△ ABC has area 1 unit, with |AB|≤|AC|<|BC|
Determine the minimum length of |AC|.
clc, clearvars;
mn=999;
for a=1:200
for b=a:201
for c=b+1:202
if c<a+b % make sure it's a triangle
s=(a+b+c)/2;
a0=sqrt(s*(s-a)*(s-b)*(s-c)); % Heron's formula
r=sqrt(a0); % from area to linear ratio
a1=a/r; b1=b/r; c1=c/r; % scale it
if b1<mn
mn=b1; % new minimum
disp([a1 b1 c1])
end
end
end
end
end
tries unit triangles of various shapes, up to a ratio of 1:200 of its sides. The initially chosen triangles have larger (or possibly smaller) than unit area, so they are scaled to unit-area size. This assures that a very wide variety of proportions of triangles are tried.
Lower and lower minimum values are found for the length of the middle side:
1.41979242240402 1.41979242240402 2.12968863360603
1.41861241350476 1.41861241350476 1.89148321800635
1.41435501909528 1.41435501909528 1.98009702673339
1.41421782496056 1.41421782496056 2.00347525202746
1.4142177955125 1.4142177955125 1.9965427701353
1.41421368734233 1.41421368734233 1.99940555796675
1.41421356605441 1.41421356605441 2.00010204341981
1.41421356605366 1.41421356605366 1.99989797219709
As the puzzle is not limited to lengths reducible to rationals, it's apparent that the dimensions are sqrt(2), sqrt(2), 2, an isosceles right triangle. That's the minimum length, sqrt(2), one of the legs of the isosceles right triangle with hypotenuse = 2.
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Posted by Charlie
on 2022-10-05 09:15:23 |
computer-aided solution
