# Can you Avoid the Spear-Wielding Gladiator?

You are trapped in a circular coliseum, and a gladiator with a spear is chasing you. You can't defend yourself, but you can run faster than the gladiator.

• You run at 11 feet per second, and the gladiator runs at 10 feet per second.
• The gladiator's spear has a range of 10 feet. You must stay at least 10 feet away from the gladiator, or else you will be killed.
• The gladiator is not smart enough to take a strategic path that optimally chases you down. They only head straight in your direction.
• You control the starting position of yourself and the gladiator.

How big does the coliseum need to be for you to be able to avoid the gladiator indefinitely?

Generalized, if your speed is V, the gladiator's speed is W, and the spear length is X, how big does the coliseum need to be?

• Feels like this would have been asked before in this site. Oct 2 at 7:56
• @justhalf I found no duplicates of this one. Oct 2 at 8:28
• @justhalf However the style of the question title suggests it might be a 538 Riddler problem. Oct 2 at 8:48
• Similar to puzzling.stackexchange.com/questions/8140/…, but with area of capture instead of point of capture. Oct 2 at 10:47
• Should we assume you never need to sleep? Oct 2 at 16:18

Normalise units so the gladiator runs at speed $$1$$, you run at speed $$s>1$$ and the spear range is $$d$$. The idea is to have the gladiator $$G$$ run in a circle $$\Gamma_1$$ of some radius $$r$$ centred in the coliseum's origin $$O$$. You, $$Y$$, put yourself at the end of the vector of length $$d$$ pointing away from $$G$$ in the direction of his instantaneous velocity (tangent to $$\Gamma_1$$) and run in a larger circle $$\Gamma_2$$ concentric with $$\Gamma_1$$.
Then $$\triangle OGY$$ is right-angled at $$G$$ with $$OG=r,GY=d$$ and hence $$OY=\sqrt{r^2+d^2}$$. You want to maintain the right-angled triangle's shape and keep it spinning around $$O$$; since speed is proportional to radius, to match you and $$G$$'s maximum speeds you need$$\frac{\sqrt{r^2+d^2}}r=\sqrt{1+(d/r)^2}=s$$ $$r=\frac d{\sqrt{s^2-1}}$$and a coliseum of size $$rs$$. For the numerical values here with $$s=1.1$$ and $$d=1$$ this gives $$r=\frac{10}{\sqrt{21}}=2.1821789\dots$$ ($$21.821\dots$$ feet) and the coliseum radius as $$\frac{11}{\sqrt{21}}=2.4003967\dots$$ ($$24.003\dots$$ feet).