If there is a paper that is split into $2$ sections (not to be assumed equal) with one section being laterally divided into $4$ equal subsections and the other $5$ (lateral means horizontal, or perpendicular to the original split mark first mentioned), and the section with the $4$ lateral sections is longitudinally- (vertically and parallel to the original split) divided into $2$ equal sections and the one with $5$ has $3$, my question is what distance do I put the original split from the left and right sides so that the longitudinal sections can be distributed equally between this distance and be timesed by the equal individual heights formed by the cross sections of the lateral divisions and be equal to that of the other side.

I know this most likely makes no sense to you at all but here is a diagram; oh I can't post it because it says I need 10 reputation to post it and given the way the word "reputation" is used in its context I don't understand it. So just try and figure it out without one.

  • $\begingroup$ i want to get the area of the sections formed equal $\endgroup$ – somebody Jul 10 '13 at 22:54
  • 2
    $\begingroup$ Welcome to MSE! If you post your diagram to a file sharing site, someone can add it. Also, please clean up your question some more, I fixed about a dozen errors. Regards $\endgroup$ – Amzoti Jul 10 '13 at 22:54

Let the area of the rectangle be $A$. The original split divides it into $a$ and $b$. If I understand correctly, $a$ gets divided into $4 \times 2=8$ sections and $b$ gets divided into $5 \times 3=15$ sections. Then we want all the small sections to have area $\frac A{23}$. So make the cut with $\frac 8{23}$ on one side, $\frac {15}{23}$ on the other. Below is a rough sketch. The first cut is the vertical heavy line. Is that what you were thinking?

enter image description here


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