| 1 384 | 9 17/32 |
| 2 26 | 10 600 |
| 3 198 | 11 85 |
| 4 32 | 12 182/729 |
| 5 986 | 13 401 |
| 6 315 | 14 25 |
| 7 757 | 15 864 |
| 8 61 |
1. x1x2 = 2, x3x4 = 4, x5x6 = 6, x7x8 = 8. Multiply.
2. AB·BC2/3 = 800, AB·BC2/3 = 1920, hence AB·BC = 240. So BC = 10, so AB = 24. Pythagoras gives AC.
3. n(3m2 - n2) = 107, which is prime, so n = 107 or 1. Only n = 1 gives m positive integer. So m = 6. N = 216 - 3·6 = 198.
4. AP = √(1-x2), so by similar triangles, x/1 = √(1-x2)/(1 - 1/n). Hence x = 1/√(1 + (1 - 1/n)2), so the side of the small square is x/n = 1/√(n2 + (n-1)2). It is easy to check that n2 + (n-1)2 = 1985 for n = 32.
5. The sequence is a1, a2, a2-a1, -a1, -a2, a1-a2, a1, a2, ... , so it is periodic with period 6. Also the sum of a complete period of terms is 0. 1492 = 4 mod 6, 1985 = 5 mod 6, 2001 = 3 mod 6, so the sum of the first 1492 terms is a1 + a2 + a3 + a4 = 2a2 -a1, the sum of the first 1985 is a1 + ... + a5 = a2 - a1. Hence a2 = (2a2 -a1) - (a2 - a1) = 493, and the sum of the first 2001 = 2a2 = 986.
6. We have area APD/area CPD = area ABD/area CBD = area ABP/area CBP. Hence x/84 = 70/(x+35). Similarly, y/35 = (x+84)/70. Solving the simultaneous equations gives x = 56, y = 70.
7. We must have A = m4, B = m5 for some m, and C = n2, D = n3 for some n. Hence n2 = m4 + 19. The difference between n2 and the next smallest square (n-1)2 is 2n-1, so n is at most 10. Checking n = 1, 2, ... , 10, we find that only n = 10, m = 3 works.
8. The sum of the 7 numbers is 19. Clearly all 7 integers must be 2 or 3 (because that gives E < 1, whereas any other integer gives E > 1). There must be five 3s and two 2s. To minimise E we must take 2 for the smallest numbers, giving errors 0.56, 0.61, 0.35, 0.29, 0.21, 0.18, 0.14, so E = 0.61 and 100E = 61.
9. Chords of the same length subtend the same angle, so we can take AB = 2, BC = 3, then AC must be 4, and ∠ACB = x/2. By cosine formula, cos x/2 = 7/8, so cos x = 2 cos2x/2 = 17/32.
10. Put f(x) = [2x] + [4x] + [6x] + [8x]. We find f(1/8) = 1, f(1/6) = 2, f(1/4) = 4, f(1/3) = 5, f(3/8) = 6, f(1/2) = 10, and f(x + n/2) = f(x) + 10n. So we realise 6 out of 10, or 600 out of 1000.
11. We need the following fact about the ellipse: for a point P on the ellipse the sum of the distances PF, PF' to the two foci is constant (and hence equal to 2a, the length of the major axis). If follows immediately that for a point P' outside the ellipse P'F + P'F' > 2a. Take F (9,20), F' (49, 55). Reflect F' in the x-axis to G (49, -55). Let FG cut the x-axis at P' and let the x-axis touch the ellipse at P. Then since P' is not inside the ellipse FG = FP' + F'P' ≥ FP + F'P. But FG ≤ FP + FP'. Hence equal. So (2a)2 = (49-9)2 +(55+20)2 = 852.
12. Let pn be the prob of return after distance n. Then pn+1 = (1-pn)/3, because to return after n+1, it must have been at a different vertex after n, and then chosen the correct edge to return. Obv p1 = 0, so p2 = 1/3, p3 = 2/9, p4 = 7/27, p5 = 20/81, p6 = 61/243, p7 = 182/729.
13. Let gcd be d. d must divide difference 2n+1. Hence also 2n2+n and 2(n2+100), hence n-200. Hence also (2n+1) - 2(n-200) = 401. So certainly d ≤ 401. But taking n = 601 gives 401 divides 100+n2 and 2n+1 and hence d ≥ 401.
14. n+10 players. The 10 worst played 45 games amongst themselves. They must have got between them 45 points from these games and hence 90 points in total. The top n players played n(n-1)/2 games amongst themselves, giving them n(n-1)/2 points. Hence they got n(n-1) points in total (the rest from the 10 worst). So the total points scored by everyone were 90 + n(n-1). But total is (n+10)(n+9)/2. Hence n2 - 21n + 90 = 0, n = 6 or 15. But the top n get n(n-1) points in total, an average of n-1 each, and the bottom 10 get an average of 9 each. Hence n ≥ 10. So n = 15.
15. A little thought shows that gluing the hexagons of two such polyhedra together gives a cube. Hence volume 123/2
© John Scholes
jscholes@kalva.demon.co.uk
18 June 2003
Last updated/corrected 18 Jun 03