I have a rectangle of width $w$ and length $l$, measured as shown:


I want to divide the rectangle into $n$ parts, where $n > 1$. Each part is created by the space between two rays originating from the center of the rectangle, and all segments are to be of equal area. How would this be possible for any rectangle and any value $n$?

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    $\begingroup$ Welcome to Math.SE. What is your approach while trying to solve this problem? $\endgroup$ – Ertxiem - reinstate Monica Apr 8 at 20:27
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    $\begingroup$ Hint: Think about the perimeter. Hint: think about the formula for area of triangle. $\endgroup$ – fleablood Apr 8 at 20:49

A naive mistake would be to think the you divide the interior angle into equal parts, $\frac {2\pi}n$.

A simple drawing will show you this fails. Assuming $n$ is significantly large this will divide one the the sides of the rectangles into segments. This triangles have equal height and equal vertex angle will have varying base angles and different base lengths.

(Example: Consider $\triangle ABC$. If $AB= \frac {\sqrt 3}2; m\angle ABC = 90^\circ; m\angle BAC = 30^\circ$ then $BC=\frac 12$. As $\triangle ABC$ is $30-60-90$. Now consider $\triangle ACD$ where $B,C,D$ are colinear and $m\angle CAD = 30^\circ$. Then $\triangle DBA$ is $30-60-90$ so $BD = \frac 32$. And $CD= \frac 32-\frac 12 = 1 \ne BC$. And if we tried to put in a third $30^\circ$ angle we'd get $90^\circ$ which will never intersect the line $\overline {BC}$ at all!)

But note if $n$ is significantly large we must divide a side into segments and we'd get triangles all of equal height. So the segments must be equal.

So we don't solve this by thinking about angles; we solve this by thinking about segments and perimeters!


Think about this. Divide each side into $n$ equal length segments. This will created $n$ triangles for each side and as the bases and heights of this $n$ triangles are equal the areas are equal. You have $4n$ triangles for wish the left and right are all the same area and the top and bottoms are the same size.

The triangles along the left and right sides will each be $\frac 12 base*height = \frac 12 (\frac ln)*(\frac 12 w)$.

The triangles along the top and bottom will be $\frac 12 base*height = \frac 12(\frac wn)*(\frac 12 l)$.

So all $4n$ of the triangles are equal area.

So glue $4$ of the triangles together to get wedges of equal size.

Ignore the angles totally. (Unless you want to express them as arc trig functions).


For a square it is particularly simple. You just divide the perimeter into $n$ parts of equal length, so if the square has side $s$ each part has length $\frac {4s}n$. The point is that in the diagram below, the altitudes of triangles $BEF$ and $EGH$ are equal, so if the bases are equal so are the areas. enter image description here
This suggest the answer to the rectangle case. A unit of perimeter on the narrow end of the rectangle makes a triangle of area $\frac l4$ while a unit of perimeter on the long side of the rectangle makes a triangle of area $\frac w4$. The area of each triangle should be $\frac {lw}n$ so if the base is along the long side it should have length $\frac l{4n}$ while if the base is along the short side it should have lentgth $\frac w{4n}$. If the piece of perimeter turns the corner, the ratio of the length on the long side to $\frac l{4n}$ and the ratio of the length on the narrow side to $\frac w{4n}$ should add to $1$.

  • $\begingroup$ Thanks for the response! My only confusion would be about the last part, for segments that cross over one or more corners. What are the ratios you described comparing to? For example, if I had $n = 3$ then one of the segments would cross over two corners, but I can't figure out how these ratios would be defined so when added together it would equal 1. $\endgroup$ – JohnSmith Apr 8 at 21:02
  • $\begingroup$ In fact, if $n=3$ and you start from a corner at least two of the segments will cross corners. For a square it will be all three. Probably the easiest algorithm is to focus on the fact that each region should have area $\frac {lw}n$. If you start one segment along a long edge, its length should be $\frac l{4n}$ as I said. If you have that much of the long edge available, mark it and you are done with that piece. If not, compute the area the remaining part of the edge gives and carry the rest onto the other side. $\endgroup$ – Ross Millikan Apr 9 at 2:37
  • $\begingroup$ Now with the altitude $\frac l2$ compute how much of the short edge is needed. Keep going around. $\endgroup$ – Ross Millikan Apr 9 at 2:37

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