Let's start with our very first puzzle.For the few weeks puzzles will be relatively easy.Then we will move on to some challenging puzzles.
Fifty accurate watches lie on a table. Prove that there exist a moment in time when the sum of the distances from the center of the table to the end of minute hands is more than sum of the distances from the center of the table to the center of the watches.
Hello friends,you are now welcome to post your solutions.
ReplyDeletehint? savaal ka meaning nahi samajh aaya :-/
ReplyDeleteisn't it obvious that php will be used?
ReplyDeleteThis comment has been removed by a blog administrator.
ReplyDeleteas more hint, php=pigeon hole principle. Now, it should be obvious.
ReplyDeleteit's very obvious.
ReplyDeleteOne, it IS possible to arrange the watches symmetrically around the centre. A CIRCLE.
Two, Arrange the watches so that they are all placed in a circle or some radius around the centre of the table. Also, orient them such that the 6-12 line is on a radius. Let us assume that the time is 12:00
then, arrange the watches such:
Have half of them point the minute hand radially outwards, and the other half radially inwards.
The problem should be changed to more than or equal to, because you can have 25 watches showing 12:00 when the other 25 "show" 6:00, hence making the distance always equal to the distance from the centre of the watches.
I think there needs some more clarification. You don't need to provide arrangement and orientation of watches. They are totally random. And you have to prove for any arrangement that at some time, the claim will be satisfied. And question is correct, I mean there is no need for changing it to 'more than or equal to'
ReplyDeleteThe two distances can never be equal at all moment.
ReplyDeleteSuppose at some moment the distances between centers of clocks and the center of table is larger than between end of minute hands and center of table. Then just after 6 hours (180 deg rotation) we will get the desired result(the order of the distances will be reversed).
let Centers are (R_i,alpha_i)
ReplyDeleteminute hand ends (R_i*cos(alpha_i)+r_i*cos(theta_i),R_i*sin(alpha_i)+r_i*sin(theta_i))
{theta being angle of minute hands with x axis, Center of the table is origin for coordinate axes, r_i= length of minute hand and R_i=distance of center of the watch from origin}
now what we need is this function to be positive at any moment.
{E=sigma}
F(theta)=E(R_i^2)+E(2*R_i*r_i*cos(alpha_i-theta))
this gives=> F(theta)''=-F(theta)+p
where p=E(R_i^2)
solution to this differential equation is=>
solution to this equation being..
F(theta)=cos(theta)+sin(theta)+p-p*cos(theta-a)
where a= const
now this where I'm stuck. what i need to prove is that for some theta ; F(theta)>0
correction: F(theta)= A*cos(theta)+B*sin(theta)+p-p*cos(theta-a)
ReplyDeleteand hence this approach sucks :(
correction : cos(theta)+sin(theta)+p-p*cos(theta-a)
ReplyDeleteand hence this approach leads no where :(
hello friends, i am glad to see your positive and enthusiastic responses. first of all, please give your name while commenting.
ReplyDeleteand, response to the just above commentator. Your approach is correct but this will involve very rigorous maths. here we are looking for some smart solution, which one of you has given already.
Congrates to him !! :)
But there is a catch here. if you say for any one clock that, at time t, if minute hand has lesser distance than center of the clock by an amount m1 and after 30 mins, minute hand has a greater distance than center of the clock by an amount m2, then you need to prove that m2 > m1 otherwise solution will be incomplete.
ReplyDeletethink again...
the above, what u said, can be proved by simple geometry. Then the sol. will be complete.
ReplyDeleteyes, now that is a complete proof :)
ReplyDelete