| 1 7/52 | 9 (3,3,3), (4,2,4) |
| 2 7(√3-1)/2 | 10 148 |
| 3 25 | 11 12 √218 |
| 4 28,811 | 12 1 + 274i |
| 5 183 | 13 8√55 |
| 6 12 | 14 30 |
| 7 428 | 15 128-16√19 |
| 8 748 |
1. Three letters form a palindrome iff the last matches the first, so the prob that the letters are not a palindrome is 25/26. Similarly, the prob that the digits are not is 9/10. So the prob of a palindrome is 1 - (25/26)(9/10) = 7/52.
2. Let the circles have radius 1. Then the long side is 14. The two outer rows of centers are a distance 2√3 apart, so the short side is 2 + 2√3. Hence ratio 7/(1+ √3) = 7(√3 - 1)/2.
3. Dick's age must be one of: 43; 53, 54; 62, 63, 64, 65; 72, ... , 76; 82, ... , 87; 92, ... , 98, giving 1 + 2 + 4 + 5 + 6 + 7 possibilities, each corresponding to a unique n. Hence 25.
4. xk = 1/k - 1/(k+1), so the sum is 1/m - 1/(n+1) = 1/29. Put N = n+1. Then mN = 29(N-m), so (m-29)(N+29) + 292 = 0. But N+29 is positive, so m-29 must be negative. But m is positive, so m-29 is not divisible by 29. Hence N+29 = 292 and m = 28. So n = 28·29 - 1 = 811.
5. Let S be the set of vertices of D. There are 66 pairs of points from S. Each pair gives 2 squares with the pair as adjacent vertices. But the 3 squares with all their vertices in S, so they are each overcounted 3 times, giving 66·2 - 9 = 123. In addition each pair has one square with that pair as opposite vertices, but each of the 3 squares with all vertices in S is again counted twice, an overcount of 6, so 123 + 60 = 183.
6. Put a = log225x, b = log64y. Then logx225 = 1/a, logy64 = 1/b, so equations become 1/a - 1/b = 1, a + b = 4. Solving (a, b) = (3+√5, 1-√5) or (3-√5, 1+√5). Hence if a1 corresponds to x1 etc, we have log225x1x2 = a1+a2 = 6, so x1x2 = 1512, and log64y1y2 = b1+b2 = 2, so y1y2 = 212. Hence answer 12.
7. So (102002 + 1)10/7 = 102860 (1 + 10/7 1/102002 + 3/49 1/104004 + ... ) = 102860 + 10859/7 + (3/49) 1/101143 + ... . Now 1/7 = 0.142857 142857 ... , period 6 and 859 = 1 mod 6, so 10859/7 = ... 1.428... and answer = 428.
8. We have a9 = 13a1 + 21a2, so if the two sequences have (a1, a2) = (a, b) and (A, B), then k = 13a + 21b = 13A + 21B. So 13(A - a) = 21(b - B). So A - a = 21n and b - B = 13n for some positive integer n. The obvious a = 1, A = 22, b = 14, B = 1 does not work because A > B. The sequences are non-decreasing iff A <= B and a <= b. So we take a = 1, A = 22, B = 22, b = 35, giving k = 748.
9. Label the posts 1, 2, 3, ... . C cannot paint 4 (or C paints all the posts thereafter). Suppose A paints 4. C cannot paint 5, or A and C both paint 7, so B paints 5. Hence C paints 6 and (a, b, c) = (3, 3, 3). Suppose B paints 4. C cannot paint 5 (or there is nothing left for A to paint). So A paints 5. Hence C paints 7 and (a, b, c) = (4, 2, 4).
10. We have AB = 12, AC = 37, so BC = 35 (Pythagoras). So area ABC = 210. Area GDCF = area ABC - area ABD - area AGF. We now use the angle bisector theorem twice. Area ABD = (12/49) area ABC = 51 3/7. Area AGF = (3/13) area AEF = (3/13) (1/4) area ABF = (10/13) (1/4) (10/37) area ABC = 10 440/481, so area GDCF = 210 - 51 3/7 - 10 440/481 = 148 - (3/7 - 41/481).
11. This needs a trick. Call the cube c. Reflect c in the face BCEF to get cube d. Then the continuation of the ray reflects into a continuation of the segment AP. If the continuation strikes face X of d at Q, then reflect d in the face X to get cube e and so on. Take A as the origin, x-axis along AB, y-axis parallel to BC, z-axis parallel to BF. Then P is the point (12, 7, 5) and we get the line (12t, 7t, 5t). The question is when this first hits a vertex (12q, 12r, 12s). Evidently at t = 12, a distance 12 (122 + 72 + 52)1/2 = 12 √218.
12. We find z --> (z+i)/(z-i) --> i(z+1)/(z-1) --> z. So z0 = z3 = ... = z2001. Hence z2002 = (1/137 + 2i)/(1/137) = 1 + 274i.
13. Let the medians meet at G. Then GE = CE/3 = 9, GC = 2GE = 18, and AG = 2AD/3 = 12. Also AE = BE = 12. Since AB, CF are chords of a circle, we have FE·27 = 12·12, so FE = 16/3. Now area AFB/area ABC = EF/EC = 16/81. The height of AEG is √(122 - (9/2)2) = 3(√55)/2, so area AEG = 27(√55)/4. Hence area AEC = 3 area AEG = 81(√55)/4, and area ABC = 81(√55)/2. So, finally, area AFB = 8√55.
14. Let S have k+1 elements with sum N. Then k divides N-1 and N-2002 and hence also 2001 = 3·23·29. If m is any element of S, then k divides N-m and hence m-1. In other words m = 1 mod k. So the largest element is at least 1+k2. Since 2002 is the largest element, k < 45. Thus the largest possible k is 29. For example: 1, 30, 59, 88, 117, 146, 175, 204, 233, ... , 813, 2002.
15. The trapezoid ABFG is clearly determined. We can regard it as hinged about AB. Similarly, the triangle CDE is determined and hinges about CD. Moreover, since the distance of E from the plane ABCD is fixed, the angle at the hinge is fixed. Why then is the angle at the hinge AB fixed? Because ADEG must be planar.
© John Scholes
jscholes@kalva.demon.co.uk
4 Aug 2003
Last updated/corrected 4 Aug 03