From The Sunday Times, 26th April 1970 [link]
Near my house is an impressive block of flats. It stands 16 storeys high, and each story is a lofty 15 feet from floor to floor. My old friends Bob and Horace operate the two lifts, and both start their shift each morning leaving the ground floor at 8 a.m. precisely.
The two lifts make automatic compulsory stops at the ground level (there is no basement), the 12th and top floors. But below the 12th, Bob serves only the odd-numbered floors, and Horace the even numbers; these stops are also compulsory (up and down). All stops take just 10 seconds, except at ground level where both lifts wait for 20 seconds.
Above the 12th both Bob and Horace stop as desired, going up and coming down. Both lifts travel between floors at identical speeds.
Every morning the two attendants make a rendezvous to exchange newspapers and collect their coffee when their arrivals at a certain floor coincide exactly at two minutes after 11 a.m.
In feet per second what the speed of each lift between stops?
Note: In the UK the floors in buildings are: ground floor, first floor, second floor, etc.
Also, as stated this puzzle does not appear to have a unique solution.
An answer was published (with Teaser 467) stating:
“This clever trap, with its nice reasoning, outweighed the slight mathematical bias and kept down the entry to a mixed handful.”
But later, with Teaser 471 the following statement was made:
“[465: Regretfully a wrong transposition led to a false calculation of time.]”
In the comments I give a variation on the puzzle that does have a unique answer.
This puzzle was originally published with no title.
[teaser465]
GeoffR and 
S2T2 
Jim Randell 7:50 am on 17 September 2019 Permalink |
As presented this puzzle has multiple solutions. However if the perimeter of the original rectangle is required to be a whole number of feet, then only one of the solutions remains.
This Python program finds all the solutions to the puzzle numerically using the [[
find_zero()]] function from the enigma.py library. It runs in 63ms.Run: [ @replit ]
from enigma import (irange, cb, cbrt, sq, find_zero, catch, printf) # consider values for t, where T = t^3 is the total volume for t in irange(1, 100): T = cb(t) # calculate y (for x in [0, t]) def Y(x): return cbrt(T - cb(x)) # how close is x^2 + y^2 - xy to 243? def f(x): y = Y(x) return abs((sq(x) + sq(y) - x * y) - 243) # find a zero for f r = catch(find_zero, f, 0, cbrt(0.5 * T)) if r is None: continue x = r.v y = Y(x) p = 2 * (x + y) # output solution printf("t={t}: x={x:.3f}, y={y:.3f}, p={p:.3f}")Solution: The (integer) perimeter of the original rectangle was 48 feet.
Allowing non-integer solutions for the perimeter we get four solutions:
Graphically we get a solution when the ellipse x² + y² − xy = 243 intersects the curve x³ + y³ = t³ for integer values of t.
For positive x and y we get solutions for t = 16, 17, 18, 19, as shown in the graph below:
The original rectangle has sides of length x and y, so the perimeter of the original rectangle is 2(x + y). Only t=18 gives an integer value for the perimeter (although the values for x and y are not integers). In this case the exact values are: x, y = 12 ± √33.
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Jim Randell 8:39 am on 19 September 2019 Permalink |
Here is an analytical solution.
Given the equations:
If we consider the product (x + y)(x² + y² − xy) we have:
We also have:
So for a given value of t we can determine values for x + y and xy, say:
These have positive real solutions for x and y when a² ≥ 4b and b ≥ 0
So t is in the range 16 to 19. And the required perimeter is given by p = 2(x + y) = 2t³ / 243.
We can consider the 4 candidate values manually and look for an integer solution, or use a short program:
Run: [ @replit ]
from enigma import (irange, cb, div, printf) for t in irange(16, 19): p = div(2 * cb(t), 243) if p is None: continue printf("t={t}: p={p}")LikeLike