I drew all the regular polygons in a circle of radius one. I decided to take one side of the equilateral triangle, one side of the square and one side of the regular hexagon to form a right triangle. Is it possible to form other right triangles from other regular polygons in this circle?

  • $\begingroup$ Did you draw a regular 2-gon as well? $\endgroup$
    – Bass
    Aug 14, 2020 at 5:31
  • $\begingroup$ No, it’s not considered as a polygon. $\endgroup$ Aug 14, 2020 at 13:18

3 Answers 3


This is WIP! An attempt at showing that @tehtmi's


are all solutions.

Update: New and elementary(-ish) but still unfinished approach using

Ptolemy theorem and Chebyshev polys of 2nd kind.

With this approach we can leave the sides of the triangle inscribed in the unit circle:

Let $\alpha<\beta<\gamma$ the edge counts of the regular polygons forming a solution. If $m$ is the lcm of $\alpha,\beta,\gamma$ let us write $m = a\alpha = b\beta = c\gamma$ and $S(k) = 2 \sin \frac {k \pi} m$ such that the sides of the polygons become $A = S(a) > B = S(b) > C = S(c)$ Take two of them, $A,C$, say and place four points on the unit circle such that they form a trapezoid with two sides $C$ and diagonal $A$. Now we can use Ptolemy's formula for inscribed quadrilaterals to translate $A^2-C^2=B^2$ into (*) $S^2(b) = S(a-c) S(a+c)$.

Now it is time to introduce Chebyshev's polyonomials of the second kind $U_n(\cos x) = \frac {\sin [(n+1)x]}{\sin x}$ These come with a handy product formula $U_n(x)U_m(x) = \sum_0^n U_{m-n+2k}(x)$. Applying this to (*) we get $\sum_0^{b-1} U_{2k}(x) = \sum_{c}^{a-1} U_{2k}(x)$ or, equivalently, $\sum_0^{c-1} U_{2k}(x) = \sum_{b}^{a-1} U_{2k}(x)$.

And that's where I'm stuck.

Stuff below is probably obsolete but some may find it interesting.

Let $m$ be the lcm of the vertex counts $a,b,c$ and $\omega = e^{\frac {2\pi i} m}$ the standard $m$th root of unity. Write $m = aA = bB = cC$. Then $\sin^2 \frac \pi a + \sin^2 \frac \pi b = \sin^2 \frac \pi c$ can be rewritten as a polynomial $P(X) = X^A + X^{m-A} + X^B + X^{m-B} - X^C - X^{m-C} - 2$ with a zero at $\omega$.

Now, my field theory is more than a bit rusty but IIRC because this polynomial has rational coeffs it follows that it must be zerod by any primitive $m$th root of unity i.e. $\omega^k$ for any relative prime $0<k<m$. It follows that $P$ must be a multiple of the $m$th cyclotomic polynomial.

Let us check this at the two solutions we know. The lcm of $(3,4,6)$ is $12$ with cyclotomic poly $C_{12}(X) = X^4-X^2+X$ In this case $P(X)=-X^4-X^8+X^3+X^9+X^2+X^{10}-2 = C_{12}(X)C_3(X)C_1(X)[X^3+X^2+2]$.

The lcm of $5,6,10$ is $30$ with cyclotomic poly $C_{30}(X)=X^8+X^7-X^5-X^4-X^3+X+1$ and $P(X)=-X^6-X^{24}+X^5+X^{25}+X^3+X^{27}-2 = C_{30}(X)C_1(X)[X^{18} + 2X^{16} + 2X^{14} + 2X^{13} + X^{12} + 3X^{11} + X^{10} + 3X^9 + 2X^8 + 2X^7 + 2X^6 + 2X^5 + 2X^4 + X^3 + 2X^2 + 2]$. Not pretty but it works

I didn't think deeply about it but I think that because all primitive roots of unity solve $P$ it means that, for example, not only the regular pentagon works but also the regular pentagram (as long as one chooses the right corresponing 10-thingy and 6-thingy configurations).

Possible strats from here

Perhaps this could be used for a proof that other solutions do not exist. For example one might succeed in because different primitive roots of unity should for larger numbers (using Euler $\Phi$) produce all orderings $\sin^2 \frac {k\pi} {a} \lessgtr \sin^2 \frac {k\pi} {b} \lessgtr$ etc. which would be a contradiction if $c$ comes out smallest.

