|
||
|
Title: Two Bees ... or Not Two Bees? Post by rloginunix on Jul 13th, 2016, 5:50pm Two Bees ... or Not Two Bees? When, is the question. At sunrise a bee B1 took off from the flower F1 and flew with constant velocity toward the flower F2 from which, at the same time (at sunrise), a bee B2 took off and flew with constant velocity toward the flower F1. After both bees met at high noon, B1 touched down on F2 at 16:00 while B2 landed on F1 at 21:00. At what time did the sun rise that day? |
||
|
Title: Re: Two Bees ... or Not Two Bees? Post by towr on Jul 14th, 2016, 11:17pm [hide] By construction in Cinderella, it seems to be about 6 hours before noon. By the law of tidy math solutions it's probably exactly that. So let's say the distance between F1 and F2 is 30 units Then B1 flies at a speed of 30u/10h = 3u/h and B2 at 30u/15h = 2u/h Then after 6 hours B1 is at a distance of 3*6=18 from F1, and B2 is at a distance of 2*6=12 from F2 and thus 18 from F1. So that confirms the solution. [/hide] |
||
|
Title: Re: Two Bees ... or Not Two Bees? Post by Grimbal on Jul 15th, 2016, 2:46am I agree. Assuming F1 and F2 are in the same time zone. :-) [hide] B1 flew 4 hours after the meeting. B2 flew 9 hours after the meeting. The flight times before the meeting and after the meeting for a bee are in a ratio matching that of the distances flown before and after the meeting. That ratio is reversed between B1 and B2. If the flight time before the meeing is x, 4/x = x/9. You see that x is the geometric mean between 4 and 9, thats is 6. [/hide] |
||
|
Title: Re: Two Bees ... or Not Two Bees? Post by rloginunix on Jul 15th, 2016, 8:50am Thank you for keeping me on my toes, Grimbal: - F1 and F2 are in the same time zone - |F1F2| <<< Re - vi <<< c towr is correct. Grimbal is correct. (my solution path resembled that of Grimbal's: [hide]if |F1F2|=d, M is a meeting point, |F1M|=x, then from flight time to M equality: x/v1 = (d-x)/v2. After the meeting, B1 covered B2's distance to meeting, d-x, and conversely: d-x=4v1 and x=9v2. Hence, in closed form: t=9v2/v1=4v1/v2[/hide]) Peculiar: [hide]the answer is a product of two smallest primes whose squares[/hide] are used in the interim. |
||
|
Title: Re: Two Bees ... or Not Two Bees? Post by alien2 on Jul 15th, 2016, 9:24am Two bees or not two bees - that is the question: Bees are fearsome in the minds of those who suffer, With stings like arrows of outrageous fortune; Should we take arms against a sea of wasps, And, by opposing, end them? |
||
|
Title: Re: Two Bees ... or Not Two Bees? Post by rmsgrey on Jul 15th, 2016, 12:52pm on 07/15/16 at 08:50:57, rloginunix wrote:
That's really not that peculiar - the general solution, with bees meeting at t=0 and arriving at their final destinations at times t=a and t=b, has them starting at t=-SQRT(ab). In order for all times to be integers, either a=b or a,b are both perfect squares. To fit within a day, {a,b} has to be a subset of {1,4,9} (unless you measure time in smaller units than hours) and {4,9} gives the most interesting answer of the simple answers. |
||
|
Title: Re: Two Bees ... or Not Two Bees? Post by JiNbOtAk on Aug 24th, 2016, 5:49pm on 07/15/16 at 02:46:34, Grimbal wrote:
I remember facing this type of questions during my high school physics. If only my teacher had used Grimbal's approach, it would had been far easier to understand (rather than brute-forcing through with the mathematical approach) :-/ |
||
|
Title: Re: Two Bees ... or Not Two Bees? Post by alien2 on Sep 26th, 2016, 8:37am on 08/24/16 at 17:49:22, JiNbOtAk wrote:
If your teacher used my approach he wouldn't be a teacher. |
||
|
Powered by YaBB 1 Gold - SP 1.4! Forum software copyright © 2000-2004 Yet another Bulletin Board |