A square formation of soldiers 50 m on each side move perpendicular to a side a distance of 50 m.

In the same amount of time, a dog moves *at a constant speed* around the formation, ending at the same place on the square it started.

How far does the dog travel?

So I undertook to find the solution. It’s easy to imagine the path the dog follows – something like this:

- but the “constant speed” part of the puzzle means you can’t make simplifying assumption like the dog crossing each side of the moving square in the same amount of time (which give an easy answer of [math] \frac{50}{4} ( 2 \sqrt{4^2+1} + 8 ) \dot= \, 203.08 \,\mbox{m}[/math]). Instead, I wrote an equation for the time [math]t[/math] it takes the square to move 50 m and the dog to around the square,

[math]\frac{L}{v_1} = t = \frac{2L}{\sqrt{v_2^2 -v_1^2}} +\frac{L}{v_2-v_1} +\frac{L}{v_2+v_1}[/math]

where [math]L[/math] is the length of the sides of the square, [math]v_1[/math] square’s speed, [math]v_2[/math] the dog’s speed

, substituted [math]L=1[/math] and [math]v_1=1[/math] for ease, solved for [math]v_2[/math], then scaled it up from [math]L=1[/math] to [math]L=\, 50 \,\mbox{m}[/math]

After way more than 30 min, I wound up with a mathematical beast, a quartic equation, which after looking up and use the general solution to quartic equations, gave me I think the correct answer,

My colleague has a less-than-10-year-old a PhD in Physics vs my 35-year-old BS in Math, so maybe he’s just more clever and faster than me, but I’m thinking the solution should be much simpler to find than this. Thoughts?