| 1 29 | 9 (√255)/2. |
| 2 99/19 | 10 16/473 |
| 3 38 | 11 tan(175o/2) |
| 4 86/99 | 12 345 |
| 5 223 | 13 40!/2780 |
| 6 314 | 14 168/295 |
| 7 650 | 15 408 |
| 8 11/18 |
1. Best is 5,11,17,23,29
2. The line through (0,0) and (10,45+x) must pass through (28,153-x), so x = 270/38. Hence slope (45 + 270/38)/10 = 99/19.
3. We have n2 - 19n + 99 > n2 - 20n + 100 = (n-10)2 for n ≥ 1. Also < (n-9)2 = n2 - 18n + 81 for n > 18. So cannot be a square for n > 18. Also if f(n) = n2 - 19n + 99, then f(19-n) = f(n). So we just have to check the values 1, 2, ... , 9. We find n = 1,9 and hence also 10,18 work. Sum 38.
4. Let the sides of the triangle be x, y, √(x2+y2). It is clear from the diagram that the perimeter of the triangle is the side of the square 1. Solving, y = (1-2x)/(2-2x) and octagon side = 1-x-y = (2x2-2x+1)/(2-2x) = 43/99. Hence 198x2 - 112x + 13 = 0. Product of roots is 13/198. So area octagon = area square - 4 x area triangle = 1 - 2xy = 1 - 26/198 = 86/99.
5. If there are k carries in going from n to n+2, then the digit sum of n+2 is 9k-2 smaller than that of n. So possible values of t(n) are 2 and 7+9k, for k = 0, 1, 2, ... . Largest < 2000 is k = 221: 7 + 221·9 = 1996. Hence 223 values < 2000.
6. Side joining (900,300) and (1800,600) lies along line y=x/3. So corresponding side of T(Q) is also a straight line, namely y = x/√3, which makes angle 30o with x-axis. Reflecting in y = x, we see that side joining (300,900) and (600,1800) goes to line along y = 3x, which makes angles 30o with y-axis. The vertices (1800,600) and (600,1800) lie on x + y = 2400, so corresponding side of T(Q) is arc of circle radius √2400. Similarly 4th side is arc of circle radius √1200. Angle between y=x/3 and y=3x is 30o, so area = (π/12)(2400 - 1200) = π/100.
7. 2a3b5c divides 2A3B5C iff a ≤ A, b ≤ B and c ≤ C. So there are (10-a)(10-b)(10-c) multiples of 2a3b5c. Thus it ends up in position A iff 4 divides (10-a)(10-b)(10-c). So we have: a = 2,6, any b, c, 200 poss; a = 0,4,8 b,c not both odd 3(100-25) = 225 poss; a odd, b = 2,5, any c, 100 poss; a odd, b odd, c 2 or 6, 50 poss; or a odd, b 0,4,8, c even 75 poss. Total 650.
8. T is an equilateral triangle as shown. S is the yellow region. The three small equilateral triangles that make up T-S have area T/22, T/32, T/62, total (14/36) area T, so area S/area T = 11/18.
9. f(z) is equidistant from 0 and z iff a = 1/2. So need 1/4 + b2 = 65, b=(√255)/2.
10. There are 10C3 ways of choosing a triangle, then 42 ways of choosing a 4th segment, whereas there are 45C4 ways of choosing 4 segments. Hence prob = (10·9·8·42·24)/(6·45·44·43·42) = 16/473.
11. 2 sin 5 sin 5k = cos(k-1)5 - cos(k+1)5. So (2 sin 5) sum = cos 0 + cos 5 - cos 175 - cos 180 = 2 + 2 cos 5. Hence sum = (1+cos 5)/sin 5. We still have to put it into the required form. cos 2x = 2 cos2x - 1, and sin 2x = 2 sin x cos x, so sum = cot(5/2) = tan 87.5o = tan(175/2).
12. Put r = inradius, s = semiperimeter. So s-a = 46, s-b = 54, c = 50. Area = rs = √(s(s-a)(s-b)(s-c)). So s = 345/2, perimeter = 345.
13. Team totals must be 0, 1, 2, ... , 39. So we must be able to order the teams as T1, T2, ... , T40, so that Ti loses to Tj for i < j. In other words, this order uniquely determines the result of every game. There are 40! such orders and 780 games, so 2780 possible outcomes for the games. Hence prob = 40!/2780.
14. P is a Brocard point. We have cot PAB = cot A + cot B + cot C (*). Using cosine formula we find cos A = 33/65, so sin A = 56/65 and cot A = 33/65. Similarly, cos C = 3/5. We recognize 3,4,5 triangle, so cot C = 3/4. Finally cos B = 5/13. We recognize 5,12,13 triangle, so cot B = 5/12. Hence cot PAB = 295/168.
To prove (*) take D so that BDA is similar to ABC. Then P lies on CD. So cot PCA = CK/DK = CA/DK + AK/DK. If the altitude from B is BE, then CA/DK = CE/BE + EA/BE = cot C + cot A, whilst AK/DK = cot B.
15. The sides of the midpoint triangle are 15, 17, 4√13. Its vertices are A (17,0,0), B (8,12,0), C(25,12,0). It is not hard to check that the 4th vertex is D (16,12,12). Then DA = 17, DB = 4√13, DC = 15 as required. Using Heron or considering the altitude from A to BC, we find that the midpoint triangle has area 102. Hence vol = (1/3) 12·102 = 408.
© John Scholes
jscholes@kalva.demon.co.uk
8 Dec 2003
Last updated/corrected 1 Feb 04