So the bounded region is the parallelogram shown, area 20·40 = 800.
| 1 25 | 9 60-12√15 |
| 2 480 | 10 100+50√2 |
| 3 800 | 11 525 |
| 4 3/14 | 12 7+3√5 |
| 5 40 | 13 -352+16i |
| 6 308 | 14 130 (3,11) |
| 7 19600 | 15 761 |
| 8 618 |
1. 1212 = 224312, 66 = 2636, 88 = 224. We must have k = 2a3b (since 2, 3 are the only primes dividing the lcm). a can be any of 0, 1, ... , 24, and b must be 12.
2. There are 302 points (m, n) with 1 ≤ m, n ≤ 60/2. We must exclude those with 2m < n or 2n < m. By symmetry, there are the same number with 2m < n and 2n < m and none satisfy both conditions. The number of points satisfying 2m < n is 0 for n = 1, 2; 1 for n = 3, 4; 2 for n = 5, 6; ... ; 14 for n = 29, 30. So total 2(1 + 2 + ... + 14) = 210. Hence requd no. is 900 - 210 - 210.
3. For x ≥ 0, the equation is (y+20)(y+2x-20) = 0. For x ≤ 0, the equation is (y-20)(y+2x+20).
So the bounded region is the parallelogram shown, area 20·40 = 800.
4. There are 5 odd tiles and 3 even tiles. Each player must get an odd number of odd tiles, so one player must get 3 odd tiles, and the other two 1 odd tile each. There are 9C3 = 84 ways for a player to get 3 tiles from 9 and 5C3 = 10 ways to get 3 odd. Hence prob 5/42 for a given player to get all odd. There remain 6 tiles, 2 odd. So the prob the next player has prob (1/3)(4/5)(3/4) x 3 = 3/5 of getting just 1 odd (and then the last player must get 1 odd). Hence prob. (5/42)(3/5) = 1/14. But any of the three players can get all odd (and these are disjoint poss), so total prob 3/14.
5. ½n(n-1) is odd for n = 2, 3 mod 4, even for n = 0, 1 mod 4. Thus A4n-1 + A4n = -½(4n-1)(4n-2) + ½4n(4n-1) = 4n-1, A4n+1 + A4n+2 = -(4n+1), and A4n-1 + A4n + A4n+1 + A4n+2 = -2. Hence A19 + ... A98 = -20·2 = -40.
6. By similar triangles 735/(RC+112) = AP/BC, 847/RC = PD/BC = 1 + AP/BC = 1 + 735/(RC+112). hence RC2 = 847·112, RC = 308.
7. There are n ordered pairs summing to 2n. Consider triples summing to 2n+1: if 1st number is 1, then n; if 1st is 3, then n-1 etc, so n(n+1)/2. Now consider 4-tuples summing to 2n+2. If 1st number is 1, then n(n+1)/2; if 1st is 3, then (n-1)n/2 etc. So total (n2 + (n-1)2 + ... + 12 + n + n-1 + ... + 1)/2 = n(n+1)(n+2)/6. Hence 48·49·50/6 = 19600.
8. The terms are 1000, n, 1000-n, 2n-1000, 2000-3n, 5n-3000, 5000-8n, 13n-8000, 13000-21n, 34n-21000, 34000-55n, 89n-55000. For the 3rd term to be non-neg, we need n ≤1000, for the 4th n ≥500, for the 5th n < 667, for the 6th n ≥ 600, for the 7th n ≤ 625, for the 8th n > 615, for the 9th n ≤ 619, for the 10th n > 617, for the 11th n < 619, for the 12th n > 617. Thus for n ≤ 617 the 10th term is negative, and for n ≥ 619 the 11th term is negative. But for n = 618 at least the first 12 terms are positive.
9. The x-axis give the arrival time of one (in minutes after 9am), the y-axis the arrival time of the other. The arrival times can differ by up to m mins, so the shaded band gives the arrival pairs (x, y) for which they meet. Thus we need (60-m)2/602 = 0.6, giving m2 - 120m + 1440 = 0, roots 60 ± √2160. But m < 60, so m = 60 - 12√15.
