Alex owns one of those folding rulers where each segment is exactly 1 foot long. While playing with the open ruler he formed it into a triangle. Then he refolded it into a second triangle with double the area. Next he refolded it into a third triangle with triple the area.
What is the smallest possible length of the ruler?
From the fact this is considered different from merely taking the larger of the two lengths found in the original puzzle, I conclude that it's not considered proper to double over one pair of segments so as to hide one unit. We want the perimeters of all the triangles to be equal.
The program stores all the triangles for all sizes up to 40 and compares areas, making sure the same perimeters for both sizes:
clearvars,clc
tri=double.empty(0,3);
area=[];
for tot=3:40
for a=1:tot/3
for b=a:(tot-a)/2
c=tot-a-b;
if c<a+b
s=sum([a b c])/2;
A =sqrt(s*(s-a)*(s-b)*(s-c));
% if isreal(A) && A>0 && c>0
area(end+1)=A;
tri(end+1,:)=[a b c];
% end
end
end
end
end
for a=1:length(tri)
for b=1:length(tri)
if area(b)==2*area(a) && sum(tri(a,:))==sum(tri(b,:))
fprintf('%19.15f %19.15f %3d %3d %3d %3d %3d %3d %4d\n', ...
[area(a) area(b) tri(a,:) tri(b,:) max([sum(tri(a,:)) sum(tri(b,:))])])
end
end
end
disp(' ')
for a=1:length(tri)
for b=1:length(tri)
if area(b)==3*area(a) && sum(tri(a,:))==sum(tri(b,:))
fprintf('%19.15f %19.15f %3d %3d %3d %3d %3d %3d %4d\n', ...
[area(a) area(b) tri(a,:) tri(b,:) max([sum(tri(a,:)) sum(tri(b,:))])])
end
end
end
Area Double orig double perimeter
8.944271909999159 17.888543819998318 2 9 9 6 6 8 20
9.949874371066199 19.899748742132399 2 10 10 5 8 9 22
16.124515496597098 32.249030993194197 3 11 12 8 9 9 26
18.973665961010276 37.947331922020552 3 13 14 7 11 12 30
14.966629547095765 29.933259094191531 2 15 15 8 9 15 32
24.000000000000000 48.000000000000000 4 13 15 10 10 12 32
15.968719422671311 31.937438845342623 2 16 16 7 11 16 34
21.817424229271428 43.634848458542855 3 15 16 9 10 15 34
23.237900077244500 46.475800154489001 3 16 17 8 12 16 36
27.495454169735041 54.990908339470082 4 15 17 10 11 15 36
29.240383034426891 58.480766068853782 4 16 18 9 13 16 38
33.763886032268267 67.527772064536535 4 17 17 11 13 14 38
32.619012860600179 65.238025721200358 5 15 18 11 12 15 38
30.983866769659336 61.967733539318672 4 17 19 8 16 16 40
35.777087639996637 71.554175279993274 4 18 18 12 12 16 40
34.641016151377549 69.282032302755098 5 16 19 10 14 16 40
37.416573867739416 74.833147735478832 6 15 19 12 13 15 40
Area Triple orig double perimeter
11.659223816361019 34.977671449083054 2 13 14 7 10 12 29
14.966629547095765 44.899888641287298 2 15 15 9 10 13 32
16.686446595965240 50.059339787895723 3 14 16 9 12 12 33
8.996527107723290 26.989581323169872 1 18 18 8 11 18 37
15.998046755776157 47.994140267328468 2 18 19 6 16 17 39
15.998046755776157 47.994140267328468 2 18 19 9 12 18 39
18.973665961010276 56.920997883030829 2 19 19 11 11 18 40
The shortest ruler for doing both doubling and tripling is the 32-foot model, which can double or triple the area of an original 2-15-15 triangle, by making an 8-9-15 triangle and a 9-10-13 triangle respectively.
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Posted by Charlie
on 2024-07-08 08:03:52 |
solution
