Teaser 3153: Pole position
From The Sunday Times, 26th February 2023 [link] [link]
Jeb’s 25×25km square ranch had his house’s flagpole at the ranch’s pole of inaccessibility (the point whence the shortest distance to a boundary fence was maximised).
At 50, Jeb gave each of his four sons a triangular tract of his land, with corners whole numbers of km along boundary fences from each corner post (as illustrated, not to scale). Each tract’s area (in square km) equalled that son’s age last birthday (all over 19, but under 30). All the tracts’ perimeters differed, and each son set his hacienda’s flagpole at his own tract’s pole of inaccessibility.
Curiously, for Jeb’s new octagonal ranch the pole of inaccessibility and the shortest distance from this to a boundary fence were unchanged.
Give the eldest son’s shortest distance from his flagpole to a boundary fence.
This puzzle marks the 62nd birthday of the Sunday Times Brain Teaser puzzle. The first numbered puzzle, Brain-Teaser 1: Tall story, was published on 26th February 1961.
[teaser3153]
Hugo and 
S2T2 
Jim Randell 5:09 pm on 24 February 2023 Permalink |
We can calculate which of the possible triangles do not change the pole of inaccessibility, and it turns out there are exactly 4 of them (and they do have different perimeters). We can then calculate the inradius of the one with the largest area to find the required answer.
This Python program runs in 51ms. (Internal runtimes is 259µs).
Run: [ @replit ]
from enigma import (irange, subsets, div, hypot, fdiv, line_distance, Accumulator, printf) # distance to boundary from pole R = 12.5 # = 0.5 * 25 km # record inradius for maximum age m = Accumulator(fn=max) # consider (a, b) plots for the sons for (a, b) in subsets(irange(1, int(R)), size=2, select="R"): # calculate area A = div(a * b, 2) # check for viable area, and age if A is None or not (19 < A < 30): continue # check boundary does not change the pole of accessibility if line_distance((0, a), (b, 0), (R, R)) < R: continue # calculate the perimeter p = a + b + hypot(a, b) # calculate inradius; area = inradius * semiperimeter r = fdiv(2 * A, p) printf("({a}, {b}) -> A={A}; p={p:.3f}; r={r:.3f}") m.accumulate_data(A, r) # output solution (inradius for maximum age) printf("answer = {m.data} km") assert m.count == 4 # check that 4 plots were foundSolution: The shortest distance from the eldest son’s flagpole to a boundary is 2 km.
Given a plot of land a “pole of inaccessibility” is the centre of a maximal radius circle that can be placed inside that plot without exceeding the boundaries.
So for the square plot of land the flagpole is erected at the centre of the square. (For a non-square rectangle there would a line of possible positions where the flagpole could be erected). And for a triangle the flagpole is erected at the incentre (i.e. the centre of the incircle) of the triangle. (See [@wikipedia]).
The plots of land, and positions of the flagpoles, are illustrated in the diagram below:
Jeb’s flagpole is at the centre of the square, and the large red circle shows the shortest distance to a boundary. When the corner plots are made they cannot cross the large red circle, or they would reduce the shortest distance from Jeb’s flagpole to a boundary. But they could touch the circle.
There are 4 different triangular plots that can be made, and they are also shown on the diagram, along with their incentres, which correspond to the positions of the sons’ flagpoles.
The boundary of the plot on the bottom left (with an area of 20) comes closest to Jeb’s flagpole (it is 12.534 km away from it). The plot on the bottom right (with an area of 24) is the largest. The other two plots have areas of 20 and 21.
LikeLike
Hugo 9:41 am on 5 March 2023 Permalink |
Sides of the triangular plots (with area and distance of third side from the centre of the square):
4, 10 (20, 12.534) nearly touched the big circle
5, 8 (20, 12.985)
6, 7 (21, 13.07)
6, 8 (24, 12.7) third side 10
There would be a couple more with two integer sides but half-integer areas.
LikeLike