Please help solve this math problem, perhaps involving calculus?

Question

If there is a circle of radius x (for this question say x=5) drawn on a table of radius y (for this question say y=10), and you randomly drop a square of side length of length z (say z=4) onto the table with no possiblility of any part of the square overhanging the table's edge, what is the probability that the square is entirely contained by the drawn circle? (i.e: No part of the square has crossed the edge of the drawn circle. The square can TOUCH the edge of the drawn circle though.)
I've been thinking about this problem for a long time and can't come up with an answer. An answer to the problem and explanation of how you solved it using the numbers above mentioned would be greatly appreciated.
Also, if someone could perhaps come up with a formula using the above x, y and z?
Thanks.
Note: Nobody has to answer all of the above mentioned. Any partial answers would also be appreciated.

Answer 1

Great question. I'm a little stale on this kind of things, but I'll give it a try.

I think that the location of the Circle with r=5 has something to do with it. If so, is it concentric with the circle of the table? Both have center ‘O’ I'm going to make that assumption. Otherwise it even gets harder.

What you have to consider is the center of the square ‘S’ and its orientation to the radii of the r=5 circle and table. That's why the table and circle should have the same center.

If the center of the square (ABCD) is within 5-[2*sqrt(2)] of the center of the circle, the square will always be inside r=5 circle. This assumes that the diagonal of the square is on a circle radius. In other words, a circle radius passes thru S and corners A&C or B&D. Draw a circle with radius 5-[2*sqrt(2)] with center O. Color this circle BLACK.

Now is the tricky part, as you spin ABCD 45 degrees, you can slide S out further. But it's almost midnight and I've got to get up early tomorrow.

You must also do the same for the r=10 table.

When you have the areas (r=5 and r=10, with regard to the square), you will have the percentages.

Hope I'm right and that I've given you a start.

I'll check this Q tomorrow, and see if I can add anything.

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I'm back.

Up 'til now, I've mentioned the case where a diagonal of the square lies along a radius. Let's fix a corner of the square (let's say 'A') on the circumference of the r=5 circle. Now tilt corner 'C' a little bit. You’ve formed a triangle, SAO. We know the length of AO [5]. We know AS [2*sqrt(2)]. And we can find OS (my trigs a little weak here, so I’ll leave that to you). As the angle SAO goes from 0 to 45 degrees, OS get larger, but the probability that all of ABCD remains within the r=5 circle gets smaller (again a trig function and calculus applies here). Once the angle SAO exceeds 45 degrees, B or D must lie outside r=5 circle.

As S moves out from the black circle described above, until angle SAO reaches 45 degrees, color the plot of S in lighter shades of gray. This represents the probability that square ABCD is still inside r=5 circle. It’s the all black and lighter shades of gray that is the numerator of out probability fraction.

Repeat for the r=10 circle. It’s the same formula as for the r=5 circle, but if S is within 10-[2*sqrt(2)] of O, then the square must be entirely within the r=10 circle.

Compare the two areas, and you’ve got your probability.

Hope this helps.

Answer 2

I assume you mean a circular table.

The first thing to notice is that we can ignore the orientation of the small square when dropping it on the table. For example, suppose we run an experiment where we drop the small square 1,000 times with random orientations. After the drop, we can rotate the table so that the square is in a preferred orientation and then see if the square overlaps the edge of the circle.

Thus, we can now pay attention to just one corner of the square when we drop it on the table, say the upper-left corner. The upper-left corner must land within the circle, and it must land within the circle shifted left by the width of the small square and up by the height of the small square.

So the upper-left corner of the square must fall within a region that is the intersection of two circles, where the second circle is the first circle shifted by the length of the small square's diagonal.

Next, note that the upper-left corner of the square may not end up anywhere on the table, but must end up in a region that is similarily the intersection of two circles formed by shifting the table around.

So, if you can compute the area of those two regions, take the ratio, and that will generate your probability.
The region that we are interested is twice the area beneath a chord. http://mathworld.wolfram.com/Chord.html gives a formula for the area enclosed by the chord. I believe the distance of the chord from the center of the circle is radius-d/2, where d is the diameter of the small square or sqrt(2)*z.

Answer 3

Consider the locus of the center of squares contained in the circle. It is apparent that this locus is a circle, smaller than the original circle. Compute the area of the locus. Now determine the locus of the center of possible squares on the table, and compare the two areas. Bingo!

Answer 4

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Answer 5

o_O