| 1 146 | 9 29/15 |
| 2 840 | 10 532/729 |
| 3 166 | 11 (84 - 48√3)π |
| 4 159 | 12 672/5 |
| 5 128 | 13 990 |
| 6 743 | 14 384 |
| 7 383 | 15 12 |
| 8 10 |
1. Put s = m + n, p = pm. Then s, p = 55, 16. If s = 16, p = 55, then m, n = 5, 11. It is easy to see we cannot have s = 55, p = 16.
2. AC = 5. So sum = 5 + 2 (1/168 + 2/168 + ... + 167/168) 5 = 5 + 5·167.
3. ai/ai-1 = 0.2 (1000 - i)/i. It is > 1 iff 200 > 1.2 i or i < 166.7.
4. The log term is negative on (0, 1) and positive for x > 1. For x > 32 = 160/5 the log term is > 1. Note that sin(5πx) = 0 at x = 0, 1/5, 2/5, ... . For x < 1, it lies in (-1, 0) in the intervals (1/5, 2/5), (3/5, 4/5). For 1 < x > 32 it lies in (0, 1) in the intervals (6/5, 7/5), (8/5, 9/5), ... , (158/5, 159/5). Thus the equation has the root x = 1, and 2 roots in each of (1/5, 2/5) and (3/5, 4/5), and 2 roots in each of (6/5, 7/5), (8/5, 9/5), ... , (158/5, 159/5).
5. 20! has 8 distinct prime factors (2, 3, 5, 7, 11, 13, 17, 19). Each prime factor must divide just one of the numerator and denominator. Also just one of m/n and n/m lies in (0, 1). Hence 28/2.
6. 546 = 7·73 + 35 = 38·7 + 35·8. Hence [x + 0.19] = [x + 0.20] = ... = [x + 0.56] = 7, [x + 0.57] = ... = [x + 0.91] = 8. So 7.43 <= x < 7.44.
7. It is clear that simplifying will give a quadratic in x. One can slog through to find it, but it is easier to note that if x = √19 + 91/x, then x will satisfy the equation, so the quadratic must be simply x2 - √19 x - 91. A minor trap is that one root is negative: x = (√383 + √19)/2 or -(√383 - √19)/2, so sum of abs values is √383.
8. If the roots are integral, so is their sum, so b must be an integer. The roots must be real, so b2 - 24b >= 0, and b(b - 24) must be a square. b = 0, 24 clearly work. Otherwise, put B = b-24 for b > 24, or -b for b < 0. Then if d = gcd(B, B+24), d must be one of 1, 2, 3, 4, 6, 8, 12, 24 and B/d and (B+24)/d are squares. Since the gap between consecutive squares increases, it is easy to check that the only solutions are:
b = -25, x = -5, 30
b = -8, x = -4, 12
b = -3, x = -3, 6
b = -1, x = -2, 3
b = 0, x = 0, 0
b = 24, x = -12, -12
b = 25, x = -15, -10
b = 27, x = -18, -9
b = 32, x = -24, -8
b = 49, x = -42, -7
9. Put t = tan x/2, and c = 1/t. Then sec x + tan x = (1+t2)/(1-t2) + 2t/(1-t2) = (1+t)/(1-t) = (c+1)/(c-1) = 22/7. So c = 29/15. But cosec x + cot x = (1+t2)/(2t) + (1-t2)/(2t) = 1/t = c.
10. The chance x 27 that AAA, BBB are received as: AAA is 8, 1; AAB 4, 2; ABA 4, 2; ABB 2, 4; BAA 4, 2; BAB 2, 4; BBA 2, 4; BBB 1, 8. Hence required prob is (8·26 + 4·24 + 4·22 + 2·18 + 4·16 + 2·12 + 2·8 + 1·0)/272.
11. The disk centers must lie just outside the circle, as shown. In the small diagram the circle has center O, radius OM and a disk has center A radius AM. Angle AOB = 30o. So AM = tan 15o = t. We have 2t/(1-t2) = tan 30o = 1/√3. Hence AM = 2 - √3. So total disk area = 12π(2 - √3)2 = 12(7 - 4√3)π.
12. We have y = 180o - 3x, so cos y = - cos 3x = 3 cos x - 4 cos3x = 9/5 - 108/125 = 117/125, and sin y = 44/125. Hence AP = 117/5, AS = 44/5.
13. Simplifying, (m-n)2 = m+n. Put N = m+n. Then larger of m, n is (N + √N)/2, so we also want N to be as large as possible. Hence N must be largest square <= 1991, namely 442 = 1936, giving 990, 946.
14. (Thanks to Zach Abel and Vincent Chan). Put AC = x, AE = y, AD = z. Note that the chord lengths depend only on the arc lengths, so for example, CF = z. Applying Ptolemy to ABCD: 31·81 + z·81 = xy, and to ACDF: xy + 812 = z2. Hence (z-144)(z+63) = 0, so z = 144. Now Ptolemy on AFED gives 812 + 81z = y2, so y = 135. Now the first equation gives x = 105. So x + y + z = 384.
15. (Thanks to Evan Dummit for this). Take points A1 (1, a1), A2 (1 + 3, a1 + a2), ... , An (1 + 3 + ... + 2n-1, a1 + a2 + ... + an). Then the lines OA1, A1A2, A2A3, ... , An-1An have lengths as given in the question. Hence their sum has minimum value when the segments are all parallel, and so Sn2 = (1 + 3 + 5 + ... + 2n-1)2 + (a1 + a2 + ... + an)2 = n4 + 172. So (Sn + n2)(Sn - n2) = 172. Hence (Sn + n2) = 289, (Sn - n2) = 1, so Sn = 145, n = 12.
© John Scholes
jscholes@kalva.demon.co.uk
1 August 2003
Last updated/corrected 23 Mar 04