The side lengths of a
Heronian Triangle are R1, R and R+1, where R is an 8digit positive integer which does not contain any leading zero.
Determine all possible value(s) of R for which this is possible.
Possibly the hardest part of this puzzle is finding a vaguely geometrical connection to what looks mainly like algebra.
The area formula for an equilateral triangle is A = 3^(1/2)/4*a^2, where a is the length of a side.
Heron's Area Formula is A = (s(sa)(sb)(sc))^(1/2), where s = 1/2(a+b+c), the sum of the three sides.
Let a=(x1), b=x, c=(x+1); clearly s = 3x/2. Now we substitute for a,b,c:
A = ((3x/2)((3x/2)(x1))((3x/2)x)((3x/2)(x+1)))^(1/2) =3^(1/2)/4*((x^2)* (x^2  4))^(1/2). The initial term makes the geometrical connection with equilateral triangles tolerably clear.
As mentioned by others, if we want integer x, we must obviously select it so that (x^2  4))^(1/2) includes an integer multiple of 3^(1/2), while otherwise being a square, i.e. x^2  4 = 3y^2, for some y.
This leads very quickly to the recurrence x = (2 + sqrt(3))^n + (2  sqrt(3))^n, where x= {4,14,52,194,...}.
(Parenthetically, the Heron Area is then most easily defined in terms of these x's, e.g. A = x/4*(3*(x2)(x+2))^(1/2), which Excel copes with very easily.)
Now it is just a question of running through the recurrence until we find a qualifying solution: when n=13, x=27246964, the only solution for R compliant with the terms of the problem. This triangle has an area of 321467351292366; the corresponding equilateral triangle has an area of 321467351292366.866..., very close indeed to the integer solution.
Incidentally (2 + (3)^(1/2))^2 = 7+4*3^(1/2), a number relevant to today's puzzle.
Edited on December 18, 2016, 7:10 am

Posted by broll
on 20161218 07:00:32 