Teaser 3308: New towns
From The Sunday Times, 15th February 2026 [link] [link]
Three new towns (A, B and C) were being built, with straight roads joining them. Each road was a whole number of miles in length, with AB the same as BC and a total length of the three roads of 108 miles.
A new power station was needed for the three towns and two plans were considered. The Plan 1 site at point P was to be the same distance from each town (a perfect square number of miles). The Plan 2 site was to be a whole number of miles along PB such that the total length of cable connecting the power station to each town was a whole number of miles. Given the above, no shorter total cable length of a whole number of miles was possible, so plan 2 was adopted.
What was the total cable length?
[teaser3308]
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Jim Randell 9:00 am on 15 February 2026 Permalink |
This Python program runs in 73ms. (Internal runtime is 291µs).
from enigma import (Accumulator, irange, is_triangle, rcs, fdiv, is_square_p, point_dist, intr, printf) # if float <f> is within <t> of integer <i> return <i> def snapf(f, t=1e-6): i = intr(f) if abs(i - f) < t: return i T = 108 # T = AB + BC + AC # choose a value for x (= AB = BC) for x in irange(1, (T - 1) // 2): # calculate y (= AC) y = T - 2*x if not is_triangle(x, x, y): continue # calculate the positions of the towns z = 0.5 * y h = rcs(x, -z) (A, B, C) = ((z, 0), (0, h), (-z, 0)) # P = (0, h - p) is the circumcentre of triangle ABC # distance p = BP must be an integer (and a perfect square) p = snapf(fdiv(x * x, 2 * h)) if not is_square_p(p): continue printf("x={x} y={y} -> h={h} p={p} [A={A} B={B} C={C}]") # find minimum total length of cable r = Accumulator(fn=min, collect=1) # choose a distance along BP (excluding endpoints) to site Q for d in irange(1, p - 1): # calculate total length of cable required Q = (0, h - d) ds = list(snapf(point_dist(Q, X)) for X in [A, B, C]) # distance to each town is an integer if None in ds: continue t = sum(ds) # output scenario printf("-> d={d} Q={Q} ds={ds} t={t}") r.accumulate_data(t, d) # output minimal length printf("minimal length = {r.value} [d={r.data}]") printf()Solution: [To Be Revealed]
[diagram to follow]
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Frits 8:09 am on 16 February 2026 Permalink |
Based on Jim’s program.
Good chance I don’t understand anymore if I will look at it at a later date.
I didn’t check if Plan 2 used a shorter total cable length then Plan 1.
from enigma import is_square, ceil # distance metric between 2 points dist = lambda p, q: ((p[0] - q[0])**2 + (p[1] - q[1])**2)**.5 sol = 99999 # choose a value for x (= AB = BC), y < 2x so 4x > 108 for x in range(28, 54): # calculate y (= AC) y = 108 - 2 * x # calculate height (line B perpendicular to AC) h = 6 * (3 * x - 81)**.5 # calculate the distance p = BP p, r = divmod(x**2, 2 * h) if r or not is_square(p): continue A = (y // 2, 0) # derive points that are an integer distance away from point A sol = min(sol, min(2 * sq + h - n for n in range(ceil(h)) if (sq := is_square((54 - x)**2 + n**2)))) print(f"minimal length = {sol}")LikeLike
Jim Randell 11:56 am on 16 February 2026 Permalink |
@Frits: This doesn’t seem to check all integer points along PB (there should be p − 1 of them, if we exclude the endpoints).
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Frits 4:33 pm on 16 February 2026 Permalink |
@Jim, my code is not totally correct.
I had noticed that the distances from Q to A and Q to C in the optimal solution were symmetrical around d=h (fi these distances were the same
for d=h-1 and d=h+1). The distance from Q to B is not symmetrical around d=h but is increasing. This means that if d > h potential new total length of cable will always be higher than the other symmetrical value (at least if p <= 2.h).
That's why I took a shortcut to check fewer entries. That is not always allowed.
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Jim Randell 8:07 am on 17 February 2026 Permalink |
@Frits: Fair enough (although does that not assume the intersection of AC and BP is a (half-) integer distance from B and P – is this justified in general?). It just wasn’t clear to me from the code you posted that all possible positions would be considered.
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Alex.T.Sutherland 2:32 pm on 18 February 2026 Permalink |
Given:- Towns form an isosceles triangle AB = a; BC = a; AC = b;
Periphery = 2*a + b = 108. b is even;
Plan 1. P is the centre of the circumsbribed circle thro’the towns
and R is the radius (a perfect square ).
The line PB halves AC at D and is perpendicular to AC.
P is distant R from B in the appropriate direction.
Soln:- Using R = a^2 / ( sqrt(4*(a^2) -b^2).
2*a+b = 108.
R is a perfect square.
Solve for R,a,b
(Since b is even use it instead of ‘a’..less nuumber of calcs).
This gives R,a,b,and the postion of P.
The total cable length in Plan 1 = 3*R.
P = site location for Plan 1.
S is the site location on line PB ,integer from P distant x = [1:R] .
For each x calculate Length = 2*AS + BS;
Find integer pairs (x Lx).(I have found 4 such pairs…one is S=B ie x= R).
There are 2 with a minimum length (B being one of them).
The answer is either one.
This does not affect the required answer only where the site is to be located.
Answer: Significantly less than Plan_1 which is 3*R.
Lots of Pythagorean triangles.
Sum of the digits for each of [AS BS CS] are all equal.
(as is the digit product of b/2).
Time :- Aprox. 0.5 ms
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