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 5ths, 7ths, or any other fraction, just stack up that many bricks (5 for 5ths, 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