What are some good questions for this trick, if $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\dots=\alpha$ then $\alpha=\frac{a+c+e+...}{b+d+f+...}$?

Question

I need some good algebra questions that are applications of this trick, often in a non obvious and elegant way: $$\text{If } \frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\dots=\alpha \text{ then } \alpha=\frac{a+c+e+...}{b+d+f+...}$$

Answer 1

1. If $\displaystyle\frac a{b+c}=\frac b{c+a}=\frac c{a+b};$ prove that each ratio $\displaystyle=\frac12$ if $\displaystyle a+b+c\ne0$

2. If $\displaystyle\frac{a-b}{x^2}=\frac{b-c}{y^2}=\frac{c-a}{z^2}$ prove that $\displaystyle a=b=c$

3. If $\displaystyle\frac{a-b}{a^2+ab+b^2}=\frac{b-c}{b^2+bc+c^2}=\frac{c-a}{c^2+ca+a^2}$ prove that $\displaystyle a=b=c$ (for a special condition)

4. If $\displaystyle\frac{a+b}{a^2+ab+b^2}=\frac{b+c}{b^2+bc+c^2}$ and $a\ne b\ne c$ prove that each ratio $=\displaystyle\frac{c+a}{c^2+ca+a^2}$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$\begin{align} \color{#c00000}{\large\alpha}&={a \over b}=\color{#c00000}{\large{c \over d}}\ \imp\ {a \over c}={b \over d}\ \imp\ {a \over c} + 1 ={b \over d} + 1\ \imp\ {a + c \over c}={b + d \over d}\ \imp\ {a + c \over b + d} = \color{#c00000}{\large{c \over d}} \end{align}

$$ \mbox{Then,}\quad {a + c \over b + d} = \alpha $$

Now, we have $$ {a + c \over b + d}={e \over f}=\cdots=\alpha $$

We can repeat the above procedure as needed. The general result will follow.