Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

|
Options

irspow
2006-01-21 08:15:55

Graeme, I am sure that you can look back at all of those wonderful illustrations from earlier in this topic to imagine the lines to which I am referring.  Remember how we determined the distance from the center of the bowl to the center of one of the balls?  We solved for the distance between balls and the center by creating a geometric figure with each ball's center as a vertice.

In the example above there were only three, forming a triangle, but any number of balls will form some sort of polygon.  And any of those polygons will have a geometric center that is equidistant from all of the balls centers.

If you look at any side of one such polygon it will form a triangle with the center of the bowl as the apex and two of the balls centers delineating its base.  This apex will always form an angle of 360° number of balls.  Either of the base angles will thus always form an angle of [180-(360/n)] /2.  That is where the a and b from the formula above come from.

Their use is derived in this way.  Looking at one of these triangles as we did above, the distance from the center of the bowl to the center of either ball will be opposite one of the base angles and two of the ball radius' will form the base opposite the apex.

Using the law of sines;

d /sin[(180 - 360/n)/2] = 2r / sin(360/n);

d = {2rsin[(180 - 360/n)/2]} / sin(360/n);

Now, remember that the radius of the bowl equals the radius of a ball plus the hypotenuse of the triangle formed with a horizontal d and a vertical r.

That hypotenuse from the center top of the bowl to the center of one of the balls is then;

√d² + r²;

(instead of typing all of those angle formulas again I will revert to a and b as shown above)

d = 2rsina/sinb;   d² = 4r²sin²a / sin²b

The radius of the Bowl, R, is now just;

R = r + √[(4r²sin²a / sin²b) + r²]

Taking out the r² from the radical sign and simplifying yields;

R = r + r√[(4sin²a / sin²b) + 1];

R = r( 1 + √[(4sin²a / sin²b) + 1];

I hope that explained it, just ask if something is not clear.  I have done nothing here different from what was done above, but solved it in the general form.

It won't work for other problems.  In other words only the number of balls can change and not any of the parameters such as the tops of the balls being even with the top of the bowl....

Graeme
2006-01-21 03:40:22

irspow wrote:

I just worked the now easy problem in reverse.

if R = radius of bowl and r = radius of ball and C = number of balls

C = 1;  R = 2r
C = 2;  R = r(1 + √2) ≈ 2.414r
C = 3;  R = r(1 + √(7/3)) ≈ 2.527r
C = 4;  R = r(1 + √3) :asmp 2.732r
C = 5;  R = r(1 + √[(4sin²54°)/(sin²72°) + 1]) ≈ 2.973r
C = 6;  R = r(1 + √5) ≈ 3.236r
C = 7;  R ≈ 3.512r
C = 8;  R ≈ 3.798r

In general for this equation;

R = r(1 + √((4sin²a / sin²b) + 1))

where a = [180 - (360/C)] / 2   and  b = 360/C  and  C = number of balls.

Could you please explain how you worked out this formula.  Apologies for the trouble but I can't really work it out.

Maybe it's too early and i'm tired

Graeme
2006-01-21 03:27:01

irspow wrote:

Once again, I show my stupidity....I couldn't figure out how to do it in two dimensions, much less using a single sphere, all the while making it much more difficult.

Thanks for the proof and the visualization.  Now I can see exactly how Graeme was looking at the problem.

Graeme, sorry for any inconvenience that I may have caused you.

No problem at all, I appreciate all sorts of approaches to problems.  Since I am only young, I haven't even touched on the more complex maths, so alternative approaches are always interesting.

Graeme
2006-01-21 03:25:23

mikau wrote:

Come to think of it, my not noticing I could use graeme's method was so idiotic its unreal. My equation and graph skill is much higher then my geometry. In my oppinion the simplest solution is the best and graeme's was the simplest. Hats off to you graeme!

Great problem!

Thank you very much mikau.

Graeme
2006-01-21 03:14:07

mikau wrote:

Maybe. I'm not sure where you got that formula.

I think the reason I didn't see the obvious solution was I did not assume the line drawn from the small circle to the edge of the large, had the angle as the line from the center of the large circle to the point of tangency.

Lets see if I can prove it.

(edit) dang. It seems obvious enough but I can't seem to prove it. I'll keep working on it but maybe you guys can get it before me.

Prove this!

http://i21.photobucket.com/albums/b299/ … raeme2.jpg

If a small circle B is inside a large circle A, and circle B is tangent to circle A at point C, prove the line drawn through point A to point B passes through the point C. (the point of tangency). Point A and point B are the centerpoints of circles A and B.

I have an explanation/proof for the line and the points for spheres, however I presume it works in principle for circles also.

