1st AIME 1983 answers

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1   60 9   12
2   1510   432
3   2011   288
4   √2612   65
5   413   448
6   3514   130
7   11/4615   7/25
8   61

Outline solutions

1.   w = x24 = y40 = x12y12z12, so w10 = x120y120z120 = w5w3z120, so w = z60.

2.   |x - p| = x - p, |x - 15| = 15 - x, |x - p - 15| = 15 + p - x, so sum = 30 - x, minimum 15, when x = 15.

3.   Careful - the quartic has 4 real roots, but for two of them x2 + 18x + 30 is negative, whereas sqrt is defined to be positive.

4.   eg use tan(OAB + BAC) to get tan OAB = 1, hence cos OAB = 1/√2, then cosine formula.

5.   Get cubic for w+z, roots -5, 1, 4.

6.   6 = 7-1, 8 = 7+1 and binomial.

7.   For two knights, there are 3 disjoint possibilities: (1) adjacent, prob 1/12, (2) one seat between them, prob 1/12, (3) more than one seat between them, prob 5/6. In case (2), prob third sits adjacent is 3/23, in case (3) it is 4/23. Hence 1/12 + 1/92 + 10/69 = 11/46.

8.   If p is a prime such that 100> p > 200/3, then p and 2p but not 3p belong to {1, 2, ... , 200}, whilst just p belongs to {1, 2, ... , 100}, so p does not divide 200C100. But consider 61 (the largest prime < 200/3). 613 divides 200!, but only 61 divides 100!, hence 61 divides 200C100.

9.   By AM/GM we have 9y + 4/y ≥ 2√(9y·4/y) = 12, with equality iff 9y = 4/y or y = 2/3. But x sin x is continuous, 0 for x = 0 and π/2 for x = &pi/2, so it is 2/3 for some intermediate value. Hence the minimum value of 12 is realised.

10.   72 each of 1xxy, 1xyx, 1xyy, 1xy1, 1x1y, 11xy.

11.   If we take EF central and parallel to AB, then two such blocks can be put together to form a regular tetrahedron side 12 √2.

12.   Put AB = 10m+n, then OH = 3/2 √(11(m+n)(m-n)), so 11 divides m+n and m-n is square. Hence m=6, n=5.

13.   For X a subset of {1, 2, 3, 4, 5, 6} the sum for X and (X union {7}) is 7. Hence 7·64 = 448.

14.   Put QP = 2a, take perp dist A, B from QP be b, c. Take line parallel to QP through B meeting perp bisector of QP at D. Considering triangle ADB gives 4a2 + (b-c)2 = 144. Also b2 = 64 - a2, c2 = 36 - a2. Hence 4a2 = 130.

15.   If M is the midpoint of AD, then the locus of M as D varies is the circle diameter AO. We want this to touch BC.

 

1st AIME 1983

© John Scholes
jscholes@kalva.demon.co.uk
16 June 2003
Last updated/corrected 6 Oct 03