07th AIME 1989

7th AIME 19891.Find sqrt(1 + 28·29·30·31).2.10 points lie on a circle. How many distinct convex polygons can be formed by connected some or all of the points?3.For some digit d we have 0.d25d25d25 ... = n/810, where n is a positive integer. Find n.4.Given five consecutive positive integers whose sum is a cube and such that the sum of the middle three is a square, find the smallest possible middle integer.5. A coin has probability p of coming up heads. If it is tossed five times, the probability of just two heads is the same as the probability of just one head. Find the probability of just three heads in five tosses. [The original question asked for m+n, where the probability is m/n in lowest terms.]6. C and D are 100m apart. C runs in a straight line at 8m/s at an angle of 60o to the ray towardsD. D runs in a straight line at 7m/s at an angle which gives the earliest possible meeting with C. How far has C run when he meets D?7.k is a positive integer such that 36 + k, 300 + k, 596 + k are the squares of three consecutive terms of an arithmetic progression. Find k8.Given that:x1 + 4 x2 + 9 x3 + 16 x4 + 25 x5 + 36 x6 + 49 x7 = 1;4 x1 + 9 x2 + 16 x3 + 25 x4 + 36 x5 + 49 x6 + 64 x7= 12;9 x1 + 16 x2 + 25 x3 + 36 x4 + 49 x5 + 64 x6 + 81 x7= 123.Find 16 x1 + 25 x2 + 36 x3 + 49 x4 + 64 x5 + 81 x6 + 100 x7.9.Given that 1335 + 1105 + 845 + 275 = k5, with k an integer, find k.10.The triangle ABC has AB = c, BC = a, CA = b as usual. Find cot C/(cot A + cot B) if a2 + b2 = 1989 c2.11.a1, a2, ... , a121 is a sequence of positive integers not exceeding 1000. The value n occurs morefrequently than any other, and m is the arithmetic mean of the terms of the sequence. What is the largest possible value of [m - n]?12. A tetrahedron has the edge lengths shown. Find the square of the distance between the midpoints of the sides length 41 and 13.13.Find the largest possible number of elements of a subset of {1, 2, 3, ... , 1989} with the property that no two elements of the subset have difference 4 or 7.14.Any number of the form M + Ni with M and N integers may be written in the complex base(i - n) as a m(i - n)m + a m-1(i - n)m-1 + ... + a1(i - n) + a0 for some m >= 0, where the digits a k lie in the range 0, 1, 2, ... , n2. Find the sum of all ordinary integers which can be written to base i - 3 as 4-digit numbers.15.In the triangle ABC, the segments have the lengths shown and x + y = 20. Find its area.7th AIME 1989 answers1 869 9 1442 96810 9943 75011 9474 67512 1375 40/24313 9056 160m14 4907 92515 1088 334Outline solutions1. (n-1)n(n+1)(n+2) + 1 = ( n(n+1) - 1)22. There is just one convex polygon for each subset of n points for n = 3, 4, ... , 10. So no. = total no. of subsets - no. of subsets size 0, 1, 2 = 1024 - 1 - 10 - 45.3. Decimal = (100d + 25)/999, so 37n = 3000d + 750. So certainly 750 divides n. If n = 750m, then 37m = 4d + 1. That has unique solution d = 9, m = 1, for d a digit.4. Middle integer n, then 3n is square, so n = 3k2 for some k. Also 5n is cube, so 15k2is cube, so k is a multiple of 15. Smallest value 15 gives n = 3·152.5. 