Solve the following relation.

Question

Find the values for $a$, $b$, $c$ and $d$ in the following equation $$\frac{2a}{2m} = \frac{2b}{a+m} = \frac{2c}{a+b} = \frac{2d}{c+m} = \frac{2M}{c+d}.$$

(Note: $m$ and $M$ are different. In fact, $m\leq M$.)

Answer 1

Organize and number the equations as follows

\begin{align} \frac{2d}{c+m} &= \frac{2M}{c+d} \tag1 \\[1.2ex] \frac{2a}{2m} &= \frac{2d}{c+m} \tag2 \\[1.2ex] \frac{2b}{a+m} &= \frac{2a}{2m} \tag3 \\[1.2ex] \frac{2c}{a+b} &= \frac{2b}{a+m} \tag4 \\[1.2ex] \end{align}

Then, $(1)$ leads to

$$ c=\frac{d^2-Mm}{M-d}. \tag5 $$

Using $(2)$ and $(5)$ we can solve for

$$ a=\frac{2(M-d)m}{d-m}. \tag6$$

Using $(3)$, $(5)$, and $(6)$ we obtain

$$b=\frac{(M-d)(2M-d-m)m}{(d-m)^2} \tag7 $$

Finally, using $(4)$, $(5)$, $(6)$, and $(7)$ we get that $d$ must solve

$$ \frac{1}{m}\frac{(d-m)^3}{(M-d)^3}\frac{Mm-d^2}{3m-2M-d}=1.$$

If $M=m$, then

$$ a = d=-2m, \;\;\;\; \text{and} \;\;\;\; b = c=m .$$