*How
to cut a brick in thirds*

This problem came about when I was building a patio, and needed to brick up a curved section. I didn’t want to use half-bricks, since the curve was too tight and the
gaps between the outside edges of the bricks would’ve been too wide. But
cutting bricks in fourths made the pieces so small, it looked I was trying to pave using oversize rocks!
Cutting the brick in thirds was almost perfect, but I had a little delay while trying to figure out where
exactly to cut the bricks to get perfect thirds. (Well, *you* tell me off the top of your head where to measure and cut a
7-3/4” brick to get it in thirds, allowing for a 3/16” saw blade kerf!)

What happened was, I discovered this little trick, which I haven’t been able to find
anywhere else. Maybe I invented it, or maybe it’s a mason’s trick
that’s been around since the Egyptians, I don’t know. But it was
cool enough to share with people, and I’d sure appreciate some feedback from any masons out there who can tell
me that this is either cool or ho-hum.

Our problem is to take a brick of given length (*l*) and
width (*w*),

and mark it so that the mark is exactly 1/3 of the way for the
edge along the long side. What I found was that, no matter
what the width of the brick is, if you stack up three of them (lay them together on the ground, I mean), and draw
a line from corner to corner, the point where that line hits the bottom brick first is at the desired 1/3 point!

Not only that, but if you want to cut it in 5^{th}s, 7^{th}s, or any other fraction, just
stack up that many bricks (5 for 5^{th}s, see below), and draw the same corner-to-corner line.
The mark you make on the top edge of the bottom brick is exactly that fraction of the whole length. (Of course, this would work for boards as well as bricks.)

That’s it. Do I have something cool here, or is this a big fat who cares? Here’s the geometry proof, for those of you who care, but it’s really straightforward.

Consider the figure I’ve drawn below, where there are 5 bricks stacked up.
I drew 5, but I’ll do the proof in general, for *n*
bricks. I’ve drawn two triangles in bold, which you’ll see are similar. (They’re both right, and their vertex angles are congruent by alternate
interior, so AA similarity.) These triangles are shown below, drawn
by themselves, with the height of the big triangle being *nw*
to represent *n *bricks.

OR

With this picture, we’re done, because the similarity proportion is ,

and this simplifies to

.

So that’s it! The length *x*
is exactly th the original length of the brick.
What do you think? Is this something that anyone out there
could actually use? I hope so, because it sure helped me!

© 2005 **Dan McGlaun**