Re: [math-fun] Window puzzle, sort of

Here's another argument that's less geometric and more calculusy. I'll also use A and B to denote the bottom and top of the window and O and X to denote the bottom of the wall and the location of the person. You choose X so that when you walk right at unit speed, say, angle AXB has an inflection point, which means OXB is decreasing at the same rate as OXA. Now having X move right at unit speed has the same effect on OXA as moving A down at speed a/x. So moving A _up_ at speed a/x will decrease the complementary angle OAX at the same rate that moving X right will decrease OXB. Thus triangles OAX and OXB are similar, giving x^2 = ab. David P. Moulton