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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.

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    $\begingroup$ How about an equilateral triangle with side length 1? $\endgroup$
    – Franklin Pezzuti Dyer
    May 5, 2017 at 23:27
  • $\begingroup$ @FranklinP.Dyer Wow, how did I forget that. I'll edit the question so it reads differently. $\endgroup$
    – u8y7541
    May 5, 2017 at 23:27
  • $\begingroup$ Degrees are artificial. Rational multiples of $\pi$ or some other condition on the radian measures would make more sense. $\endgroup$
    – Ethan Bolker
    May 5, 2017 at 23:30
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    $\begingroup$ @EthanBolker It actually doesn't change the answer whether you take integer numbers of degrees or rational numbers of degrees (equivalently, a rational multiple of $\pi$ radians). $\endgroup$
    – Mark S.
    May 6, 2017 at 0:07
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    $\begingroup$ Euclidean geometry, right? $\endgroup$
    – chepner
    May 6, 2017 at 13:51

3 Answers 3

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As mentioned on the Wikipedia page for "integer triangle", the law of cosines forces the cosine of an angle in a triangle with integer sides to be a rational number.

Now, when does an integer angle (in degrees) have a rational cosine? We can go one step further: when does a rational angle (in degrees) have a rational cosine? This question is answered with a clever proof on page 2 of When is the (co)sine of a rational angle equal to a rational number? by Jörg Jahnel. The answer is that the rational cosines of the rational angles are just $\pm1,\pm\frac12,0$. (This is known as Niven's Theorem and an alternate proof can be found at ProofWiki.)

Since cosines of $\pm1$ wouldn't correspond to the angle of a genuine triangle, the only potential options are $\cos\theta=0,\pm\frac12$ for angles of $\theta=60^\circ,90^\circ,120^\circ$. But if there is one angle of $90^\circ$ or $120^\circ$, there isn't enough room in $180^\circ$ for two more angles of at least $60^\circ$. And the only triangles whose angles are all $60^\circ$ are equilateral.

The answer to the question "Can a non-equilateral triangle with integer sides and integer angles (in degrees) exist?" is no.

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    $\begingroup$ A cosine of 0 can correspond to the angle of a genuine triangle. $\endgroup$
    – aschepler
    May 6, 2017 at 1:24
  • $\begingroup$ @aschepler Of course, fixed. $\endgroup$
    – Mark S.
    May 6, 2017 at 1:41
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Why measured angles in degrees and call those triangles important when degrees are a convention unrelated to the natural division of angles. For example 3°, 7°, 9° and 11° are all terrible divisions of angle with little mathematical meaning. I suggest these angles are human divisions with little significance in regard to appearing as integers in our poor conversions. I suggest 180° and 90° are important as 1/2 and 1/4 of a whole turn and NOT the fact that they appear as integers in some randomly chosen divisional selection of 360 that the Babylonians dreamt up and that society procrastinated from fixing for 2000 years.

My opinion is that angular divisions should be based on real geometric divisions of a circle that can be constructed with compass and straight edge. Instead of handing our young children devices that can not be constructed from the other devices. Children can NOT make protractors on a piece of paper from the geometric tools at their disposal. Isn't that strange and almost deceitful to do to them? It's like we are training children to think that integer number of degrees have some real significance and not just happen to line up with the constructional ones because some fractions of our poor divisions line up with the more geometrically true divisions.

Angles are fractions of a whole turn. Only some angles are important when the angle is measured in integer degrees. The question is asking which triangles have angles of even divisions of 1/360th of a turn. ie: if you measured your triangles with radian angles, then their would be only 3 integer angles per turn: 1,2 and 3 rads. Which can not be summed to make triangles of pi rad degrees. And the answer to your question would be: "no triangle like that exist"

A more general question would be: which triangles have integer length sides and angles that are all rational fractions of 1/2 turn (180 degrees).

For example: what if there was a triangle like you described having internal angles of 1/7(360°), 2/7(360°) and 4/7(360°) with integer sides, wouldn't you want to know? Angles measured in degrees would miss this triangle.

Comparison of degree angle vs rational angle triangles:

60°:60°:60° = rational angles 1/3:1/3:1/3  = 1:1:1

30°:60°:90° = rational angles 1/6:2/6:3/6 = 1:2:3

36°:72°:72° = rational angles 1/5:2/5:2/5 = 1:2:2

45°:45°:90° = rational angles 1/4:1/4:2/4 = 1:1:2

But now notice:

rational angles  1/7 : 2/7 : 4/7 = 1 : 2 : 4 != degree angles ~25.71° : ~51.42° : ~205.71°

...which is a triangle (if it had integer sides) that would be excluded from the set of triangles with integer degrees. But since all the angles are proportions of a common angle (1/14th of a full turn), isn't this triangle special?

As further evidence this question is limited (because of degrees) is that: No triangle has integer sides. The length of the side of triangle is dependent on the scale of the ruler to measure it. For example, equilateral triangles almost never have integer sides unless you declare it to be so and make that side to be the ruler of measure. The important fact is really the proportion of one side to another. Because only one specific equilateral triangles really has sides of one. But all equilateral triangles have sides that divide into each other evenly to produce a ratio of 1:1

  • this argument in not an answer, but would not fit within comment bounds
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Yes. There is another integer side/angle triangle.

The degenerate triangle:

Sides: 1:1:0

Degree angles: 0°:0°:180°

It looks very much like a line segment of length 1.

Some people don't like degenerate triangles. But they are valid when calculating the points 1 and -1 on a unit circle using Eulers formula which calculates based on sides of triangles including the degenerates.

https://en.wikipedia.org/wiki/Degeneracy_(mathematics)#Triangle

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  • $\begingroup$ Euler's formula does not need to involve geometric shapes at all, but if I were to model it geometrically, I would just use the coordinates of points on a unit circle around the origin of a Cartesian plane. Triangles are something you might superimpose on the figure after the actual mathematics is done in order to relate it to the trigonometry of right triangles. So we don't need to accept degenerate triangles for this purpose. (On the other hand I don't object to considering them; this is just to point out that the particular argument in the answer is not conclusive.) $\endgroup$
    – David K
    Jul 7, 2021 at 1:40
  • $\begingroup$ If we consider degenerate triangles, a triangle with sides in the ratio $1:1:0$ must have one angle $0$, but the other two angles can be any two supplementary angles. You can also have a triangle with sides in the ratio $a:b:a+b$ for any positive $a$ and $b$, where the angles then are $0,0,\pi.$ $\endgroup$
    – David K
    Jul 7, 2021 at 1:41
  • $\begingroup$ You're correct about euler. I should have just mentioned cos or sin and not euler's formula. Truth is that cos(0) has an answer of 0. Which implies the degenerate triangle is accepted as a right triangle and not a line. If right degenerates, where the second angle is 90 to 180 and 0 to 90 for the third angle, count as answers then all of them are valid. So I guess there are 91 triangles which qualify as triangles that are not equilateral. All 91 triangles having sides of 1:1:0 and the angles being one of: 0:0:180, 0:1:179, 0:2:178 ... 0:89:91, 0:90:90 $\endgroup$
    – peawormsworth
    Jul 7, 2021 at 2:32

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