[Guest]

You have a circle of radius 'r'. In it are 'n' congruent circles of radius 'a' such that each circle (except the big circle) touches exactly two congruent circles and the large circle. find 'a' in terms of 'r' and 'n'.

it's not as hard as it sounds. It took me under five minutes, but its still interesting.

[Guest]

a=Pi*r/(n+Pi)

[Guest]

here's what i got

a= r / (sin (360/2n) + 1)

[plainglazed]

a = r sin(π/n) ?

[Guest]

Because of radial symmetry, the centers of each circle should have an angle difference from the center of the big circle equal to 2 pi/n

If we were to draw radii in each of the circles to the points where they meet their small 2 neighbors, we would have a regular polygon in the center of the big circle with each of the small circles being the edges of the polygon.

From this fact we can calculate the distance from the center of the big circle to the center of any of the smaller circles.

For a regular n-gon, with distance from center to corners h, the center and two adjacent corners form an isosceles triangle with the two equal sides being h. The angle between the equal sides is 2*pi/n. From trigonometry, the side of the regular polygon is given by 2*h * sin( (1/2) * 2*pi/n) = 2h*sin(pi/n). (Consider the bisector of the isosceles angle).

But this is equal to the distance between the centers of any of the small circles or 2*a

a = h*sin(pi/n)

h = a*csc(pi/n)

h is the distance between the center of the big circle to the center of a small circle.

Because the small circle touches the large circle,

h+a = r

a*(1+csc(pi/n)) = r

a = r/(1+csc(pi/n))

[Guest]

a = r / (1 + cosec(Pi/n))

[Guest]

a=rsin(180/n)/(1+sin(180/n))

similar triangles. Since chord length is given by 2*r*sin(c/2), where c is the central angle, a is 1/2 chord of circle passing through smaller circles & r2 (radius of circle passing through small centers) is r-a. c can be defined as 360/n. 1/2 chord of large circle is r*sin(180/n) and chord of smaller circle (also a) is r2*sin(180/n).

a/(r*sin(180/n)) = (r-a)/r - simplify for a and evaluate sin in degrees.

Now I'll read everyone else's solutions.

Edited March 26, 2010 by MacGyverish

[Guest]

Sorry, guys, but nobody has gotten it yet. However, jimmy was closest. In fact, when he sees the answer, he's probably going to hit himself...

[Guest]

math is a foreign language, but

a=sqrt(pi/n)

[Guest]

Sorry, guys, but nobody has gotten it yet. However, jimmy was closest. In fact, when he sees the answer, he's probably going to hit himself...

Jimmy's answer is very close to mine, kanth_sri's and McGyverish which are of course all the same other than notation.

To get from Jimmy's answer to our answer requires replacing sin(...) by 1/sin(...)

I'm assuming you are familiar with the notation csc(x) = 1/sin(x) = cosec(x)

as well as the relationship between radians and degrees.

I cannot see anything wrong with my derivation and it is surprising that if such an answer were wrong, 3 people which seem to have done it independently due to differences in notation could have all gotten the same answer. Especially if the problem takes 5 minutes.

Perhaps you can post the solution in a spoiler, and a derivation if possible.

[Guest]

Jimmy's answer is very close to mine, kanth_sri's and McGyverish which are of course all the same other than notation.

To get from Jimmy's answer to our answer requires replacing sin(...) by 1/sin(...)

I'm assuming you are familiar with the notation csc(x) = 1/sin(x) = cosec(x)

as well as the relationship between radians and degrees.

I cannot see anything wrong with my derivation and it is surprising that if such an answer were wrong, 3 people which seem to have done it independently due to differences in notation could have all gotten the same answer. Especially if the problem takes 5 minutes.

Perhaps you can post the solution in a spoiler, and a derivation if possible.

I got the same answer as you guys (mmiguel1, kanth_sri, McGyverish). I'm absolutely sure that a=r*sin(pi/n)/(1+sin(pi/n)) = rsin(180°/n)/(1+sin(180°/n)) = r/(1+csc(pi/n)) is the right solution.

Just in case I made some CAD draw which proved our result.

[Guest]

r(1-1/(1+sin(180/n)))

[Guest]

my apologies, you guys were right after all. I just checked and our answers are in fact the same. Sorry.

[Guest]

r(1-1/(1+sin(180/n)))

same as posted before..

edit: ok you corrected

Edited March 27, 2010 by det

[Guest]

my apologies, you guys were right after all. I just checked and our answers are in fact the same. Sorry.

That's ok.

Nice puzzle.