Whenever two spheres touch each other (one inside the other or outsides touching), their point of contact will be on the line through their centers. You can see this intuitively if you simplify the problem and picture one ball at the bottom of the bowl--clearly the contact is at the very bottom of the bowl. (You can prove this rigorously by rotating the bowl and sphere about the line connecting their centers. The bowl and
sphere will remain in place because of their symmetry, so their point of contact must as well, meaning it is on the line of symmetry, which only touches the bowl at the bottom). Back to this more complex case, if you froze the spheres in place in the bowl and rolled the bowl so that you look straight down through its center and through the center of one of the balls, it looks exactly like the one-ball case (with two extra balls in the way).

irspow
2006-01-20 15:31:22

I just worked the now easy problem in reverse.

if R = radius of bowl and r = radius of ball and C = number of balls

C = 1;  R = 2r
C = 2;  R = r(1 + √2) ≈ 2.414r
C = 3;  R = r(1 + √(7/3)) ≈ 2.527r
C = 4;  R = r(1 + √3) :asmp 2.732r
C = 5;  R = r(1 + √[(4sin²54°)/(sin²72°) + 1]) ≈ 2.973r
C = 6;  R = r(1 + √5) ≈ 3.236r
C = 7;  R ≈ 3.512r
C = 8;  R ≈ 3.798r

In general for this equation;

R = r(1 + √((4sin²a / sin²b) + 1))

where a = [180 - (360/C)] / 2   and  b = 360/C  and  C = number of balls.

mikau
2006-01-20 15:03:30

Maybe. I'm not sure where you got that formula.

I think the reason I didn't see the obvious solution was I did not assume the line drawn from the small circle to the edge of the large, had the angle as the line from the center of the large circle to the point of tangency.

Lets see if I can prove it.

(edit) dang. It seems obvious enough but I can't seem to prove it. I'll keep working on it but maybe you guys can get it before me.

Prove this!

If a small circle B is inside a large circle A, and circle B is tangent to circle A at point C, prove the line drawn through point A to point B passes through the point C. (the point of tangency). Point A and point B are the centerpoints of circles A and B.

irspow
2006-01-20 14:26:28

r(ball) = R(bowl) / [1 + √(7/3)]

mikau
2006-01-20 14:13:18

I wonder if this problem would be more difficult if the bowl size was given, but not the ball size.

mikau
2006-01-20 14:12:08

heres a drawing of how graeme did it. I take it he found the center coordinates of the small circle same as I did. Then just used the pythagorean theorem plus the radius.

irspow
2006-01-20 14:08:02

Once again, I show my stupidity....I couldn't figure out how to do it in two dimensions, much less using a single sphere, all the while making it much more difficult.

Thanks for the proof and the visualization.  Now I can see exactly how Graeme was looking at the problem.

Graeme, sorry for any inconvenience that I may have caused you.

mikau
2006-01-20 13:55:51

Come to think of it, my not noticing I could use graeme's method was so idiotic its unreal. My equation and graph skill is much higher then my geometry. In my oppinion the simplest solution is the best and graeme's was the simplest. Hats off to you graeme!

Great problem!

mikau
2006-01-20 13:50:18

Just read greame's solution and its much simpler. Pretty much the same as mine only uses geometry rather then equations. Works just as well only its faster, so I'd say his is the best way to do it. Still I'm proud of my equal slopes observation. :-)

mikau
2006-01-20 13:05:54

This one was pretty tough being that it really needs to be done in three dimensions.

Thats what I thought irspower, but with a little care and a nifty observation, it can be solved in 2d.

Take a look at my pic. Its pretty much self explanatory.

The key to this problem is the slopes of the two cirlces are equal at the intersection. This gives us an extra equation that does not contain R. Then we have two equations for x and y that do not contain R. We can then solved for y in terms of x. Substitute this value of x in equation a to solve for x, then solve for y. Once we have the values of x and y we can use equation b to find the radius.

In the end I got the same answer Greame did, so I guess thats the correct answer.

The slope observation I think is an interesting one that I never made before. Its very usefull in finding the intersection of tangent circles and the size of tangent circles (as in this case). A circle of an unkown radius, with a given centerpoint, can intersect a given circle, at an infinite number of points. But there are only two points for which they intersect, and are tangent, the points of intersection where the slopes are equal. That gives us an extra equation to find the intersection. :-)

Graeme
2006-01-20 11:10:14

Thank you very much irspow for your kind words and help with this.

I will post pictures of my possible solution as soon as possible (within 24 hours hopefully) and hopefully it can clear things up about my idea.

Thanks again