t^2-st+s^2)=0\\(p^2+q^2+pq)-(st)-\dfrac{1}{2}((p+q)^2+(s-t)^2)=0$

Using this, we can rewrite the original equations as:
$p^2+pq+q^2=s^2\\ q^2+qr+r^2=t^2\\(p+q)^2+(s-t)^2=2(st)$

Now, let's multiply the first equation by $s^2$, the second equation by $t^2$, and the third equation by $s^2t^2$. We get:
$s^2(p^2+pq+q^2)=s^4\\t^2(q^2+qr+r^2)=t^4\\(p+q)^2s^2t^2+(s-t)^2s^2t^2=2(s^2t^2)(st)$

Combining these equations, we get:
$(p^2+q^2+pq)s^2+(q^2+qr+r^2)t^2+(p+q)^2s^2t^2+(s-t)^2s^2t^2=2(s^2t^2)(st)+s^4+t^4$

Using the original equations, we can simplify this to:
$(p^2+q^2+pq)s^2+(q^2+qr+r^2)t^2+(p+q)^2s^2t^2+(s-t)^2s^2t^2=2(s^2t^2)(st)+(p+q)^2s^2+(s-t)^2t^2$

Next, we can rewrite the left side of this equation as:
$(p^2+q^2+pq)s^2+(q^2+qr+r^2)t