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
