Asymmetric Propeller of Squares II

What Is This About?

Problem

Asymmetric  Propeller of Squares II, problem

Solution

We'll use the lower case letter corresponding to the names of of the points as associated complex numbers. We also assume, $a=1,$ $b=i,$ $c=-1,$ $d=-i.$ Then, for example, $(g-a)=i(e-a)$ so that $g=a+i(e-a).$ Similarly,

$\displaystyle\begin{align} j&=b+i(h-b),\\ m&=c+i(k-c),\\ p&=d+i(n-d). \end{align}$

Further,

$\displaystyle \begin{align} Q&=\frac{E+P}{2}=\frac{e+p}{2}=\frac{e+d+i(n-d)}{2},\\ R&=\frac{N+M}{2}=\frac{n+m}{2}=\frac{n+c+i(k-c)}{2},\\ S&=\frac{J+K}{2}=\frac{j+k}{2}=\frac{k+b+i(h-b)}{2},\\ T&=\frac{H+G}{2}=\frac{h+g}{2}=\frac{h+a+i(e-a)}{2}. \end{align}$

Further,

$\displaystyle \begin{align} 2(q-s)&=[e+d+i(n-d)]-[k+b+i(h-b)]\\ &=[e+d-k-b]+i[n-d-h+b]\\ &=[e-k]+i[n-h]+2(i-1),\\ 2(r-t)&=[n+c+i(k-c)]-[h+a+i(e-a)]\\ &=[n+c-h-a]+i[k-c-e+a]\\ &=[n-h]+i[k-e]+2(i-1)\\ &=-i\{[e-k]+i[n-h]\}+2(i-1), \end{align}$

because $(d-b)(1-i)=2(i-1)=(c-a)(i-1).$ It follows that $r-t=i(s-q).$

Acknowledgment

This problem grew out of another one and was proposed by Alexandr Skutin.

 

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