# A theorem about angles in the form of arctan(1/n)

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
• 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. Dec 9 '20 at 21:15

The statement is

true

Proof:

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).

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

• Jaap claiming the opposite...this can't be good. Dec 8 '20 at 10:39
• @GeorgeMenoutis You seem to have focused on one particular example which does not work in your answer but discounted many examples that might work. Dec 8 '20 at 10:42
• 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. Dec 8 '20 at 23:02

The statement is

true

Because

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.

Proof:

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: $$tan\alpha=\frac{n+m}{nm+n^2}=\frac1n$$ For the other case, simply swap $$\alpha$$ & $$\beta$$, and $$n$$ & $$m$$.