[k-man]

There is a unit square on a plane. A unit circle (r=1) is placed randomly, so that its center is somewhere inside the unit square. Every point inside the unit square has the same probability of being chosen as the center of the unit circle.

What is the probability that the square fits entirely inside the circle?

[Guest]

0?

[HoustonHokie]

Did some drawing and I came up with 31.51%. No time to show reasoning now, but that's my best guess. Area looks like a square made with 4 arc segments, each radius 1, with centers on corners of the square.

[Guest]

Did some drawing and I came up with 31.51%. No time to show reasoning now, but that's my best guess. Area looks like a square made with 4 arc segments, each radius 1, with centers on corners of the square.

You went further than I did, but sketching it out real quick like, that's what I get. Your prob passes my reasonability test.

[Guest]

Did some drawing and I came up with 31.51%. No time to show reasoning now, but that's my best guess. Area looks like a square made with 4 arc segments, each radius 1, with centers on corners of the square.

I came up with a similar drawing as you did but I got a slightly different answer

~34.31%

I'm working on trying to make it more accurate.

[bonanova]

1 - [sQRT(3) - Pi/3] = ~ .3151467436

[k-man]

HoustonHokie and bonanova got it! Extra credit to bona for the exact formula!

[Guest]

1 - [sQRT(3) - Pi/3] = ~ .3151467436

could you explain?

[Guest]

probability is (3-2*SQR2)2

[Guest]

could you explain?

Refer to the attached image

spvu is the unit square

Capital letters denote area of the enclosed region and small letters are references for points

Now, the area you want to calculate for probability is O; ie as long as the centre of cirle is in this region, the square is inside the cricle completely

Now, consider triangle pqr.

pq = 1 and rp = 1/2

this means qr = root(3) /2

Area of triangle pqr = root(3) / 8

Also angle subtended by arc qst at p is 120° as angle qpr = 60°

so area of region qstp = pi/3

Now area of qstr = area of arc qstp - 2*area of trianlge pqr

So, area of qstr = pi/3 - root(3) / 4

And by symmetry, this is also the area enclosed in region by points spt

Now area of arc sptu = pi/4

And area of square spuv = 1

So, area of region puv = G+C+H = 1 - pi/4

Now area of D+O+B = area of spu - (E+A+F)

By symmetry, E+A+F = G+C+H

So, area of D+O+B = pi/4 - (1-pi/4) = pi/2 - 1

Now, A+O+B+F = area of spt

D+O+B = A+O+B

So, F = spt - (A+O+B)

F = (1 - root(3) /4) - (pi/6)

Now, A = pvu - 2F

Solving it you get

A = pi/12 - 1 + root(3) / 2

Now area of O = AOB - 2A

O = pi/2 - 1 - 2A

Solving it you get

O = 1 - root(3) + pi/3

O = 0,3151

And your probability is 31,51%

Square and circles.bmp

Edited May 6, 2011 by DeeGee

[Guest]

There is a unit square on a plane. A unit circle (r=1) is placed randomly, so that its center is somewhere inside the unit square. Every point inside the unit square has the same probability of being chosen as the center of the unit circle.

What is the probability that the square fits entirely inside the circle?

38.45 %. The shadowed area in the attached draw, is the square region that accomplish condition. It is easy calculate the proportion of total doing it in a quarter. I could send a detailed reasoning.

Edited May 7, 2011 by Jalegre

[Guest]

38.45 %. The shadowed area in the attached draw, is the square region that accomplish condition. It is easy calculate the proportion of total doing it in a quarter. I could send a detailed reasoning.

Sorry the solution is the former posted, I had a calculus error.