We have $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$. 
If $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$ , find the sum of all the possible values of $r$ .

What I Tried :- First of all, when $a = b = c = d$ , the $4$ equations hold and we get $r = 3$ as $1$ solution .
For other solutions I simplified to get $6$ expressions as :-
$b^2 + bc + bd = a^2 + ac + ad$
$bc + c^2 + cd = a^2 + ab + ad$
$bd + cd + d^2 = a^2 + ab + ac$
$ac + c^2 + cd = ab + b^2 + bd$
$ad + cd + d^2 = ab + b^2 + bc$
$ad + bd + d^2 = ac + bc + c^2$
Now I have no idea how to start finding solutions for $r$ from here . Can anyone help?
 A: First of all $$\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$$ gives you $$b + c + d=ar\\a + c + d=br\\a + b + d=cr\\a + b + c=dr$$ summing all these we get $$3(a+b+c+d)=r(a+b+c+d)$$ Now if $(a+b+c+d)\neq0$ then we must have $r=3$.
Now if $(a+b+c+d)=0$, then $r=\frac{a+b+c}{d}\implies r=\frac{a+b+c+d-d}{d}=\frac{-d}{d}=-1$ (For clarity: since we are writing $a,b,c,d$ in the denominators, we must need $a,b,c,d$ to be non-zero.). Therefore the sum of all possible values of $r$ is $3-1=2$.
A: First of all, the equation only makes sense when $a,b,c,d\neq 0$ so you can use this: If
$$\frac{x}{y}=\frac{a}{b}$$
Then
$$\frac{x}{y}=\frac{x+a}{y+b}$$
So your problem solves proving this property, and the only value will be
$$\frac{3(a+b+c+d)}{a+b+c+d}=3$$
But if you get the case $a+b+c+d=0$, then WLOG, $a+b+c=-d$, then you achieve that $a+b+c=-dr$, implying $r=-1$.
Then the only solutions are $r\in \{-1,3\}$, so the solution to the problem would be $3-1=2$.

So to achieve the identity, notice that
$$x(y+b)=y(x+a)$$
$$xy+ab=yx+ya$$
And substracting $xy$ to both sides you get:
$$xb=ya$$
$$\frac{x}{y}=\frac{a}{b}$$
The other side is the same way.
A: Here is the second thing I thought of (@iam_agf has the other). Add $1$ to all the terms and set $a+b+c+d=E$ and you get$$\frac Ea=\frac Eb=\frac Ec=\frac Ed=r+1$$ from which either $E=0$ or $a=b=c=d$
In the first case $r=-1$ and in the second $r=3$
Just putting $a+b+c+d=E$, which should be an early thing to try, also gets to the same place.
A: We obtain: $$\frac{b+c+d}{a}+1=r+1$$ or
$$\frac{a+b+c+d}{a}=r+1.$$
Now, $a+b+c+d=0$ gives $r=-1$, but for $a+b+c+d\neq0$ we obtain:
$$\frac{a}{a+b+c+d}=\frac{1}{r+1}$$ and
$$\frac{4}{r+1}=\sum_{cyc}\frac{a}{a+b+c+d}=1,$$ which gives $$r=3.$$
A: If you happen to know eigenvalues/eigenvectors, your issue can be written :
$$\underbrace{\begin{pmatrix}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{pmatrix}}_A\begin{pmatrix}a\\b\\c\\d\\\end{pmatrix}=r\begin{pmatrix}a\\b\\c\\d\\\end{pmatrix}$$
meaning that $r$ must be an eigenvalue of matrix $A$.
The eigenvalues of $A$ are $3$ and $-1$ ($-1$ being a
triple eigenvalue, but the text allows to take it only once) associated to resp. eigenvectors:
$$\begin{pmatrix}a\\b\\c\\d\\\end{pmatrix}=\begin{pmatrix}1\\1\\1\\1\\\end{pmatrix},\begin{pmatrix}1\\1\\1\\-3\\\end{pmatrix}$$
fulfilling the condition that neither $a$, nor $b,c,d$ are zero (being in the denominator of the initial equation)
Therefore the answer is $3+(-1)=2$.
