OEIS A102341: areas of 'close-to-equilateral' integer triangles. To quote from OEIS:
The corresponding sequence of areas for these triangles is 12, 120, 1848, 25080, 351780. So the 'close-to-equilateral' integer triangle with sides 241, 241 and 240 has an integer area of 25080. Here is the image that I posted to my Twitter account, taken from WolframAlpha:A close-to-equilateral integer triangle is defined to be a triangle with integer sides and integer area such that the largest and smallest sides differ in length by unity. The first five close-to-equilateral integer triangles have sides (5, 5, 6), (17, 17, 16), (65, 65, 66), (241, 241, 240) and (901, 901, 902).
Here is further detail from the OEIS entry:
Heron's formula: a triangle with side lengths (x,y,z) has area A=√s∗(s−x)∗(s−y)∗(s−z) where s=(x+y+z)/2. For this sequence we assume integer side-lengths x=y=z±1. Then for A to also be an integer, x+y+z must be even, so we can assume z=2k for some positive integer k. Now s=(x+y+z)/2=3k±1 and A=√(3∗k±1)∗k∗k∗(k±1)=k∗√3∗k2±4∗k+1. To determine when this is an integer, set 3∗k2±4∗k+1=d2. If we multiply both sides by 3, it is easier to complete the square: (3∗k±2)2−1=3∗d2. Now we are looking for solutions to the Pell equation c2−3∗d2=1 with c=3∗k±2, for which there are infinitely many solutions: use the upper principal convergents of the continued fraction expansion of √3.Here are links supplied in the OEIS entry:
- Danny Rorabaugh, Table of n, a(n) for n = 1..875
- Steven Dutch, Almost 30-60 Triples
- ProjectEuler, Problem 94: Almost equilateral triangles
- Eric Weisstein's World of Mathematics, Heronian Triangle and Pell Equation
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