5p(1-p)4= 10p2(1-p)3, so 1-p = 2p, and p = 1/3. Prob of 3H is 10p3(1-p)2.6. Applying cosine rule to velocity triangle, we get that D's vel relative to C is k where 72 = 82 + k2 - 8k. So k = 3 or 5. Hence largest possible k is 5. Time 100/5 = 20s.7. Let the ap be a-d, a, a+d, squares a2-2ad+d2, a2, a2+2ad+d2. So d(2a - d) = 264, d(2a + d) = 296. Hence d = 4, a = 35, so 1225 = 300 + k.8. Label equations given (1), (2), (3). Then (1) - 2 (2) + (3) = 2(x1+ ... + x7), (3) - (2) = 5x1+ 7x2+ 9x3+ ... . Hence 7x1+ 9x2+ 11x3+ ...= (1) - 3 (2) + 2 (3) and required value = (1) - 3 (2) + 3(3).9. The last digits of n and n5 are the same. Hence last digit of lhs is same as that of 3 + 0 + 4 + 7 = 4. Hence last digit of k is 4. Also 133 = 1 mod 3, 110 = -1 mod 3, 84 = 0 mod 3, 27 = 0 mod 3, so lhs = 0 mod 3. Obviously k > 133. So smallest possibility is 144, next is 174. Now 115 = (10 + 1)5 = 105 + 5·104+ 10·103+ 10·102+ 5·10 + 1 = 161051, so 1105 < 2 1010. Obviously, 27 and 84 < 100, so 275 and 845 < 1010. Similarly, 1335 < (1331/10)5 = 1115/105 < 5 1010. Hence lhs < 1011. But 1702 = 28900 > 28000, 1704> 780000000 > 7 108and 1705> 1011. So only possibility is 144.10. wlog c = 1. Obviously a, b must be about 31.5 (remember |a - b| < 1). Suppose angle B = 90o. Then a2 = b2 + c2, so a = 9951/2, b = 9941/2. Hence cot B = 0, cot C = 1/cot A = 9941/2. Hence expr = 994. So if cot C/(cot A + cot B) is independent of a, then it must have value 994. Proving that it is independent is harder, but not strictly necessary.11. Optimum is evidently n+1 terms = 1, n terms = 1000, 999, 998, ... (as far as necessary). n = 1 gives 2 of 1, 1 each of 881, ... , 1000, mean 924.9, diff 923. n = 2 gives 2 each of 942, ... , 1000, mean 946.9, diff 945. n = 3 gives 3 each of 962, ... , 1000, mean 948.6, diff 947. n = 4 gives 4 each of 972, ... , 1000, mean 945.3, diff 944. In the general case the mean is (1001 - (120-n)/n + 1000) (120 - n)/2 + n+1)/121, so we wish to maximise (2002 - 120/n)(120 - n)/2 + n and hence to maximise -1000n - 7200/n, or equivalently to minimise 36/n + 5n. Easily seen this is at n = 3.12. Using the cosine formula twice we find that if a triangle has sides a, b, c and median length m to the middle of the side length a, then 4m2 = 2b2 + 2c2 - a2. So the square of the median from vertex 7, 18, 41 to the side 13 is 577/4 and the square of the median from the vertex 27, 36, 41 to the side 13 is 3881/4. Hence 4 x requd square = 577/2 + 3881/4 - 412.13. can anyone see a neat way to do this?14. We have (i - 3)3 = -18 + 26i, (i - 3)2 = 8 - 6i. The imaginary partmust vanish, so 26a3 - 6a2+ a1= 0. If a3= 1, this implies a2= 5, a1=4; if a3= 2, it implies a1= 9, a1= 2; and we cannot have a3> 2. The 4-digitnumber 1540 in this base = 10, and 2920 in this base = 30. So required sum is 10 + 11 + ... + 19 + 30 + 31 + ... + 39.15. If we let BE vary, the configuration is uniquely determined by the angle APC. Guess that it is 90o. Take DX = 6. Then by similar triangles, Bx = 9. Area ACD = 54. Area ABX = 81, but area BDX = 27, so area ABC = 108. It remains to show that BE = 20. BF = 15 (3, 4, 5 triangle). Line BFE meets three sides of triangle ACD, so (-1/2)(1/1)(CE/EA) = -1. CE = 3 sqrt(13), so EA = sqrt(13), CE = 2 sqrt(13). Applying cosine formulato FEA and FEC, 81 = 52 + y2- 2y sqrt(13) cos CEF, 36 = 13 + y2+ y sqrt(13) cos CEF. Hence 75 = 3y2, y = 5 and BE = 20 as claimed.。