User:KEVIN LIU

So for Question #20 Amc 10b, we can see that we can create a square from the edges of each diameter because each semicircle is congruent. From there, we realize that because the diagonals are 4, the side lengths of the square are 4/sqrt(2). We realize that we can draw a right triangle connecting the center of the circle to the edge of the sphere and the center of the diameter of one of the semicircles. With this information, we realize that we must set one of the lengths as half of the length of the side of the square so we get 4/2sqrt(2). using Pythagorean theorem, we get 2+x^2 = 4(the radius squared of the sphere which happens to be the hypotenuse). we solve for x and we get sqrt2 which is the radius of each semicircle. The circumference of each is pi*sqrt(2). we multiply that by 4 and get 4sqrt2*pi and that equals sqrt(32)pi so we get A, 32. ~KEVIN_LIU

So for question #16 on AMC 10B, we realize the only way to get a different amount of upnos and downos is due to the number 0 as 012 = 12 but 120 =/ 12. We also realize that their is a limit on how many numbers there are because of the fact that we can only have a number up to 10 digits of only increasing/decreasing digits. Therefore, we consider our cases. We can try cases and we just realize that as the amount of digits increases, the number of ways of choosing them increase then decrease. In our first case with 2 digit numbers, we have 9 numbers and we can only choose 1 because 0 must be the last digit. In a 3 digit number, we have 9 choose 2 because there are no 2 places to put numbers because once again, 0 must be in the last spot.. We do this until we have no more cases. 9C1 + 9C2+....+9C9. We can simplify this to the form of 2(9C1+9C2+9C3+9C4)+9C9 = 510 + 1 = 511 cases. ~KEVIN_LIU