9oh no still confused can u pls exp ur prev fig ;l(
How thick should a coin be to have a 1/3 chance of landing on edge?
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58 Answers
62i am waiting...
in my mind i think what i have done is correct....
But then until aditya turns it to a paradox ;)
I wud take pride in getting a paradox named after me. :P :D
9sry for bein so stupid
can u giv a som more hint on gagars exp ;)
62well what he dis is in spirit is the same as what we did...
Basically the mistake that i was doing is that there is no flipping.. it is just the coin placed randomly kind of thing.. (as i now understand)
So if u see carefully the area of the part which it should fall within the part of the edge it should fall... the probability will be given by the ratio of the area :)
WHich comes out as the ratio that we found :)
62umm.. well
can u understand the diagram that i made?
It is the coin as seen from the top.... (the circle is only to show its orientation..))
Basically celestine.. this is the same as our original method... (exactly the same)
drawing a line through the center and seeing if the CM lies within the vertical lines :)
I dont know if u are able to see how they are the same :)
9ah understood now only thnks
that circle over there misleaded me ;)
1I m posting the solution as given in the buk..........
We can think of the coin 2 b inscribed in a sphere, where the center of coin is center of sphere. The coin itself is regarded as a right circular cylinder. Then a random point on surface of the sphere is chosen.If the radius from that point 2 the center striked the edge, the coin is said 2 hav fallen on edge.
To stimulate this condition in reality, the coin might b tossed in such a way that it fell on a thicky substance dat wud grip the coin wen it touched, n the coin wud slowly settle to its edge or face.
A key thm in solid geometry simplifies this prob. When parallel lines cut a sphere, the orange peel-like band produced between them is called a zone. The surface area of the zone is proportional 2 the distance between the planes, and so our coin shud be 1/3 as thick as the sphere. How shud the thickness compare with d diameter of the coin?
Let R b the radius of the sphere n r that of coin.
R2=r2 +1/9R2
(1/3)R = (√2/4) r = 0.354r
So the coin shud b abt 35% as thick as diameter of coin.
CHEERS!
1nishant i am damn sure that if u continue ur work in this field u will get something in ur name. u r worthy of it
62The surface area of the zone is proportional 2 the distance between the planes, and so our coin shud be 1/3 as thick as the sphere.
Aditya check this statement...
This is wrong....
do u want the proof for this being wrong??
Take two parallel planes at distance R/2from the central plane on both sides..
The distance is 1/2 of the diameter...
Is the ratio of the surface area half??
btw
My proof exactly uses the same logic... that your proof does till that point where i am contradicting you.......
1name it as a nishant paradox or even aditya's paradox and tells me if u have got the final ans right now i have to take a class so bye for now
62gaurav... Aditya already posted his solution...
Which i think is wrong....
But then As I hae been in the past... I may be wrong.... :)
1the buk tells dat.......but honestly speaking even i didnt understand that.
1actually it shud hav been a doubt rather than challenge bcoz i myself m not satisfied with the solution.
62aditya to be true i am almost sure this is wrong...
but i will think more on this one...
my head is overcooked thinking aobut the question of the day i posted today :)
It took me 30 minutes to formulate it... :(
So i will see this more carefully some other time...
I hope that the flaw i have posted is correct :)
9oops allof us ve forgotten that this is an olympiad beauty ;)
the sol gvn is right we r rong ;(
where i , nishant sir,gagar ve gone rong is that we considered only rot abt 1 axis neglecting rot abt another perpendicular axis
and nishant sir that statement is indeed right , i ve a proof for that.
but itll be nice to see how u disprove that fact ;)
62i see merit in what u are saying celestine..
Even when i was thinking of the solution i knew that i was being biased.. but somehow i had convinced myself that it wont matter....
wheter it rolls whihc way..
Now indeed i am getting the same Bertrand paradox in mind...
If u understood that one, then the solution that we are giving is also correct ;)
lol... (because we have not been told what is random!)
9no our sol cant be right bcos we never considered coin spinning in both axes like a cricket ball
62Celestine: you have not understood that paradox correctly ;)
Isnt what you are saying an assumption?? (is it given in the question?)
62you can definitely say that the other solution is also correct ;)
9yes but then it should be wrong of us to assume that there is spin along only one axis cos the hands of the person who spins thwcoin might be skilled enuf to spin it in both axes.
i feel that would make a better assumption than wat we ve assumed
any way how did u disprove that statement ;)
9i think u can put
the orange peel-like band produced between them is called a zone. The surface area of the zone is proportional 2 the distance between the planes,
as QOD cos many ppl will integrate the surface area wrongly . i had to change my approach thrice to find correct method