In this puzzle, Snoopy and the Red Baron fly past each other in opposite directions, turning hard left and hard right, respectively. The first one to shine his laser pointer, fixed on a forward pointing mount, on the other wins. David Uhland observes that the key to this puzzle is that the pilot with the smaller turning radius always wins.
The smaller turning circle is, except for one tangent point, completely inside the larger turning circle. The laser ray from the inner circle always intersects the outer circle and the pilot on the outer circle will at some point fly through it. The laser ray from outer circle never even intersects the inner circle, with the trivial exception of the tangent point. If Snoopy is turning with constant speed V and constant turning acceleration A his turning radius is R = V2/A. In the first case the Red Baron’s speed is 2V and his turning acceleration is 3A, so his turning radius is (4/3)R and he is defeated. In the second case the Red Baron’s velocity is 0.9V and his turning acceleration is 0.75A, so his turning radius is approximately 1.08R and once again he is defeated by Snoopy.
Thanks to Harvey Kim for submitting a solution this month!