Or perhaps

look at degrees of field extensions: If sides $\alpha,\beta,\gamma$ form a right triangle then, any of the three is either already contained in the extension field formed by the other two or is part of a quadratic extension. This constrains the Euler $\Phi$'s of $a,b,c$ and derived quantities.

  • $\begingroup$ rot13( "orpnhfr guvf cbylabzvny unf engvbany pbrssf vg sbyybjf gung vg zhfg or mrebq ol nal cevzvgvir zgu ebbg bs havgl"; lrf, V guvax guvf vf gehr orpnhfr gur Tnybvf tebhc pna gnxr nal cevzvgvir ebbg gb nal bgure cevzvgvir ebbg jvgubhg punatvat nal pbrssf. ) I agree this seems like a fruitful way to approach the proof. $\endgroup$
    – tehtmi
    Aug 16, 2020 at 0:19
  • $\begingroup$ @tehtmi Feel free to take it from here if you have any ideas. I'm stuck. $\endgroup$ Aug 16, 2020 at 1:40

I must interpret the question to only care about the side lengths. If the triangle, square, and hexagon are simultaneously inscribed in the same circle, their sides cannot form any triangle at all, but the side lengths ($1$, $\sqrt{2}$, and $\sqrt{3}$) satisfy the Pythagorean theorem.

We can use the pentagon, hexagon, and decagon. If it is allowed to use two sides from one polygon, we can also use two sides from the hexagon and one side from the square. These should be the only solutions (see proof sketch).

In general, we can expand with the law of cosines so that if an $a$-gon, $b$-gon, and $c$-gon give a solution ($a < b \leq c$), then $1=\cos(\frac{2\pi}{b})+\cos(\frac{2\pi}{c})-\cos(\frac{2\pi}{a})$.

For our solution, these values are well-known results that can be calculated using elementary geometry and algebra: $\cos(72^\circ)=-\frac{1}{4}+\frac{\sqrt{5}}{4}, \cos(36^\circ)=\frac{1}{4}+\frac{\sqrt{5}}{4}$.

Proof sketch:

If we expand $\cos(\theta)=\frac{1}{2}\exp(i\theta)+\frac{1}{2}\exp(-i\theta)$ we can think of our cosine formula as an integral linear combination of roots of unity $2+z_1+\bar z_1-z_2-\bar z_2-z_3-\bar z_3=0$. It is known that all such relations are generated by relations of the form $1 + \zeta + \zeta^2 + ... + \zeta^{p-1}=0$ for prime $p$ (where $\zeta$ is a $p$-th root of unity) and their rotations. But, we are restricted to using only roots of unity corresponding to angles $\pm \frac{2\pi}{k}$, or their reflections (from the negative signs). There are only a small number of cases to consider. We can use $p=2$ (rotated) where we can get $z+\bar z=0$ from a square, $p=3$ where we can get $1+z+\bar z = 0$ from a triangle or $1-z-\bar z$ from a hexagon, or $p=5$ where we can get $1+z_1+\bar z_1-z_2-\bar z_2$ with $z_1$ coming from a pentagon and $z_2$ coming from a decagon. Thus, we see our three solutions.

  • $\begingroup$ You are right. There aren’t more triples of different polygons. Can you provide a proof for this fact? $\endgroup$ Aug 14, 2020 at 13:27

Lets draw a circle with R=1 and then draw two diameters perpendicular to each other (Figure 1). From this construction we mark the four vertices of a square A,B,Z,E. From the point B draw BC=R=1 and from the point A draw the straight line AC. (AC)^2=(AB)^2+(BC)^2 so AC is equal to the square root of 3 which means AC is the side of the equilateral triangle inscribed in the circle. As you can see only two vertices of the triangle ABC touch the circumference.

Now draw another circle with R=1 (Figure 2). Draw a chord AB=R. Now from the point A draw a chord AC equal to the side of a regular decagon and then draw the chord CB. The segment AB is 1/6 of the circumference and the segment AC is 1/10 so the segment CB=1/6-1/10=1/15. This result sows the chord CB is the side of a regular quindecagon.


  • $\begingroup$ What are you trying to show? I don't think figure 2 is a right triangle. $\endgroup$
    – tehtmi
    Aug 16, 2020 at 0:08
  • $\begingroup$ Figure 2 displays a non right triangle made with sides which are sides of regular polygons with the three vertices touching the circumference. $\endgroup$ Aug 16, 2020 at 1:38

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