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Overlapping Squares

Category:Probability
Submitted By:cms271828
Fun:*** (2.23)
Difficulty:*** (3.03)



Two squares each 17cm by 17cm are drawn randomly inside a square measuring 1m by 1m.
The edges of the two smaller squares are both parallel/perpendicular to the edges of the large square.
What is the exact probability the two smaller squares overlap (or touch)?

Show Answer



Comments on this teaser


Posted by Psychic_Master04/19/06
Yay! First to comment! Well, pretty good teaser, and pretty hard too. Good job. :wink:

Posted by vlerma04/20/06
:oops: Sorry too complicated for this old brain. Suspect there are seom brainiacs out there who can do these, but I am not one of them. But even so, it appears you v\gave a very good expanation of how you arrived at your answer so all in all It was a good job.

Posted by BrownEyes04/21/06
Grrrr that teaser wasn't easy... And y didn't you share your chocolate with me in the chat room.?? *waaahh!!!* :cry: :cry:

Posted by Moondancer04/22/06
So very confusing! :(

Posted by cms27182804/22/06
Its not really, just follow the solution on paper and everything will become clear

Posted by Jimbo04/26/06
It's a very interesting problem and a well constructed solution but I am not going to vote on this until I have thought about the solution. Certainly if the squares fall in the designated area, they will touch/overlap, but I am struggling to connect the specific areas with a general solution. I like the problem and the way the solution has been tackled! :D

Posted by garul04/27/06
Superb problem. I got it though by a round about method. Assuming the side of the small sq is A meters (0.17 in this case) and the big sq is 1 M, the answer I got is (A*(2-3A)/(1-A)^2)^2 which matches the solution. I solved it by splitting the bigger square into three areas (a center sq of side 1-3A, 4 corner squares of side A and 4 rectangles of side A & 1-3A). The probability of the squares overlapping when the first square is in each of the above areas are 4AA/(1-A)^2, 9AA/4/(1-A)^2 and 3AA/(1-A)^2. Adding these after multiplying by first square probabilities, we can get the answer as (A*(2-3A)/(1-A)^2)^2. The solution given by the author is definitely more elegant. Here in the xy grid of side (1-A), overlap will happen if the X coordinates of the two squares are within the center rectangle of sides sqrt(2)*A and sqrt(2)*(1-2A) or the two corner triangles. So the probability of one coordinate say X suitable for overlapping is sqrt(2)*A * sqrt(2)*(1-2A) /(1-A)^2 + AA = A(2-3A)/(1-A)^2. Square of this will give the answer.

Posted by Jim102905/14/06
In being one of the editors who accepted this teaser, I think it is really hard. Even I couldn't get it! :D

Posted by kayleeskitties05/16/06
All I can say is Wow.....very hard....very good teaser though! :D

Posted by cms27182805/16/06
Thanks, I'm glad everyone liked it, my first solution was difficult, using double integrals, and was tough to understand. But while I was lying in bed, the solution just came to me, and in the middle of the night, I got up and grabbed my calculator, tested it, and it matched my other solution. Just to make sure it was correct, I did a simulation using a java program. I think I ran the experiment a billion times, counted the successes, and divided by a billion. The result was practically identical (to about 4sf) to the theoretical probability. Numerical verification isn't an absolute proof, but its good enough for me :D

Posted by watup9030605/18/06
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Posted by udoboy05/26/06
Thank you. My brain exploded while contemplating this question.

Posted by love_____xoxo09/10/06
whoa. That was fun! :D

Posted by McBobby121203/01/07
my.....brain.....is.....turning......to.......mush :-?

Posted by gnarldroot02/04/10
Are you kidding me? this is probably the most basic probability riddle ive ever seen.

Posted by bug-e12/04/11
The solution is nice and there are many more problems that can be solved using the same technique. This is a set method for problems like these. Another problem could be like probability of whether bf and gf meet each-other in 1 hr window if the other one can only wait for 15 mins and they dont care much abt each other :P.

Posted by rachit13806/08/12
The squares can overlap diagonally as well in which case the distance between their centers will be \sqrt(2) times 17.




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