10. The centers of the 8 spheres form an octagon. Let the distance of the center O of this octagon from each sphere-center be d. Let the central sphere have radius r. Then the triangle shown has sides r+100, d, r-100, so d2 = 400r. Two adjacent sphere centers and O form a triangle with sides d, d, 200, and angle 45o opp the 200, so by the cosine rule 2002 = 2d2(1 - 1/√2), or d2 = 2002/(2-√2). Thus r = 100/(2-√2) = 50(2+√2).
11. We are given the points A, F, E below. Let B be the midpoint of its edge. Then BE is parallel to AF, so B also belongs to the plane. Hence the center of the cube, O, belongs to the plane, and so CD, obtained by rotating AF about O belongs to the plane.
AF = 15√2. BE = 20√2, so AC/2 = √(AB2 + ((5√2)/2)2) = √(125 + 25/2) = 15/√2. So AC = 15√2. Hence area hexagon = AC·AF + AC(5/√2) = 450 + 75 = 525.
12. Compare the figures APRE and CRQD. If PE > RD, then RE > DQ, so QF > RD. Similarly that implies QF > PE, and that implies PE < RD. Contradiction. Similarly if PE < RD. So PE = RD. Similarly PE = QF. Put PF = QD = RE = x, AF = 1. Then BQD, PQF are similar, so x/1 = (1-x)/x or x2 + x - 1 = 0, so x = (√5 - 1)/2 (we discard the negative root). By cosine rule PQ2 = x2 + (1-x)2 - x(1-x) = 3x2 - 3x + 1 = 3(x2+x-1)+4-6x = 7-3√5. Area ABC/area PQR = 22/x2 = 7 + 3√5.
13. There are 8Ck subsets A of {1, 2, ... , 8} with k elements. If we adjoin 9 to one of these we add 9ik+1 to f(A). So adjoining 9 to each of the subsets of {1, 2, ... , 8} increases the sum by ∑ 8Ck 9ik+1 = 9i (1+i)8. The subsets of {1, 2, ... , 9} are the subsets of {1, 2, ... , 8} and the subsets of {1, 2, ... , 8} with 9 adjoined to each. Hence S9 = 2S8 + 9i(1+i)8. Note that (1+i)2 = 2i, so (1+i)8 = 16. Hence S9 = -352 - 128i + 144i = -352 + 16i.
14. c = 2(a+2)(b+2)/(2ab-(a+2)(b+2)) = 2(a+2)(b+2)/(ab-2a-2b-4) = 2 + (8a+8b+16)/(ab-2a-2b-4). So (c-2)/8 = (a+b+2)/(ab-2a-2b-4). To make progress we need to realise that it helps to put c'=c-2, b'=b-2, a'=a-2, so c'/8 = (a'+b'+6)/(a'b'-8). Now we cannot have a or b = 1 or 2, because (1+2)/1 and (2+2)/2 are already ≥ 2 (so multiplying by (c+2)/c gives > 2). So a', b' ≥ 1, so a'+b' ≤ a'b'+1. Hence c'/8 ≤ (a'b'+7)/(a'b'-8) = 1 + 15/(a'b'-8) ≤ 1 + 15. Hence c ≤ 130 and it is easy to check this is realised.
15. There are 39 elements [m, 40]. Unless 39 appears as the first or last number in the line, it must occur an even number of times. Similarly for the other numbers. So 38 numbers and hence 19 dominos [m, n] cannot appear. So the sequence cannot be longer than 780 - 19 = 761. We claim that a sequence of 1 + (4 + 8 + ... + 4n-4) with first number 1 and last number 2 is possible for the set of dominos with largest number 2n. Induction. n = 1. [1,2] gives 1. [1 2n-1][2n-1 2][2 2n][2n 3][3 2n-1][2n-1 4] ... [2n-2 2n][2n 1] prefixed to the line for 2n-2 gives 4n-4 more dominos than the line for 2n-2. (The extra dominos are the pairs [2n-1 a] and [2n a] for a = 1, 2, ... , 2n-2.) Putting n = 40 shows that 761 is realised.
© John Scholes
jscholes@kalva.demon.co.uk
7 Oct 2003
Last updated/corrected 7 Oct 03