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How would you divide an octagon into 5 equal parts?

This is a question that we are working on in 2nd grade. Do you have an answer for us?

Thanks, Mrs. Parsons Class West View Elementary Burlington, Wa

Parsons 2nd Grade
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    I assume you mean regular octagon. How are those 5 parts equal? Same area? Same size and shape? Can they be mirrored? – flawr May 03 '16 at 19:46
  • Divide the octagon into eight triangles, then each triangle into 25 sub-triangles (divide each side into five parts, so you get $1+3+5+7+9$ sub-triangles in each triangle). There are then 200 equal subtriangles. Since $200 = 40\cdot5$, divide the subtriangles into groups of 40 each. – Nominal Animal May 04 '16 at 00:31
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    This is surprising. And your students won't believe it maybe. But if you measure the perimeter (let's say it's 40 inches-- 5 inches to a side). Divide into 5. (say 8 inches). Start anywhere and measure that distance 5 times. Each wedge will be equal. – fleablood May 29 '16 at 00:02

2 Answers2

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If your intention is to divide a regular octagon into five regions of equal area, see the figure below:

OctagonDivFive At first divide side $AB$ (length 5 units) into five equal segments ($BO, OP, PQ, QR$ and $RA$).

Then mark points $T, U, V$ and $W$ on the sides of the octagon, so that $AT=BQ$, $GU=BO$, $FV=BR$ and $DW=BP$.

At last draw the segments $ST, SU, SV, SW$ and $SB$, where $S$ is the center of the ocatagon.

RicardoCruz
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Let $ABCDEFGH$ be the octagon. Go $2/3$ of the way from $A$ to $B$ on side $AB$ and construct an auxiliary line segment from that point to the midpoint of $DE$. Then construct successive $45°$ rotations of this segment around the center of the octagon.

You create a smaller octagon in the center, whose area measures $1/5$ that of $ABCDEFGH$. That is one of your parts. To get the other four parts use symmetry to divide the area between the octagons into four pieces.

Oscar Lanzi
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