Origami regular octagon

Task

Lisa makes an octagon by successively folding a square piece of paper as follows. First, she folds the square in half vertically and horizontally and also along both diagonals leaving these creases:

Next, Lisa makes four more folds, identifying each pair of adjacent lines of symmetry for the square used for the folds in the first step:

Finally Lisa folds the four corners of her shape along the red creases marked below:

Explain why the shape Lisa has made is a regular octagon.

IM Commentary

The goal of this task is to study the geometry of reflections in the context of paper folding. The folding instructions have been provided and students will need to model the folds with reflections and then explain why the construction produces a regular octagon. Once one fold (vertical, horizontal, or diagonal) is performed, the rest of the folds can be viewed as bisections of different angles made by lines containing the center of the square: it will therefore be vital for students to keep track of the different angle measures at each step. In addition, students will need to apply prior knowledge of triangle congruence in order to establish that all sides of the octagon are all congruent. There are many steps and a lot of information to keep track of so students will need ample time and the task is well suited for group work.

The same procedure could be applied again, bisecting each of the 22.5$^\circ$ angles formed by adjacent lines meeting at the center, to construct a regular polygon with 16 sides. So this method can be used to fold a square into a regular polygon with $2^n$ sides for any $n \geq 3$. Since the key to this problem is calculating angle measurements, the teacher may wish to have students calculate interior angle measurements as they add more and more creases.

Many of the standards of mathematical practice are relevant for this task. First, students will ''Use Appropriate Tools Strategically'' (MP5) if they are given paper squares to fold and experiment with before writing down a formal argument. They will ''Look For and Express Regularity in Repeated Reasoning'' (MP8) as the folds, and formal arguments, they make need to repeated multiple times to each side and angle of the octagon. Students will also ''Reason Abstractly and Quantitatively'' (MP2) as they need to represent the paper folding geometrically in order to calculate angles and use triangle congruence to establish that the shape is a regular octagon.

Solution

The key geometric result which will be used repeatedly here is that if $\ell$ and $m$ are lines in the plane meeting at a point $P$, then a reflection of the plane which interchanges $\ell$ and $m$ is the reflection about a line which bisects one of the two pairs of vertical angles made by $\ell$ and $m$. The relationship with origami folds comes from the fact that in this problem the folds are designed to interchange pairs of lines: so we conclude that the crease made by the fold bisects the angles formed by those lines.

We begin by labeling lines for the different creases and analyzing the angles that these lines form where they meet at the center of the square. We label five of the lines only $\ell_1, \ell_2, \ell_3, \ell_4, \ell_5$, focusing on the bottom left corner of our square:

We will show that the angles made by each of the adjacent pairs of lines in the picture are 22.5 degrees. Note that $\ell_1$ and $\ell_2$ meet in four right angles: this can be seen by noting that $\ell_1$ is a horizontal line while $\ell_2$ is vertical or by observing that the vertical fold along $\ell_2$ takes the horizontal crease made by $\ell_1$ to itself. Next we examine the fold of the square along the diagonal. Reflection about $\ell_3$ interchanges $\ell_1$ and $\ell_2$ and so this means that $\ell_3$ must bisect the right angles made by $\ell_1$ and $\ell_2$. We saw that these are 90 degree angles so the (acute) angles made by $\ell_1$ and $\ell_3$ are 45 degree angles and the same is true for the (acute) angles made by $\ell_2$ and $\ell_3$. Next we analyze the angles made by $\ell_2$ and $\ell_5$. Note that reflection about $\ell_5$ interchanges $\ell_2$ and $\ell_3$ because $\ell_5$ is the ''crease'' made when $\ell_2$ is folded over to $\ell_3$. This means that the (acute) angles made by $\ell_2$ and $\ell_5$ are half of 45 degrees or 22.5 degrees and the same holds for the (acute) angles made by $\ell_2$ and $\ell_5$. The same folding argument shows that the (acute) angles made by $\ell_1$ and $\ell_4$ and also the (acute) angles made by $\ell_3$ and $\ell_4$ are also 22.5 degrees.

We now label some vertices at the center and bottom left of our octagon and use what we have shown in the previous paragraph to show that this octagon is regular.

We claim that $\triangle aoc$ is isosceles with $\overline{oa}$ is congruent to $\overline{oc}$. To see why note that $m(\angle aob) = m(\angle cob) = 22.5$ by construction. We also know that $\overleftrightarrow{ob}$ is perpendicular to $\overleftrightarrow{ac}$ so by AA $\triangle aob$ is similar to $\triangle cob$. But these triangles share side $\overline{ob}$ and so they are congruent. Therefore $\overline{oa}$ is congruent to $\overline{oc}$. Next we consider $\triangle doc$. We claim that reflection about $\ell_3$ maps $\triangle doc$ to itself. To see why, note that reflection about $\ell_3$ is a symmetry of the square: since our picture was created by bisecting the four lines of symmetry of the square, reflection about $\ell_3$ must map the eight lines in the picture to themselves and in particular it interchanges $\ell_4$ and $\ell_5$. This means that $|oc| = |od|$ and by SAS $\triangle aoc$ is congruent to $\triangle doc$. This means that $|ac| = |dc|$. Applying the same argument repeatedly, or using the eight symmetries of the square, we can conclude that all 8 sides of the octagon are congruent.

Finally, to check that each interior angle of the octagon measures 135$^\circ$, we know that $m(\angle cao) + m(\angle aco) + m(\angle aoc) = 180$. We have already seen that $m(\angle aoc) = 45$ and $\triangle aoc$ is isosceles. From this we deduce that $2m(\angle aco) = 135$. We have $m(\angle(dco) = m(\angle aco)$ and so

Applying the same argument repeatedly, or using the symmetries of the square, shows that the octagon has equal sides and equal angles so it is a regular octagon.