tag:blogger.com,1999:blog-8070362.post114884156335277925..comments2022-03-18T11:24:32.869+05:30Comments on Death Ends Fun: Priest on a hillDilip D'Souzahttp://www.blogger.com/profile/08221707482541503243noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-8070362.post-1149430957374971592006-06-04T19:52:00.000+05:302006-06-04T19:52:00.000+05:30they will meet at a time that is the harmonic mean...they will meet at a time that is the harmonic mean of the times to go up and come down.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8070362.post-1148966368598070752006-05-30T10:49:00.000+05:302006-05-30T10:49:00.000+05:30Thanks for your interesting responses, guys. Anony...Thanks for your interesting responses, guys. <BR/><BR/>Anonymous 846 and Suresh (nothing undesirable!), you have the answers that I had in mind. It's best to think of it as two simultaneous journeys -- then the two men have to meet somewhere. The rope helps make it even clearer.<BR/><BR/>Arnold, like Suresh, I'm not sure why you would need Brouwer's theorem (a thought-provoking theorem by itself, of course). Please do explain.<BR/><BR/>Varun, thanks too. Uma, that's right! Damn. We'll have to do it another time.Dilip D'Souzahttps://www.blogger.com/profile/08221707482541503243noreply@blogger.comtag:blogger.com,1999:blog-8070362.post-1148941442164803702006-05-30T03:54:00.000+05:302006-05-30T03:54:00.000+05:30doing the undesirable of following up my own post,...doing the undesirable of following up my own post, i should note that no fixed point theorem is required - all it needs is something far more elementary, known I think as the intermediate value theorem in real analysis. Brouwer's theorem is more deep and is not needed here.<BR/><BR/>S.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8070362.post-1148918552219122742006-05-29T21:32:00.000+05:302006-05-29T21:32:00.000+05:30Imagine a rope between the house and the temple. ...Imagine a rope between the house and the temple. Now imagine the priest at the bottom of the hill and a "clone" at the top. If they walk towards each other at daybreak holding onto the rope, then there will be a point where the priest and his clone meet. Note that this does not require the priest and the clone to start simultaneously: the priest or the clone could continue to be at his starting point for some time after the other has started to move. Nonetheless, since both hold onto the rope, there is a point where they meet.<BR/><BR/>As Arnold noted, there is a fixed point theorem hidden somewhere. However, Brouwer's theorem concerns a mapping from a domain D to itself. I am not sure what the relevant function is here: care to fill in the details, Arnold? Clearly, however, continuity (the "rope") is necessary for the result.<BR/><BR/>Suresh.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8070362.post-1148915770385256712006-05-29T20:46:00.000+05:302006-05-29T20:46:00.000+05:30The problem of the priest going upone day and down...The problem of the priest going up<BR/>one day and down another day is the same as if two people started at the same time, one from the top and one from the bottom. The question now is whether these two people will meet somewhere....Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8070362.post-1148890716341849172006-05-29T13:48:00.000+05:302006-05-29T13:48:00.000+05:30This is of course beside the point (ha), but you d...This is of course beside the point (ha), but you didn't do the magic trick on Saturday!umahttps://www.blogger.com/profile/01578048005614408340noreply@blogger.comtag:blogger.com,1999:blog-8070362.post-1148849180859889392006-05-29T02:16:00.000+05:302006-05-29T02:16:00.000+05:30Answer yes. Simple application of Brouwer's Fixed ...Answer yes. Simple application of Brouwer's Fixed Point Theorem.FifthBeatlehttps://www.blogger.com/profile/10401668749535983431noreply@blogger.comtag:blogger.com,1999:blog-8070362.post-1148848373623419092006-05-29T02:02:00.000+05:302006-05-29T02:02:00.000+05:30Lets plot a graph (Position=x vs Time=y). While cl...Lets plot a graph (Position=x vs Time=y). While climbing, priest starts with - t=0 & d=0<BR/>when he reaches the top, t=12 hrs (assume) & d=D.<BR/>When he starts from the top, t=t1 & d=D<BR/>When he reaches bottom, t=t2 & d=0.<BR/>t1,t2 < 12 hrs<BR/><BR/>The Y positions grow from 0 to D while going up and come down from D to 0 while coming down. The graph for the upward journey would go from (0,0) to (t=12, D), it need not be linear. t=12 & D are the maximum values. The graph for the downward journey would always intersect the first graph at one point (and *only* one point if we assume it to be linear)Varun Singhhttps://www.blogger.com/profile/09677910077835835502noreply@blogger.comtag:blogger.com,1999:blog-8070362.post-1148844039454635282006-05-29T00:50:00.000+05:302006-05-29T00:50:00.000+05:30Vish, you gotta prove it. (Yes, same path both way...Vish, you gotta prove it. (Yes, same path both ways, should have made that clear).Dilip D'Souzahttps://www.blogger.com/profile/08221707482541503243noreply@blogger.comtag:blogger.com,1999:blog-8070362.post-1148843798128179382006-05-29T00:46:00.000+05:302006-05-29T00:46:00.000+05:30If the priest climbs down the same path he climbs ...If the priest climbs down the same path he climbs up, and if the path is a continuous function of time, then the answer is yes.Anonymousnoreply@blogger.com