"Critical" Triangle Problem

[DanteKennedy]

Homework Statement

Let triangle ABC be a right-angled triangle with angle C = 90°. Let a, b, and c denote the lengths of the sides opposite to vertices A, B, and C, respectively, and let alpha and beta denote the measures of the interior angles at A and B.

Relevant Equations

A right triangle is called "critical" if it simultaneously satisfies the following conditions:
1. All side lengths a, b, and c are strictly positive rational numbers

2. The measures of the acute angles alpha and beta, when expressed in pure degrees, are rational numbers (ex. 90°, 68.4°, etc.)

If such a triangle exists, provide an example; if not, prove its non-existence.

No attempts to solve it by myself yet. It would be very nice if someone can help me

 

[.Scott]

This is a big clue.
Don't click this unless you're ready for a spoiler.

 

[DanteKennedy]

Thanks, so it seems like no such triangle exist

 

[PeroK]

Here's my thinking. We need an angle $\theta = q\pi$ (in radians), where $q$ and both $\cos \theta$ and $\sin \theta$ are rational. We can write this using Euler's identity:
$$e^{iq\pi} = e^{i\theta} = \cos \theta + i\sin \theta = a + ib \ \ (q, a,b \in \mathbb Q)$$Now, if $q = k/m$, then $2mq = 2k$ and:
$$(a + ib)^{2m} = e^{i2k\pi} = 1$$I don't know if that helps.

 

[PeroK]

I think you can finish it off by looking at the imaginary part and using the rational root theorem. This would show that either $a = 0$ or $b = 0$.

There's a key factor in this particular problem that means you don't need to use the full power of Niven's theorem. That is that all sides of the triangle have rational lengths.

Or, perhaps not?

 

[.Scott]

According to Wiki, the case against rational tangent of rational angle in degrees is provided in Nivens "Irrational Numbers" - a book that I do not have.
But that is the specific case required by the OP.

 

[willyengland]

You can find the pdf online, e.g.
https://webhomes.maths.ed.ac.uk/~v1ranick/papers/niven.pdf

 

[.Scott]

You can find the pdf online, e.g.
https://webhomes.maths.ed.ac.uk/~v1ranick/papers/niven.pdf

Thanks @willyengland !

Section 5 (page numbers 38 to 41; pdf sheets 49 to 52) is titled "Extension to the Tangent" and provides a proof that refutes the general case of a rational tangent of a rational angle in degrees. The result is state in his Corollary 3.12 which reports the only exception being tan(0).

 

[Hill]

In the Niven's book referenced above, is there an error in the last line of the following excerpt?

Shouldn't it rather be
$F^{(2)}(x) \sin (x)+F(x) \sin (x)=f(x) \sin (x)-f^{(4p)}(x) \sin (x)$
?

 

[Hill]

In the Niven's book referenced above, is there an error in the last line of the following excerpt?

View attachment 371939

Shouldn't it rather be
$F^{(2)}(x) \sin (x)+F(x) \sin (x)=f(x) \sin (x)-f^{(4p)}(x) \sin (x)$
?

Ah, perhaps it should, but as $f(x)$ is a polynomial of degree $4p-2$, $f^{(4p)}(x)=0$.

 

[pbuk]

According to Wiki, the case against rational tangent of rational angle in degrees is provided in Nivens "Irrational Numbers" - a book that I do not have.
But that is the specific case required by the OP.

No, we do not need to look for rational tangents. We can actually ignore condition 1 altogether:

1. All side lengths a, b, and c are strictly positive rational numbers

All we need is condition 2:

2. The measures of the acute angles alpha and beta, when expressed in pure degrees, are rational numbers (ex. 90°, 68.4°, etc.)

Niven's Theorem gives us that the only possible value for both alpha and beta are 30°, and $30° + 30° + 90° \ne 180°$

Wikipedia gives a link to a freely available proof as Lemma 12 in "Fermat's last theorem for rational exponents".