Sources page probability recreations

Ex. 8, p. 135. "A can row a mile in ¾ of a minute less time than B. In a mile race, B gets 250 yards head start, and lose by 14 yards." Determine their times to row a mile, assuming uniform speed.

Prob. 37, pp. 161 & 467, If B gets a 12 sec headstart in a mile race, he loses by 44 yards. If he gets a 165 yards headstart, he wins by 10 sec. Determine their times for a mile.

Lewis Carroll. Letter of 8 Apr 1897 to Enid Stevens. = Carroll-Wakeling, prob. 25: Handicaps, pp. 33 & 71. A loses 10 yards in every 100 against B, while B gains 10 yards in every 100 against C. What handicaps should B give A and C in a quarter mile (= 440 yards) race? Mentioned in Carroll-Gardner, pp. 50-51 Cohen's comment is that is that this is too ambiguous to have a precise answer, but such problems were common in late Victorian times and Wakeling gives an answer (but no method) which seems correct to me and in line with other problems of the time. However, Gardner does not comment on Cohen's remark and does not give an answer. Letting A, B, C denote the uniform speeds of the runners, Wakeling and I would interpret the problem statements as: A = 90/100 · B and B = 110/100 · C. The ambiguity is that one could read the statements as saying A = 100/110 · B and B = 100/90 · C, but I think the first interpretation is more natural. Either case permits a clear answer. Cohen says it is clear that A = 90/100 · B, but is unsure whether B = 100/90 · C or B = 110/100 · C, but the first case makes C = A, which makes the problem much less interesting.

Charles Pendlebury & W. S. Beard. A "Shilling" Arithmetic. Bell, London, 1899 -- my copy is 59th ptg, 1960 and says it was reset for the 49th ptg of 1944. Examination Papers XIV, prob. 11, p. 187. A runs at 12 1/2 mph and B runs at 12 7/16 mph. If A gives B ten yards headstart, when will he overtake him. Who is ahead when A has run a mile?

Hummerston. Fun, Mirth & Mystery. 1924. Speed, Puzzle no. 18, pp. 49 & 174. Cook can run 100 yards in 12 sec. Brown can give Cook a 10 yd headstart and finish even with him. Adams can give Brown a 5 yd headstart and finish even. Adams gives Cook a 15 yd headstart and Cook loses 1/20 of a sec in a bad start. Who wins and by how much? He notes that one must assume uniform speeds.

Collins. Fun with Figures. 1928. The 100-yard dash, pp. 24-26. = Hummerston with Adams, Brown, Cook replaced by Bob, Jack, Bill.

Doubleday - 2. 1971. Running commentary, pp. 59. In a 100 yard dash, A gives B a head start of 25 to make an even race and B gives C a head start of 20. How much head start should A give C to make an even race?
10.A.6. DOUBLE CROSSING PROBLEMS
New section. I recall Loyd and/or Dudeney have versions. Also, the problems in 10.A with the faster meeting the slower on the return trip are related.
Mittenzwey. 1880. Prob. 296, pp. 54 & 105; 1895?: 326, pp. 57 & 106; 1917: 326, pp. 52 & 100. Field in the shape of a right triangle. Runners whose speeds are in the ratio 13 : 11 start at the right angled corner along different legs. They first meet at the midpoint of the hypotenuse and then at 60m from the starting point. As far as I can see, this leads to a quadratic.

Wood. Oddities. 1927. Prob. 61: The errand boys, p. 47. Alan goes from A to B and back while Bob goes from B to A and back, both starting at the same time. They first cross at 720 from A and then at 400 from B. How far is it between A and B? Gives a general formula: If the two crossings occur at distances a, b, then the width is 3a - b. He asserts that Alan goes faster, but the ratio of velocities is a/(2a-b) which is 9/13 in this case.

Philip Kaplan. More Posers. (Harper & Row, 1964); Macfadden-Bartell Books, 1965. Prob. 73, pp. 76 & 106. Two swimmers in a pool, like above with distances 40, 20.
10.A.7. TRAINS PASSING
New section. If a fast train, going at velocity V, takes time T to overtake a slow train, going at velocity v, and it takes time t for the trains to pass when meeting, then the lengths of the trains cancel out and one gets V/v = (T + t)/(T - t).
Haldeman-Julius. 1937. No. 28: Speed problem, pp. 6 & 22. T = 35, t = 3, V = 38.
10.A.8. TOO SLOW, TOO FAST
New section. If I go at rate v₁, I arrive t₁ too late; if I go at rate v₂, I arrive t₂ too soon. I do not recall seeing anything like this before.
Charles Pendlebury & W. S. Beard. A "Shilling" Arithmetic. Bell, London, 1899 -- my copy is 59th ptg, 1960 and says it was reset for the 49th ptg of 1944. Examination Papers XI, prob. 185, p. 185. If a man goes at 4 mph, he arrives 5 min late; if he goes 5 mph, he arrives 10 min early. How far is he going?

