Can a non-equilateral triangle with integer sides and integer angles (in degrees) exist?

Question

I'm wondering if a triangle with integer sides and integer angles in degrees exists, which is not equilateral. Could someone come up with a proof for why or why not? *Edit:* Please do not answer with a triangle which has integer sides only. The triangle must have integer angles in degrees; this means they must actually be integers, not rounded off.

Answer 1

No I don't believe so. Here is the best solution I can give you:

Use Excel to create a spreadsheet.

First column numbers 1-90. I put "1' in block A1 then in A2 put "=1+1" and then copied and pasted down to 90

Second column convert from degrees to radians. I put "=pi()*A1/180" in B1, then copied like above.

Third column find sin of second column. I put "=sin(B1) in C1 then copied like above.

I then formatted column C to be a number to 15 decimal places.

In reviewing the outputs, the only angles that appeared to have rational values (ratio of integers) was 0, 30, 90.

0 and 90 don't make a triangle. Sin of 30 is rational, but the third side would then equal the hypotenuse times sin(60) which is not rational.

After 15 decimal places, they all appeared to just have zeros after that, suggesting they might be rational. But I believe this is just a limitation of Excel.

I would suggest doing the same on a calculator that you can switch between rational (exact) and decimal approximation answers. If for the sine of any integer angle X, you get an exact rational answer, you should then check the sine of its Complimentary Angle (90-X). If both answers are rational, you then have the 2 angles of a right triangle that would meet your requirement.

In writing this answer, I just realized that there was no stipulation that it had to be a right triangle. However, the above proof still works for the most part. If, as I suspect, the only rational value of sin is sin(30) = 1/2 (0,90 don't help as mentioned earlier), then 30-60-90 doesn't work and 30-30-120 doesn't work.

Hope that helps and makes sense.

Answer 2

Proof by contradiction using a pigeon hole argument:

Suppose there exists such a triangle, with integral side measures a,b, and c and integral angle A,B, and C,

with angles A,B, and C NOT ALL EQUAL.

sin A/ a = sin B / b = sin C / c by Law of Sines and

These equations can be rewritten as: b = a*sinB/sinA, c = b*sin C / sin B, and a = c * sin A/ sin C

So the sine in the numerator MUST be an integral multiple of sine in the denominator.

So we must find angles A.B and C such that sin B = k * sin A, sin C = n * sin B, and sin A = m * sin C

for integers, k,m, and n.

Then, B = inv_sin(k * sinA) , C = inv_sin( n * sin B) and A = inv_sin(m * sin C)

Plugging the first into the second, C = inv_sin( n * k * sin A)

Plugging that into the third:

A = inv_sin(m * n * k * sin A)

THen, taking the sine of both sides:

sin A = m*n*k * sin A

1 = m*n*k

this forces m=n=k = 1, since m,n, and k are integers, unless two of the three are both -1.

Then, in the former case, the sine functions are all identical, per the statement above in bold.

This forces the angles to be the same, which is a contradiction, since the triangle is given to

be not equilateral.

Now , if any two of the three fixed integers of n,m, and k are -1. then two pair of these trig functions have

OPPOSITE values, since their quotient is -1. In that case, sin T = -sin T.

Then 2*sin T = 0 ---> sin T = 0 ----> T = 0 or 180 which cannot be angle measures of a triangle.

This completes the proof.

A cleaner argument feature law of cosines and sines, stating the cosine and sine of any angle other

than 30 and 60 degrees respectively are irrational. Specifically, sin 30 = cos 60 = sin 150 = 1/2 and

sine and cosine trig functions are irrational everywhere else. A triangle with two angles of degree 30 each, forces the 3rd angle to be 120 which has irrational trig functions. 60 degree angles force the equilateral, and

the 150 degree angle will not fit at all.