By only drawing two lines on the rectangle, separation the rectangle right into 4 components where the areas are in the ratio 1:2:3:4.

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How would one go roughly doing this? i tried because that a solid hour or two; yet to no avail.

Disclaimer: This question is NOT developed by me.


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First, division the rectangle same on its long side, as argued in the comments. In the left rectangle, mark the point that is $\frac45$ that the means up and centred horizontally; perform the exact same for the best rectangle, but mark the suggest only $\frac35$ of the way up. Then attract the heat connecting the two marked points. Due to the fact that this line crosses the left and right edges of the totality rectangle, the four little triangles shown above (bounded between the purple and darker grey lines) have equal area, for this reason the left rectangle is divided 1:4 and also the ideal rectangle divided 2:3. Overall, the rectangle has actually been separated into four parts with ratio 1:2:3:4.

This equipment does no require relocating the pieces as in Donald Splutterwit"s answer.


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answer Feb 9 "18 at 11:11
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Parcly TaxelParcly Taxel
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An different solution very comparable to Parcly"s (image lovingly produced in Excel):

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answer Feb 9 "18 at 16:30
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aPaulTaPaulT
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You require to move the two pieces right into the ideal positions prior to you perform the 2nd chop.

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answered Feb 9 "18 at 10:59
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Donald SplutterwitDonald Splutterwit
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There is a one-parameter infinite family of solutions. Below is an algorithm to uncover them all. Together a spoiler, I"ll suggest out increase front the this algorithm functions equally well for any convex subset that $\rewildtv.combbR^2$--not just the certain 5 through 6 rectangle in question.

To start, let"s brand the areas in the dissected rectangle by number 1,2,3,4, in correspondence through their share of the area. If we look at the bespeak of the four regions together we move clockwise around the crossing point of the two lines, starting at region 1, there room 6 possibilities: (1,2,3,4), (1,2,4,3), (1,3,2,4), (1,3,4,2), (1,4,2,3), (1,4,3,2). The figure below shows one example, whereby n,p,q are any type of permutation the 2,3,4.

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Algorithm step 1: pick an setup of regions from the 6 possibilities.

Let"s currently examine what happens if we pick a certain orientation because that the heat $l$ (as mentioned by e.g. A unit vector, or one angle v respect to some recommendation line) the runs follow me the anticlockwise side of an ar 1, as presented in the number above. Let $n$ be the variety of the region that is the clockwise ar of region 1. Then we want $l$ to divide the rectangle right into two regions, one with portion $\frac1+n10$ that the full area. If us imagine scan $l$ across the rectangle, we see that all the area moves from one next of the line to the various other in a monotonic way. By the intermediate worth theorem, there is a unique line $l$ the does the job.

Algorithm action 2: choose an orientation for line $l$. By the over remarks, this states the heat entirely.

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Now I insurance claim that the place of the second line $l"$ is uniquely specified by the options we have currently made. This adheres to from 3 observations:

For any kind of intersection suggest between the two lines, there is a distinctive orientation of the second line $l"$ that gets the proportion of locations 1 and also $n$ right. This again complies with by the intermediate worth theorem. The orientation found in allude (1) varies monotonically with the place of the crossing suggest along heat $l$. If the crossing point is at an end of line $l$, then choosing $l"$ as above would make one of the areas $m$, $q$ ~ above the various other side have actually area zero. Therefore again by the intermediate value theorem, there is a distinct position the the crossing allude along $l$ that gets the ratio of the remaining two areas right.

Summary: The division of the rectangle deserve to be mentioned uniquely by 2 choices: The plan (1,n,p,q), and also the orientation the the line $l$.

Commentary:

The examples given through Parcly Taxel and also aPaulT exchange mail to one orientation parallel to the quick side that the rectangle and also an setup (1,2,3,4) and (1,3,2,4), respectively. In enhancement to working for any convex subset of $\rewildtv.combbR^2$, this algorithm likewise works perfect well for ratios other than 1:2:3:4.It is possible to filter this algorithm into an explicit formula for the currently $l$ and $l"$ together a role of the setup of regions and also orientation the $l$. The reason I"ve no done this is that the formula has an annoying piecewise definition. The source of the trouble is that as you sweep a heat with solved orientation end a rectangle, the area on either side changes in a non-smooth way each time you hit a corner. To placed this algorithm to an excellent use, listed below is a brand-new solution come the problem. I"ve restricted myself to the special situation where areas 1 and 4 room adjacent, because in this situation the line dividing them from areas 2 and 3 must split the rectangle right into two regions of same area. The unique way to perform this is because that the corresponding line to pass through the rectangle center, and also this simplifies the algorithm.

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The equations for the lines in this case are $y = \frac56 x$ because that $l"$ (a diagonal line of the rectangle) and also $x = \frac6(\sqrt2+1)5 - \frac12(\sqrt2-1)25 y$ because that $l$.