David K Butler
@davidkbutler
Maths Learning Support lecturer at Adelaide Uni (my views here). Grad Dip Ed & PhD Finite Geom. Love maths and helping people learn. he/him
A combinatorics problem inspired by this little project my wife just finished: How many ways are there to arrange the drawers so that drawers with the same design aren’t next to each other vertically or horizontally?
Found this in my scrap paper during a meeting with my manager and had to suppress a guffaw.
These two triangles are *congruent*! Both have an edge of length c and a purple edge. They also both have an angle of 120°, though not between the two matching edges. But the angles that touch vertices are each 60° minus the same angle.
Ooh I’ve seen another pair of similar triangles. I suppose whenever you have two meeting chords of a circle you’d be guaranteed to make two pairs of similar triangles like this, hey?
I’m going to relabel everything and I reckon similar triangles/ lines cut by parallel lines will relate things in pairs.
Those first two similar triangles I noticed have two matching edges that are edges of the yellow and pink hexagons. If I label the longest edge of the smaller triangle x then the longest edge of the bigger triangle is x/b*c, because /b*c is how you get from one triangle to the other.
So let me give the edges of the hexagons names. a for the small orange one, b for the middle-sized pink one and c for the big yellow one. Because regular hexagons can be cut into equilateral triangles, their main diagonals are twice their edges.
And here’s another pair. I know they’re similar because of the vertically opposite angles and one of the other angles is 60°.
And here’s another pair. They’re similar because of the parallel edges created by the edge and main diagonal of the yellow hexagon.
Here’s one pair. I know they’re similar because they have a pair of vertically opposite angles and one of the other angles is definitely 120°.
Wait if I draw in the other side of that triangle it has to be equilateral because the leftmost corner is subtended by the opposite edge and I know it subtends an angle of 60°. How satisfying. So satisfying I could stop there and be ok with it if I never came back.
Hmm these two chords both subtend an angle of 60° at the circumference and so they have to be the same length. Don’t care if it’s helpful because it’s pretty cool.
And another one from that same corner of the yellow hexagon along the top edge and through the middle of the orange hexagon.
Ooh there’s one here from one corner of the yellow hexagon through the opposite corner and along the edge of the pink hexagon.
#MathSky Eighth and final post on two-sided ruler construction is done! I've saved my favourites for last. In this one I show how to construct an equilateral triangle and a regular pentagon, and I give myself some stars.
#MathSky The penultimate post on two-sided ruler constructions is done! This one is about finding where circles meet without actually drawing the circles. davidkbutler.xyz/2026/07/03/t...
The next blog post in the series about two-sided ruler constructions is finally done! This one is about constructing parallel lines. davidkbutler.xyz/2026/07/02/t...
#MathSky Fifth blog post on two-sided ruler constructions is now complete! This one is on constructions that make perpendicular lines. davidkbutler.xyz/2026/06/28/t...