Equation

[Muffet]

A numerical 'crossword' of a while ago based on 60 and 120 degree triangles..
a sq +/- ab + b sq = c sq. c is theside facing the angle.
It was not known whether 60 or 120 deg.
a b and c are all integers, possibly 6 figures in length. Various values for c were given . The values for a and b had to be obtained.
Ideas on how to soilve would beappreciated. If lengthy, is there a library text book that would inform?

[TeeGee]

The formula should read c=sqrt(a*a+_2*a*bcosC+b*b) and is a derivation of the cosine rule. Bearing in mind that the cos of 60deg is +- 0.5 gives you c=sqrt(a*a+-a*b+b*b).

GCSE Maths.

[gen2]

TeeGee, That is precisely the equation that Muffet has given. Let me explain what I understand Muffet is asking:

For any given value of c there will be an infinite number of possible values for a and b. The problem is that Muffet is only looking for solutions where both a and b are integers and may want answers up to 999999. I initially thought of doing it with a spreadsheet but you would need a huge spreadsheet with 999999 rows. I cannot think of a simpler way yet.

[TeeGee]

I think I better understand your question, but not your logic.
Side c is opposite angle C, side a is opposite angle A etc.

If C is either 60 or 120 degrees ( which gives a cos of +-0.5 ) then Muffet's formula is as stated. Surely therefore, for each value of c there will be an finite number of values each for a and b ( these are interchangable ) depending upon C being 60 or 120deg. With C at 60deg incidentally, one set of results is always a=b=c ( eqilateral triangle )
My solution would be to write a small Pascal programme that would test for each given value of c ( from 1 to 999999 ) and each produced value for a and b were indeed integers, using Muffet's formula.

[Muffet]

Many thanks, TeeGee and gen2. This was one of the Listener puzzles. Sadly, I haven't any of the values of c that I can offer to you for experimentation.
Just on the matter of finite or infinite, I would think there must be an infinite number of possible values for a and b in relation to each c, in keeping with the infinite number of possible other angles. Presumably only one combination for each consists of integers.
I hope you'll not have sleepless nights over this!!

[TeeGee]

`Yes, infinite is the right answer, but there would only be a finite number of integer answers. Values of c from 1 to 50 yield 156 combinations, many of them duplicated of course. ( 6,10,14 : 10, 6,14 : 10,14,6 : 14,6,10 : 14,10,6 for example )
I havn't bothered to search for unique sets yet, but could do if requested.

Night, Night.

[Muffet]

Fantastic, TeeGee. Just by way of explanation, I've only just become aware of Answerbank. The question of how to solve that particular puzzle has been bugging me for a long time, and I thoughjt I would give it an airing. Glad I did !! I'll see if I can get hold of some of the higher values of c. If so, I'll try again. Thanks for your interest.