The general solution is easily found to be D = v₁v₂(t₁+t₂)/(v₂-v₁). However, I wondered what was the correct rate, v, so that I arrive on time. This leads to the following.

(t₁+t₂)/v = t₁/v₂ + t₂/v₁ or 1/v = t₁/(t₁+t₂)·1/v₂ + t₂/(t₁+t₂)·1/v₁, so v is a weighted harmonic mean of v₁ and v₂. On the other hand, we find the time to arrive is t = D/v = v₁/(v₂-v₁)·t₁ + v₂/(v₂-v₁)·t₂, which is a weighted mean of t₁ and t₂.

The only type of problem that I can recall that leads to similar means is the classic problem: If I make a journey at 30 mph and return at 20 mph, what is my average speed. In general, if I travel d_i at speed v_i, what is my average speed? Letting t_i be the time at speed v_i, we have t_i = d_i/v_i, so the average speed is given by

(d₁+d₂)/v = t₁ + t₂ = d₁/v₁ + d₂/v₂ or

1/v = d₁/(d₁+d₂)·1/v₁ + d₂/(d₁+d₂)·1/v₂, so v is again a weighted harmonic mean of v₁ and v₂.

10.B. FLY BETWEEN TRAINS
There are two trains d apart, approaching at rates a, b. A fly starts at one, flies to the other, then back to the first, then back to the second, etc., flying at rate c. How far does he go?

NOTATION. We denote this problem by (a, b, c, d). For overtaking problems, we let b be negative. I need to check for details.

Laisant. Op. cit. in 6.P.1. 1906. Chap. 53: Le chien et les deux voyagers, pp. 132-133. A is 8 km ahead and goes at 4 km/hr. B starts after him at 6 km/hr. A dog starts at one and runs back and forth between them at 15 km/hr until they meet. I.e. (4,  6, 15, 8). Notes that the distance the dog travels is independent of where he starts and that the travellers could be meeting rather than overtaking.

Dudeney. Problem 464: Man and dog. Strand Mag. (Jul 1919). ??NX.

Dudeney. Problem 643: Baxter's dog. Strand Mag. (1924?). ??NX. A goes at rate 2 and has an hour's head start. B goes at rate 4 and dog starts with him at rate 10. I.e. (4,  2, 10, 2).

G. H. Hardy. Letter of 5 Jan 1924 to M. Riesz. In: M. L. Cartwright. Manuscripts of Hardy, Littlewood, Marcel Riesz and Titchmarsh. Bull. London Math. Soc. 14 (1982) 472 532. (Letter is on p. 502, where it is identified as Add. MS. a. 275 33, presumably at Trinity College.) Says it defeated Einstein, Jeans, J. J. Thomson, etc. Fly between cyclists (10, 10, 15, 20). "One thing only is necessary: you must not know the formula for the sum of a geometrical progression. If you do, you will take 15-20 minutes: if not, 2 seconds."

Ackermann. 1925. Pp. 116 117. Couple walking up a hill. Their dog, who is twice as fast as they are, runs to the top and back to them continually. (a, 0, 2a, d) -- he only says the couple start 1/4 of the way up the hill.

Dudeney. Problem 754: The fly and the motor cars. Strand Mag. (Jun 1925) ??NX. (?= PCP 72.)

H&S 53, 1927, says this is 'a modern problem'.

M. Kraitchik. La Mathématique des Jeux, 1930, op. cit. in 4.A.2, chap. 2, prob. 17, p. 30. (15, 25, 100, 120). (Identified as from L'Echiquier, 1929, 20, ??NYS.) I can't find it in his Mathematical Recreations.

M. Adams. Puzzles That Everyone Can Do. 1931. Prob. 181, pp. 71 & 156: Little Red Riding Hood. Little Red Riding Hood with her dog and Grandma set out at the same time to meet: (2, 2, 8, 6).

Dudeney. PCP. 1932. Prob. 72: The fly and the motor cars, pp. 28 & 86. = 536; prob. 86: The fly and the cars, pp. 26 & 243. (50, 100, 150, 300).

Phillips. Week End. 1932. Time tests of intelligence, no. 38, pp. 21 & 193. Fly between cyclists. (10, 15, 20, 60).

Abraham. 1933.