There is a famous classical geometry puzzle about the angles formed by integer coordinates:

What is the sum of angle A and B in the following image? Do not use any advanced mathematics such as trigonometric formulas or complex numbers.

with an elegant answer using pure geometry (spoiler-ed for those who want to try out themselves):

Now, we consider a general problem involving angles in the form of arctan(1/n), which is the smallest angle formed by a right triangle of height 1 and width n (an integer). For example, the angle A in the above image is equal to arctan(1/3), and B is arctan(1/2).

Prove or disprove the following statement:

Given two angles α=arctan(1/n) and β=arctan(1/m) where n and m are integers, it is possible to draw three triangles ABC, DEF, and GHI on a Cartesian plane, such that all vertices have integer coordinates and the following conditions are satisfied for each triangle:

  • the edge AB is parallel to the x-axis, angle A = α, angle B = β
  • the edge DE is parallel to the x-axis, angle E = α, angle F = β
  • the edge GH is parallel to the x-axis, angle I = α, angle G = β
  • $\begingroup$ FWIW, here is an SVG of an alternative demonstration of the fact in your spoiler. And if you like this kind of thing, here's a huge collection of integer (and some half integer) arccot identities. $\endgroup$
    – PM 2Ring
    Dec 9 '20 at 21:15

The statement is



The 𝐴𝐵𝐶 triangle can be simply constructed by joining two right-angled triangles of height 1 together, leading to the points 𝐴=(0, 0), 𝐵=(𝑚+𝑛, 0), 𝐶=(𝑛, 1).
Note that once you have the right-angled triangle 𝐴𝑃𝐶 with 𝐴=(0, 0), 𝑃=(𝑛, 0), 𝐶=(𝑛, 1), then you can simply extend the side 𝐴𝑃 by a factor (𝑚+𝑛)/𝑛 to get point 𝐵 of the triangle 𝐴𝐵𝐶 that we want.

So let's construct a triangle similar to 𝐴𝑃𝐶 such that its hypotenuse is horizontal. We can do this by scaling it by a factor 𝑛: (0, 0) (𝑛2, 0) (𝑛2, 𝑛), which allows us to join a rotated similar unscaled triangle (𝑛2, 0) (𝑛2, 𝑛) (𝑛2+1, 0). This results in the triangle 𝐸𝐷𝑄 with 𝐸=(0, 0), 𝐷=(𝑛2+1, 0), 𝑄=(𝑛2, 𝑛) where the top vertex 𝑄 is the right angle. We can now extend the side 𝐸𝑄 by a factor (𝑚+𝑛)/𝑛 to find point 𝐹. This gives the triangle 𝐸𝐷𝐹 we want: 𝐸=(0, 0), 𝐷=(𝑛2+1, 0), 𝐹=(𝑛(𝑚+𝑛), 𝑚+𝑛).

enter image description here

The 𝐺𝐻𝐼 triangle is the same as 𝐷𝐸𝐹 but with 𝑚 and 𝑛 swapped.

  • 1
    $\begingroup$ Jaap claiming the opposite...this can't be good. $\endgroup$ Dec 8 '20 at 10:39
  • 1
    $\begingroup$ @GeorgeMenoutis You seem to have focused on one particular example which does not work in your answer but discounted many examples that might work. $\endgroup$
    – hexomino
    Dec 8 '20 at 10:42
  • $\begingroup$ I'm accepting this answer because it is the first fully correct one. My approach was to apply the scaling and work with the product of two-square sums to find the coordinates of the point F, which gives the same coordinates as this answer. $\endgroup$
    – Bubbler
    Dec 8 '20 at 23:02

The statement is



Here is a constructive proof.

I am using the fact that in the space of Gaussian integers (complex integers) multiplication represents an angle-preserving transformation.
Triangle ABC can be created easily by setting A=(0,0), B=(n+m,0), C=(n,1).

To find triangle DEF, consider CAB in the complex plane and multiply it by -n+i.
DE ends up aligned with x or the real axis by construction.
Likewise, to find triangle GHI, multiply BCA by -m-i.

As an example, n=2, m=3.

A = (0,0), B=(5,0), C=(2,1)

D: $C(2-i) = (2+i)(-2+i) = -5$.
E: $A(2-i) = 0$.
F: $B(2-i) = 5(-2+i) = -10+5i$.

This gives D = (-5,0), E = (0,0), F = (-10,5).

G: $B(-3-i) = 5(-3-i) = -15 -5i$.
H: $C(-3-i) = (2+i)(-3-i) = -5 + -5i$.
I: $A(-3-i) = 0$.

Giving G = (-15,-5), H = (-5,-5), I = (0,0),

Triangle ABC obviously has the required angles α at A and β at B. The other triangles are just ABC rotated and scaled, preserving the angles. So it is easy to see that A, E and I all have the same angle, α, and B, F, G have the same angle, β.


Yes, the statement is true.


Let the unnamed angle in the three triangles be $180°-\phi$. That is, $$\phi=\alpha+\beta$$ Thus, $$\tan\phi=\frac{n+m}{nm-1}$$ The $ABC$ case is obvious. We have $AB$ of length $n+m$, and the perpendicular height of the triangle is 1. Eg, $$A=(-n,0), B=(m,0), C=(1,0)$$ The other two cases are symmetrical to each other, so I'll just illustrate the case with $\alpha$ on the $X$ axis: arccot triangle $$tan\alpha=\frac{n+m}{nm+n^2}=\frac1n$$ For the other case, simply swap $\alpha$ & $\beta$, and $n$ & $m$.


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