[Math] Subdivide a rectangle into N equal parts by area radially

geometry

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

here

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 $n$?

Best Answer

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!

So...

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).

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