We have $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$.

Question

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?

Answer 1

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 $3-1=2$.

Answer 2

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 $3-1=2$.

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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.

Answer 3

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 $3+(-1)=2$.

Answer 4

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.

Answer 5

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 $$r=3.$$