Blacknell.net

Arlington’s Got a Gang Problem

A *math* gang problem, that is.  Found these along the W&OD trail yesterday.

So who’s going to solve it?

11 comments

1. Morning Notes | ARLnow.com says:

   [...] Marauding Mathletes Deface W&OD Trail — Someone has been creating mathematical and geometric graffiti (in chalk) on the W&OD trail. Mark Blacknell, who may or may not be organizing a posse of rogue road bicyclists to find the culprit(s), has the photos. [...]

   July 8th, 2010 at 4:10 am

2. freewheel says:

   That’s wonderful.

   July 8th, 2010 at 8:21 am

3. ontarioroader says:

   That’s great. On a similar note, I saw some Shakespeare graffiti last week in Palisades DC – I think it was from Henry IV.

   July 8th, 2010 at 9:58 am

4. Suzette says:

   Loves it – awesome!

   July 8th, 2010 at 12:47 pm

5. Marauding Mathletes Deface W&OD Trail | ARLnow.com says:

   [...] Someone has been busy creating mathematical and geometric graffiti (in chalk) on the W&OD trail. Mark Blacknell, who may or may not be organizing a posse of rogue road bicyclists to find the culprit(s), has more photos. [...]

   July 8th, 2010 at 12:53 pm

6. John says:

   Nice mention by GGW: http://greatergreaterwashington.org/post.cgi?id=6481

   July 8th, 2010 at 1:14 pm

7. MB says:

   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.

   July 8th, 2010 at 2:03 pm

8. Rick says:

   Can’t see questions 1 & 2 very clearly; but the answer to 4 is 72˚ and I believe the only possible answer to 3 is 7.

   July 8th, 2010 at 2:47 pm

9. thm says:

   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.

   July 8th, 2010 at 5:50 pm

10. SJ says:

    There was similar nerd chalk graffiti along 3rd street in Silver Spring a week or two ago. Not sure if it’s persisted since then.

    July 8th, 2010 at 9:09 pm

11. Max says:

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

    July 15th, 2010 at 7:34 pm

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