How to find the solution given below through this system?

Question

Knowing that $a^2 + b^2 = c^2 + d^2 = e^2 + f^2$ the following equalities are given

$$ac + bd = ec + df = ae + bf$$

A solution to this equality is given if $a=-c-e$ and $b=-d-f$.

From these equalities can I find these solutions? I tried a lot and I could not. Did someone solve something like that and used some trick? I'm grateful for any help.

Answer 1

In general, this will not follow from your equations. For example, another solution is given by $c=0$, $f=b$ and $$ e^2 + b^2 = d^2, a^2 + b^2 = d^2. $$ This need not satisfy $a=-c-e=-e$, but could be $a=e\neq -e$.

Still another solution is $$ a^2 + b^2 = c^2 + d^2,e= a,f = b. $$

Answer 2

Hint: Substitute $$ac+bd=t$$ $$ec+df=t$$ $$ae+bf=t$$ and solve this system.

Answer 3

Viewing $(a,b)$, $(c,d)$, $(e,f)$ as vectors in $\Bbb R^2$, the conditions can be inerpreted as:

• The three vectors have equal length
• The angle between two of them is always the same

So geometrically, it is clear that we either have three identical vectore or three vectors (of same length) in stepd of $120^\circ$, that is

• $a=c=e$ and $b=d=f$, or
• $a,b$ arbitrary, $\begin{pmatrix}c\\d\end{pmatrix}=\begin{pmatrix}\cos 120^\circ&\sin 120^\circ\\-\sin 120^\circ&\cos 120^\circ\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}$, $\begin{pmatrix}e\\f\end{pmatrix}=\begin{pmatrix}\cos 120^\circ&-\sin 120^\circ\\\sin 120^\circ&\cos 120^\circ\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}$
• as before with $c,d\leftarrow e,f$ swapped