This is probably simpler that I'm making it (it would have to be), but I'm stuck and don't even quite know how to language this.

Imagine I have two boxes; one has a width of 3/4 of the whole, and the other is the remaining 4th. They are side by side and initially have 0 height.

I fill box-1 (3/4) box with an object that is 100% wide and who's height is determined based on aspect ratio. So box-1 stretches to the height of this object and keeps it same width. Box-2 still has the same width (1/4th of the whole) and zero height. Using aspect ratio I want to create an object that will fill box-2 and match the height of box-1, but I can't figure out how to calculate said aspect ratio.

Specifically - the aspect ratio of the item in box-1 is currently 4.5:1 and using trial and error I got the heights to match with ~1.6:1 in box-2; but I have no idea if that is reliable or if it will work consistently.

Any help?

A Drawing: enter image description here

  • 2
    $\begingroup$ It would help a great deal if you drew a picture and labeled the various lengths with letters like $a,b,c,\ldots$. The question as written is very difficult to understand. $\endgroup$
    – vadim123
    Feb 21, 2015 at 0:55
  • $\begingroup$ I understand completely. Hopefully the drawing helps. $\endgroup$
    – dgo
    Feb 21, 2015 at 1:21
  • 2
    $\begingroup$ According to your drawing, you need to set $y=1$. Also, $4.5$ is three-quarters of 6, so you need to set $x=1.5$. $\endgroup$
    – vadim123
    Feb 21, 2015 at 1:22
  • $\begingroup$ Although your answer was, in retrospect, hilariously obvious, it unfortunately isn't working in practice. I think the reasons why may involve other variables that fall outside of the purview of this math problem. The logic of your answer is what I was looking for though; so thank you. $\endgroup$
    – dgo
    Feb 21, 2015 at 17:55
  • 1
    $\begingroup$ My pleasure. This was still a useful exercise, as it uncovered that the real issue is not the math but the "other variables". $\endgroup$
    – vadim123
    Feb 21, 2015 at 17:57


You must log in to answer this question.

Browse other questions tagged .