| 1 1279/1024 | 9 11+11√5 |
| 2 19952 | 10 215 |
| 3 3/64 | 11 40 |
| 4 √224 | 12 -3 + 2√2 |
| 5 51 | 13 400 |
| 6 589 | 14 294π - 81√3 |
| 7 13/4 - √10 | 15 3/34 |
| 8 85 |
1. Square area 1, ¼, ¼2 etc. ¾ of each square does not overlap earlier squares. So total area = 1 + ¾(¼ + ¼2 + ¼3 + ¼4) = 1 + ¾(64/256 + 16/256 + 4/256 + 1/256) = 1 + 255/1024.
2. Take logs (all to base 1995). We get ½ + log2x = 2 log x. So log x = 1 + √½ or 1 - √½. Hence x = 19951+√½ or 19951-√½, both positive reals, product 19952.
3. It needs either 4 steps (N, N, E, E in some order) or 6 steps (also a pair of steps in opposite directions). The first has prob 6/44. The second has NNEENS or NNEEEW in some order. There are 6!/3!2!1! = 60 possibilities for NNEENS. But there is a trap. In some of those 60 we already reach (2, 2) on step 4, so we are double-counting. In fact it is easy to see there are 12 such possibilities. So we have 48 possibilities. Similarly for NNEEEW, so 96 in all, prob 96/46 = 6/44, so total prob 3/64.
4. XAC and XBD are similar, so XA = 9. Let O be the center of the large circle. Then XOM is also similar and XO = 15. Hence OM = (15/9) AC = 5. Hence if the chord length is 2x, we have x2 + 52 = 92, so x = √56.
5. Let roots with sum 3 + 4i be α, β. Their sum is not real, so they are not complex conjugates. But the roots occur in complex conjugate pairs (since the polynomial is real), so the other roots γ, δ must be the complex conjugates of α and β. Hence γ + δ = 3 - 4i and αβ = 13 - i. Hence b = (α + β)(γ + δ) + αβ + γδ = 25 + 26 = 51.
6. n2 has 63·39 factors. Apart from the factor n, they form pairs d, n2/d, with one factor of each pair < n. Hence there are (63·39-1)/2 = 1228 factors of n2 less than n. There are (32·20-1) = 639 factors of n less than n, all of which are obviously factors of n2. Hence there are 1228 - 639 = 589.
7. Put k = (1 - sin t)(1 - cos t). Then 5k/4 = cos2t sin2t. But 5/4 + k = 2 + 2 sin t cos t, so 5k = (k - 3/4)2. Solving k = 13/4 ± √10. But k ≤ 2, so k = 13/4 - √10.
8. x/y - (x+1)/(y+1) = (x-y)/(y(y+1)) must also be an integer, so x must be y plus a multiple of y(y+1). Thus we get:
y = 1, x = 3, 5, 7, ... , 99 (49 solutions)
y = 2, x = 8, 14, ... , 98 (16 solutions)
y = 3, x = 15, 27, ... , 99 (8 solutions)
y = 4, x = 24, 44, 64, 84 (4 solutions)
y = 5, x = 35, 65, 95 (3 solutions)
y = 6, x = 48, 90 (2 solutions)
y = 7, x = 63
y = 8, x = 80
y = 9, x = 99
9. Let tan BAM = x. Then 11x = tan BDM = (3x - x3)/(1 - 3x2), so x = 0, ½ or -½. Hence x = ½, and BM = 11/2. Hence AB = (11/2)√5 and perimeter = 11 + 11√5.
10. We need only consider 42a + 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41. Note that 84+1, 42+2, 42+3, 42+7, 84+11, 42+13, 126+17, 126+19, 42+23, 126+29, 84+31, 84+37, 84+41 are composite, so the tricky case is 42n + 5. We have to go to 5·42 + 5 to get a composite number. (So 6·42 + 5 = 42 + (5·42+5), but 215 = 5·42 + 5 cannot be written in the required form).
11. We must cut the longest edges, so the similar piece has dimensions a x 1995 x k for some 1 ≤ k < c. The shortest edge of this piece cannot be a, so it must be k. Thus a x 1995 x c and k x a x 1995 are similar. Hence c = 19952/a, k = a2/1995. Now 1995 = 3·5·7·19, so 19952 has 34 factors, of which (34-1)/2 = 40 are < 1995.
12. Let AX be the perpendicular from A to OB. Let AX = 1, then OX = 1, OA = √2, so BX = √2 - 1. Hence AB2 = 1 + 3 - 2√2, and AC2 = 8 - 4√2. But AC2 = AX2 + CX2 - 2·AX·CX cos θ, so cos θ = -3 + 2√2.
13. f(k) must take one of the values 1, 2, ... , 7. The boundary values are (n + ½)4 = n4 + 2n3 + ½(3n2 + n) + 1/16, giving 5 1/16, 39 1/16, 150 1/16, 410 1/16, 915 1/16, 1785 1/16. So the sum is 5 + 34/2 + 111/3 + 260/4 + 505/5 + 870/6 + 210/7 = 400.
14. OY2 = OA2 - AY2, so OY = 9√9. XY2 = OX2 - OY2, so XY = 9 and ∠OXY = 60o. Hence DX = AX = AY + XY = 48, BX = CX = 30. Also ∠AXB = 180o - 2·60o = 60o. So area triangle AXB = ½AX·BX sin 60o = 360√3. ∠ACB = ∠XBC and their sum is ∠AXB = 60o, so ∠ACB = 30o. Hence ∠AOB = 60o. So area sector OAB = 422π/6 and area between AB and minor arc AB = 294π - 422(√3)/4. Hence required area = 294π - 81√3.
15. Let p = prob that first toss is H and 5H occurs before 2T, q = prob that first toss is T and 5H occurs before 2T. We have q = ½p, because the T must be followed by an H. Also p = q/2 + q/4 + q/8 + q/16 + 1/32 (there are either 1, 2, 3, 4, or 5 Hs in the initial run) = (15/16)q + 1/32 = (15/32)p + 1/32. So p = 1/17, q = 1/34.
© John Scholes
jscholes@kalva.demon.co.uk
3 Oct 2003
Last updated/corrected 3 Oct 03