Two questions

First:

Sally has five red cards numbered 1 through 5 and four blue cards numbered 3 through 6. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the nummbers on the middle 3 cards?


Two:

A semicircle of diameter 1 sits at the top of a semicircle of diameter 2. The shaded area inside the smaller semicircle and outisde the larger semicircle is called a lune. Determine the area of this lune.

Question two is quite easy:

Suppose r1 is half of diameter 1 and r2 is half of diameter 2. Extend the large semicircle to a full circle, since it doesn't really matter in this situation, but is important to understand my solution. The arc of the large semicircle covered by the small semicircle forms a sector of the large circle. Through this sector runs the bottom line of the small semicircle, with length 2 * r1 (= diameter 1). This line divides the sector into a segment above the line and a triangle under the line. The height of the triangle is sqrt(r2²-r1²). The area of the triangle is h * r1 (since r1 is the half basis as well).
The half angle of the sector, alpha, can be calculated: sin alpha = r1 / r2. Now it's easy to calculate the whole area of the sector (which is pi * r2² * 2 alpha / (2 * pi) = r2² * alpha). The area of the segment should be subtracted from the area of the small semicircle, and presto!

I didn't solve question 1, since I don't really understand your question.

Hmm, I was assuming that when you mentioned 'dividing' you meant integers. Then it would be impossible, so I guess you didn't.

By 'divides evenly' i guess you mean 'is a factor', not 'divides leaving an even number.

In that case, some observations:
4 (red) must be next to 4 (blue)
5 (red) must be next to 5 (blue)

You ca kinda work your way from that to:
4 4 2 6 3 3 1 5 5
(red 4 - blue 4 - red 2 - ...etc)

=> sum of middle 3 numbers = 12

Yeah, I got 12 too.

But then I had doubt because 1 doesn't divide evenly with 5, and 2 does not divide evenly with 6.

Quote:

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Originally posted by riis
Question two is quite easy:

Suppose r1 is half of diameter 1 and r2 is half of diameter 2. Extend the large semicircle to a full circle, since it doesn't really matter in this situation, but is important to understand my solution. The arc of the large semicircle covered by the small semicircle forms a sector of the large circle. Through this sector runs the bottom line of the small semicircle, with length 2 * r1 (= diameter 1). This line divides the sector into a segment above the line and a triangle under the line. The height of the triangle is sqrt(r2²-r1²). The area of the triangle is h * r1 (since r1 is the half basis as well).
The half angle of the sector, alpha, can be calculated: sin alpha = r1 / r2. Now it's easy to calculate the whole area of the sector (which is pi * r2² * 2 alpha / (2 * pi) = r2² * alpha). The area of the segment should be subtracted from the area of the small semicircle, and presto!

I didn't solve question 1, since I don't really understand your question.

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1 - alpha? What is alpha, bud.

Alpha is one of the angles of the described triangle. Alpha is located at the center of the large circle (where r=r2). The triangle's opposite side is r1, and the diagonal side is r2, hence sin(alpha) = r1/r2.