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• Print publication year: 2006
• Online publication date: January 2011

# 3 - The Solutions

## Summary

The Lion and The Christian

A lion and a Christian in a closed circular Roman arena have equal maximum speeds. Can the lion catch the Christian in finite time?

Solution. At the first sight the Hint gives an elegant and very simple solution. Indeed, writing O for the centre of the arena, L for the lion, and M (‘man’) for the Christian, if L keeps on OM and approaches M at maximal speed then we may simplify the calculations by making M run along the boundary circle of radius 1. Then if L starts at the centre (which may clearly be assumed) then L will run along a circle of radius 1/2, so L will catch M in the time it takes to cover distance π. This assertion is instantly justified by Figure 22 which shows that if the arc length MM′ on the outer circle of radius 1 is φ then OPL′ is also φ and hence OSL′ is 2φ, where S is the centre of the inner circle (of radius 1/2) touching the outer circle in P and the line OM in O. Consequently, in the time it takes the man to get from M to M′ on the boundary circle, the lion gets from L = O to L′. Hence L catches M in P. (Equivalently, the angles marked ψ and 2ψ show that the arc length M′ P on the boundary circle is precisely the arc length L′P on the inner circle.)