How to obtain a list of random integers satisfying the Law of Cosine?

Hello everyone,

Can anybody help me in obtaining random integers, “a”, “b” and “c” ranging from 0 to 100, which satisfy the Law of Cosine c^2=a^2+b^2+a*b, for the fixed corresponding angle equals 120 degrees.

[darova]

Can you show what you have tried?

[John D'Errico]

Not hard, but an interesting homework assignment, but surely your homework. If you want help, then make an effort. Show how you might try to solve it, and then you might get help.

Answer 1

[Image Analyst]

So I guess the triple nested for loop won't work because they need to be obtained from a random draw and not sequentially? Have you heard of the randi() function?

Do you have to find them ALL? Since to make sure you've tested them all, and you're using randi() instead of a triple-for-loop, you're going to have to ask for lots of them. Let's see, a 100x100x100 would take a million test, so taking them randomly might take a few hundred million, which is certainly within MATLAB's power. Then use a single for loop to test them all

N = 200000000; % or whatever.
r = randi(..........
for k = 1 : N
    if r(1) + ....................do the test
end

[John D'Errico]

There are 122 solutions, such that a and b are no larger than 100. That reduces to 84 solutions such that c is also no larger than 100.

If we then consider that [a,b,c] is equivalent to the solution [b,a,c], then we can recognize that we need only consider solutions such that a<=b. The set now reduces to 42 solutions.

Finally, the interesting solutions, like those of Pythagorean triples, are only those cases where [a,b,c] are all relatively co-prime.

Given all that, there appear to be exactly 14 such co-prime solutions where none of a,b,c exceed 100, and a<b.

So the simple approach to this would be to just list all 14 such solutions.

abc =
     3     5     7
     5    16    19
     7     8    13
     7    33    37
     9    56    61
    11    24    31
    11    85    91
    13    35    43
    16    39    49
    17    63    73
    19    80    91
    32    45    67
    40    51    79
    55    57    97

Finally, you could just choose randomly from that set.

Deriving the set abc is slightly less trivial, but it can be done easily enough using brute force.

Of more interest yet, is the question of whether there is a Euclidean-style formula for all such triples in lowest form (much like the classic formula for Pythagonean triples)? A solution might be impossible, IF one exists. (I won't answer that question, at least not yet, since it has nothing to do with MATLAB.)

[Guillaume]

I get a few more solution that satisify the equation:

>> abc
abc =
     3     5     7
     5    16    19
     6    10    14
     7     8    13
     7    33    37
     9    15    21
     9    56    61
    10    32    38
    11    24    31
    11    85    91
    12    20    28
    13    35    43
    14    16    26
    14    66    74
    15    25    35
    15    48    57
    16    39    49
    17    63    73
    18    30    42
    19    80    91
    20    64    76
    21    24    39
    21    35    49
    22    48    62
    24    40    56
    25    80    95
    26    70    86
    27    45    63
    28    32    52
    30    50    70
    32    45    67
    32    78    98
    33    55    77
    33    72    93
    35    40    65
    36    60    84
    39    65    91
    40    51    79
    42    48    78
    42    70    98
    49    56    91
    55    57    97
>> all(abc(:,3).^2 == abc(:, 1).^2 + abc(:, 2).^2 + abc(:, 1).*abc(:, 2))
ans =
  logical
   1

[John D'Errico]

Yes, but they are not reduced to lowest form. Thus, we see sets like [3 5 7], and [6 10 14], and [30 50 70], all of which are essentially the same set. I said in my comment there are 122 sets with a and b no larger than 100.

If I reduce that to have also c less than 100, then we see 84 sets. If we further reduce it to have a less than b, and then sets where all are coprime, it looks like 14 such irreducible sets.

The actual question by the OP asked for nothing more than that all are no larger than 100, so my additional filtering is over the top, except that it helps to understand how one might derive a formula for the actual triples, much like a Euclid style formula for Pythagorean triples.

[John D'Errico]

By the way, better than testing all 1e6 sets of numbers, one can simply tst all 10000 combinations of a and b. Compute c from, that, then if it is an integer, you have found a solution.

[Guillaume]

@John, understood, but as you said, for the OP problem I'm not sure that filtering homothetic triangles is what they're after.

[NaviKan]

Thanks, everyone. Especially thanks to John D'Errico, Image Analyst and Guillaume for their useful suggestions, efforts, and guidance. I have solved the problem in some other ways.

[Bruno Luong]

The reduction has one advantage : make a generation of the table faster (migh be) since it reduce the number of cases if brute force method is used.

Now if some one can propose a method that is not a bruteforce check for a^2+b^2+a*b is a perfect square (with optional assumption of a and b prime to each other) I'll be very interested.

[Image Analyst]

I'd be interested too. So Naveed you did not use random numbers and then check, nor did you use brute force to check all permutations -- these were the two methods that we suggested. So what method did you use? Did you even use randomness at all?

[NaviKan]

@Image Analyst In my case, I have used two for loops along with several conditional statements, such that “a” ranges from 1:n and “b” ranges from n:m. Whereas, on “c” I put the condition <=100. I have also used “gcd(a,b)~=1” and gcd(b,c)~=1 to avoid the similar angles repetition. Because my main target was to obtain a tangent angle “tan(theta)=b*sqrt(3)/(2*a+b)” at which some properties are repeated.

[Bruno Luong]

I don't think you need to check gcd(b,c)~=1, it is redundant to gcd(a,b)~=1.

For the record here is the code generate 14 primitive triples (brute force)

a=1:100;
b=(1:100)';

c=sqrt(a.^2+b.^2+a.*b);
[j,i,c]=find(mod(c,1)==0 & c<=100 & bsxfun(@gcd,a,b)==1);
abc=unique(sort([a(i)',b(j),c],2),'rows')

EDIT: optimization following Guillaume's suggestion

[Guillaume]

Note that you can do:

[j, i, c] = find(mod(c,1)==0 & c<=100);
a = a(i)'; b = b(j);

No need for the ind2sub.

[Bruno Luong]

Thanks Guillaume.