Geometry Journal, #2

The problem:

My process:

I put a rectangle into Geogebra. It actually took me a few tries to really nail down what the problem was saying, but once I did I started dragging the points around. I noticed that the angle made by the two perpendiculars was constant also. Cool! After noodling around with the diagram I understand why:

I started drawing a lot of lines. This is another one of those diagrams that has a bajillion similar triangles. I was playing with this while my kids were going to sleep, so it was about 30 minutes of lazy drawing and redrawing. I found a lot of congruent triangles, some I was pretty sure would help me, but I couldn’t make them work. Here was my favorite congruent triangle pair:

After skipping to the solution of the previous problem (and seeing Benjamin Leis’ solution) I was feeling motivated to find a solution to this one before checking the solution.

I grew frustrated with the computer and (following my takeaway from the previous post, actually!) I took out some paper and started writing equations. I was thinking about how I would get an expression for EF + EG (the two perpendiculars) and I had the idea that you could do that using similar triangles.

I took the two triangles above and set them similar to each other. Awful quality I know, sorry:

Here is that picture, cleaned up a bit:

The point is that if those triangles are similar, then those perpendiculars are in ratio also. Some fiddling tells us that if that’s so, the sum of the perpendiculars is equal to:


Is EF a function of x, so that the x’s cancel out? If so, we would be done. Call that yellow angle in the top right corner theta.

\Large \sin{\theta} = \frac{EF}{x}

\Large x \dot \sin{\theta} = EF

And that works. The sum of perpendiculars is equal to l \dot \sin{\theta}.


Here is the book’s solution:

Here’s the point:

It’s a nice solution. When I was looking for a constant I was trying to do something like this — slide a segment over using a rectangle. The problem was I couldn’t get the right structure when I was trying this, I always ended up stuck with a trapezoid.

The big strategy here is “when you’re looking to prove something is constant, you can try to construct the thing out of both pieces.” I did try that first, but ended up stumped so I moved to a different approach.

Otherwise, I’m not sure what else to take away from this solution. The move of using an isosceles triangle to prove both sides are congruent is sweet.

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