Find |K| of six congruent circles forming a ring

[p53ud0 dr34m5]

Six congruent circles form a ring with each circle externally tangent to two circles adjacent to it. All six circles are internally tangent to a circle C with radius 30. Let K be the area of the region inside C and outside all of the six circles in the ring. Find |K|. (The notation |K| denotes the greatest integer that is less than or equal to K.)

I've already worked out the problem and my work yielded 942. Show your work and good luck. I have more problems after this one.

 

[Justin Lazear]

I'm in the mood for one of these, so I'll bite...

Draw the r = 30 circle and divide it into 6 equal parts by drawing six lines to the middle. The angle between each of these radial lines is naturally 60 degrees. Additionally, we can position the six circles such that each of these lines is tangent to two of the circles.

We need only figure out the radius of one circle, so consider only one of them. We then have a situation where we have an external point of this circle and two tangents of this circle intersecting at this point. The angle between the lines from this point is 60 degrees, as before. Tangents intersect the circle perpendicularly, so draw the radii from the tangent intersections to the two points. Each of these radii naturally have radius r. Draw the line connecting the center of the circle and the exterior point, which is also a bisector of the 60 degree angle. This gives us two 30-60-90 triangles, with short side r. The hypotenuse of this triangle is the segment connecting the exterior point to the center, and it's easy to determine (since the triangle is 30-60-90) that its length is 2r. Extend the line connecting the exterior point and the center of the circle so that it intersects the far edge of the circle. This adds an additional distance r to this segment, so the distance between the exterior point and the far edge of the circle is then 3r.

Next, note that the interior circle must intersect the exterior circle as far away from the center of the exterior circle as possible, and that this segment with length 3r connects the center of the exterior circle to the farthest point on the interior circle, and therefore is a radius of the exterior circle. This gives the equation 3r = 30, easily solved for r, r = 10.

Finally, compute the area of the exterior circle and compute the area of the interior circle (easy, since we know r = 10), and subtract 6 of the latter from the former. This gives us the final equation \pi (30^2 - 6\cdot 10^2) = 300\pi = 942.478, so \lfloor 942.478 \rfloor = 942.

--J

 

[p53ud0 dr34m5]

thats exactly how i approached the problem earlier today. nice, JL.

heres another one for everyone:

For each positive integer k, let S_k denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is k. For example, S_3 is the sequence 1,4,7,... For how many values of k does S_k contain the term 2005?

i think the answer is 12, but I am not sure.

 

[Euclid]

For each positive integer k, let S_k denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is k. For example, S_3 is the sequence 1,4,7,... For how many values of k does S_k contain the term 2005?

i think the answer is 12, but I am not sure.

The nth term in the sequence is 1 + (n-1)k. The question posed is thus for which k does there exist an n such that 2005 = 1 +(n-1)k, i.e., 2004=mk, where m=n-1. This is simply asking which k divide 2004. The factorization is 2*2*3*167. So, your answer is k=1, 2, 3, 4, 6, 12, 167, 334, 501, 1002, 2004.

 

[tongos]

no discussion of aime problems until tomorrow, thursday. Even though most of us don't even know that is. nice try.

 

[p53ud0 dr34m5]

haha, thanks. i actually looked on my packet thing to make sure, but i didnt see anything on my packet. my teacher didnt tell us when not to, so thanks for telling me! tongos, how do you think you did on the aime?

 

[tongos]

I'm taking the AIME2 on March 22, 2005. Although by the looks of this AIME, I should've taken it, but i very sick yesterday. So i came in today and got the test booklet. I'm thinking that i will get a 10 or 11 on the AIME2. So I am pretty sure that i will go all the way to USAMO. On other forums, such as the artofproblemsolving, you are not allowed to post or discuss specific problems until thursday to my understanding. How did you do?