AMC 10 2012A Review

1. AMC 10 2012AReview

2. Each non-leap year has 365 days, 365 = 1 (m 7)  1 extra day Each leap year has 366 days, 366 = 2 (m 7)  2 extra days For 200 years, there are 200 / 4 = 50 possible leap years; but year 1900 is not a leap years, so there are 49 leap years Total extra days = (200 – 49) * 1 + 49 * 2 = 249 Again, 249 = 4 (m 7)  4 extra days This means that, 2/7/2012 has 4 extra weekdays from the date Dickens was born. 4 days before Tuesday is Friday. Answer: (A)

3. Suppose A, B, C, running at speed of 4.4, 4.8, 5.0 meters/sec. Note that A lose .4 meters per second behind B C gains .2 meters per second ahead of B Thus every 500/.4 = 1250 seconds, A lose a full circle behind B and every 500/.2 = 2500 seconds, C gains a full circle ahead of B The LCM(1250, 2500) = 2500 would be the time A, B, C meet again where A falls two full circles behind B, and C gains one full circle ahead of B. Answer: 2500 (C)

4. a3 – b3 = (a – b) (a2 + ab + b2) a3 – b3 / (a – b)3 = (a2 + ab + b2) / (a2 - 2ab + b2 ) = 73/3 3a2 + 3ab + 3b2 = 73a2 - 146 ab + 73b2 70a2 - 149ab + 70b2 = 0 (7a – 10b) (10a – 7b) = 0 Hence a = 10/7 b or a = 7/10 b Since a > b > 0, we can only have a = 10/7 b  a/b = 10/7 Since a and b are relatively prime, we get a = 10, b = 7 Answer: a – b = 10 – 7 = 3 (C)

5. Note that the enclosed area is: area(hexagon) + 6 of the area(120-fan) – 3 * area(120-fan) Area(hexagon) = 6 * (area of equilateral triangle of size 2) = 6 * (1/2 * 3 * 2) = 6 3 Area(120-fan) = 1/3 *12 * =  /3 Answer = 63 +  /3 * 3 = 63 +  (E)

6. Let Vp be speed of Paula, and V be speed of helpers. Let X be the # of minutes for lunch break (480 – X) (Vp + V) = 50% ------- (1) (372 – X) V = 24% ------- (2) (672 – X) Vp = 26% ------- (3) From (2)+(3): (372 – X) V + (672 – X)Vp = 50% ----- (4) From (1), (4): (372 – X) V + (672 – X)Vp = (480 – X) (Vp + V)  372 V + 672 Vp = 480 Vp + 480 V  192 Vp = 108 V  V/Vp = 16/9 ----- (5) From (2)/(3): (372 – X)/(672 – X) (V/Vp) = 24/26  (372 – X)/(672 – X) * 16/9 = 24/26  (372 – X)/(672 – X) = 27/52  25 X = 372 *52 – 672 * 27 = 1200  X = 48 (D)

7. We know the center piece doesn’t change color. So the probability for the center to be black is ½. Consider the 4 corners: 7 ways to succeed If there are 4 blacks, always succeed, and 1 way to paint so. If there are 3 blacks, always succeed, and 4 way to paint so. If there are 2 blacks, only succeed if opposite blocks are white. And there are 2 ways to paint so. If there are 1 blacks, will never succeed. Similarly we can find 7 ways to make all non-corner blocks black. Total ways to succeed (for the 8 non-center block): 7 * 7 = 49 Total ways to paint 8 non-center block: 28= 256 Answer: ½ * 49/256 = 49/512 (A)

8. Calculate the coordinates of E, F, G, H: c E=(½, 0, 3/2); F=(½, 0, 0), G=(0,1,0), H=(0,1,3/2) G Hence |EF|=|GH|=3/2; |FG|=|HE|=5/2 H F Also EF // GH; FG // HE a b Now Vector(EF) = (0, 0, -3/2) E Now Vector(FG) = (-1/2, 1, 0) d Calculating the dot-product of EF, FG: Vector(EF) * Vector (FG) = 0 * (-1/2) + 0*1 + (-3/2) * 0 = 0 Hence EF  FG  EFGH is a rectangle Area(EFGH) = 3/2 * 5/2 = 35/4 (C)

9. Sum of the first m odd integers = 1 + 3 + … + 2(m-1) + 1 = m2 Sum of the first n even integers = 2 + 4 + … + 2n = n (n + 1) m2 = n (n + 1) + 212  n2 + n + 212 – m2 = 0 n = (-1 + (12 – 4(212 – m2))/2 = (-1 + (m2 - 847))/2 ---- (1) For n to be an integer, m2 – 847 must be the square of an odd int. Assume m2 – 847 = k2, where K is an odd integer m2 – k2 = 847 = 7 * 112 (m + k) * (m – k)= 7 * 112 Since (m+k), (m-k) are integers, and (m-k) < (m+k), only possible values for m-k are: 1, 7, 11 Solving for m – k = 1, 7, 11 m + k = 847, 121, 77 We get m = 424, 64, 44; k = 423, 57, 33 From (1): n = 211, 28, 16. Answer: 211+28+16 = 255 (B)

10. Let N be the # of friends each one can have. We have 1  N  4 If N = 1, we can form 3 pairs of friend. For the 1st person, he has 5 choices, then his friend has 1 choice. The 3rd person will then have 3 choices, & his friend has 1 choice. The 5th and 6th person will only has 1 choice Total choice for N =1: 5 * 3 = 15 If N = 2, we can form 2 rings of 3-friends or 1 ring of 6-friends. For the 1 ring of 6-friends, we can have 5*4*3*2/2 = 60 ways For the 2 rings of 3-friends, we can have 5*4/2 = 10 ways to split into 2 groups, and each group only one way to form the friend. Total choices for N = 2: 60 + 10 = 70 Note that N=3 is the same as N = 2, and N=4 is the same as N =1 Total ways = 15 + 70 + 70 + 15 = 170 (B)

11. a2 - b2 - c2 + ab = 2011 ----- (1) a2 +3b2 + 3c2 – 3ab – 2ac – 2bc = -1997 ----- (2) (1) + (2): 2a2 +2b2 + 2c2 – 2ab – 2ac – 2bc = 14 (a - c)2 + (a - b)2 + (b - c)2 = 14 Since a  b  c, (a-c)  (a-b), and (a-c)  (b-c) (a – c) is the largest compared to (a-b) and (b-c) Note that (a - c)2 , (a - b)2 , (b - c)2 are all positive integers, (a-c) can only be 3! And we have either a – b = 1 & b – c = 2 or a – b = 2 & b – c = 1 Putting: c=a-3, b=a-1 into (1)  a = 2021/7 -- invalid Solving: c=a-3, b=a-2 into (1)  a = 2024/8 = 253 (E)

12. For 0  x, y, z  n, the total possible space is n3 Assume 0  x  y  z  n, total possible space will be n3/6(since there are 6 permutations of x, y, z) For no two of x, y, z are within 1 unit, we have: y – x > 1 and z – y > 1  x < y - 1 and y < z - 1 Let y’ = y - 1  x < y’ and y’ + 1 < z - 1 Let z’ = z – 2  x < y’ and y’ < z’ Hence we get: 0  x  y’  z’  n - 2 Again, for 0  x, y’, z’  n-2, the total possible space is (n-2)3 Thus for 0  x  y’  z’  n-2, the total possible space is (n-2)3/6 Hence we need (n-2)3/6 / (n3/6) > ½  (n-2)3/n3 > ½  smallest n = 10