Arlington’s Got a Gang Problem

That’s wonderful.

Loves it – awesome!

I’m thrilled to see that I’m not the only one who appreciates this. Thanks to ARLnow, GGW, and about 11 jillion (to use a math term) Twitter users and Facebookers for the links.

As Rick says, the answer to 4 is 2 pi/5 radians or 72 degrees; I initially saw this on GGW where a commenter read the question as CD instead of (angle) D and so I initially solved for the ratio of the lengths of the sides.

For question 3, 7 is a sort of degenerate solution because the side with length 4 is at a right angle to the top and bottom. If we demand x10 but where it still looks like a trapezoid, then there are solutions also for 13 (degenerately, again with side 4 at a right angle), 14,15,16,17,18, and 19. There are also solutions of 8 and 12 if the 4 and 5 sides slant the same way. Finally, there are the truly degenerate solutions 7,9,11,and 13 if all the sides are co-linear, that is, if the shape has no thickness.

If we make triangles out of the 4/5 sides, and the common height, and a line segment along the 10 side, and call the lengths of those segments m and n, then it’s pretty easy to show that n=Sqrt(9+m^2). Plot this for -4 <= m <= 4 and plot also the solutions to m+/-n=z, where z is an integer.

Yeah, so in regards to that “Shakespeare” graffiti, did the cops catch that Henry IV dude yet?