# Is it possible to explain geometrically why $\arctan (1/2) +\arctan (1/3) = 45$ degrees?

$$\arctan(1/2)$$ seems to be some strange, irrational angle, and the same goes for $$\arctan(1/3)$$, but those two angles seem to sum up to $$45$$ degrees. This seems like a mystery to me even though I can derive the result algebraically as follows, by using the summation formula for the tangent function.

$$\tan\big(\arctan(1/2)+\arctan (1/3)\big)=\dfrac{5/6}{1 - 1/6}=1\,.$$

Can someone come up with a geometric explanation?

A remark: I started thinking about the described arctan puzzle while I was trying to solve a problem in complex analysis, namely this one:

How to find a conformal mapping which maps the region between |z + 3| < √ 10 and |z − 2| < √ 5 onto the interior of the first quadrant?

• This is the gist of the well-known "Three Squares Problem" popularized by Martin Gardner. Numberphile has a YouTube video about it; my Trigonography site has a picture proof. I'm pretty sure I've seen it mentioned here, but I can't seem to find any instances after a cursory search.
– Blue
Aug 14, 2023 at 20:21
• I like the approach here a lot. The formula $$\arctan1+\arctan2+\arctan3=\pi$$ is equivalent to yours, and easier to visualize! Aug 14, 2023 at 20:57
• @Blue Did you have in mind the same picture I had (see the link)? Aug 14, 2023 at 21:00
• @JyrkiLahtonen: That's good, too. See my own link.
– Blue
Aug 14, 2023 at 22:46
• The simplest algebraic reason is $(2+i)(3+i)=5 (1 + i)$.
– lhf
Aug 15, 2023 at 1:11

There are many good geometric renderings; my favorite invokes the familiar Red Cross symbol. This symbol consists of a central square block sharing each of its edges with an additional congruent block. We superpose right triangle $$ABC$$ and label some additional vertices $$D,E,F$$ as shown below.

$$\triangle \space ACD$$ and $$CBE$$ are right triangles with congruent legs, so congruent triangles by SAS; thus their hypotenuses which are also legs of right $$\triangle ABC$$ are congruent. So the acute angle $$BAC$$ measures $$45°$$. But the component of that angle within right $$\triangle ACD$$ measures $$\arctan(1/2)$$ and the remaining component within right $$\triangle ABF$$ measures $$\arctan(1/3)$$. Thereby $$\arctan(1/2)+\arctan(1/3)=45°$$.

• When only the best will do...Nice! Aug 15, 2023 at 1:13
• @qwr I don't think that they are 30-60-90 triangles.They are two triangles with an angle $\arctan(1/2)$. Aug 15, 2023 at 2:54
• Yes you're right. I misinterpreted the diagram
– qwr
Aug 15, 2023 at 2:56
• Nice. I foresee a few calculus students working this out when reviewing trig functions :-) Aug 15, 2023 at 8:42

Let us take $$\triangle ABC$$ with $$AB=3, BC=4, CA=5, \angle B = 90^\circ$$. Let the angle bisectors $$AI, BI, CI$$ meet at incenter $$I$$. From $$I$$ drop perpendiculars $$ID,IE,IF$$ onto the sides. The inradius is $$ID=IE=IF=1$$. Clearly $$BFID$$ is a square, so $$BF=BD=1$$, $$AF=AE=2$$, $$CD=CE=3$$. This leads to the angles as in the following diagram.

Thus $$\frac{A}{2}=\tan^{-1} \frac{1}{2}$$, $$\frac{C}{2}=\tan^{-1} \frac{1}{3}$$. But $$A+C=\frac{\pi}{2}$$ means $$\frac{A}{2}+\frac{C}{2}=\tan^{-1} \frac{1}{2}+\tan^{-1} \frac{1}{3}=\frac{\pi}{4}$$ As a bonus, adding the six angles formed at the incenter gives $$2\tan^{-1} 1 + 2\tan^{-1} 2 +2\tan^{-1} 3 =2\pi$$ $$\Rightarrow \tan^{-1} 1 + \tan^{-1} 2 +\tan^{-1} 3 =\pi$$

Similar dissections of other right triangles $$\{(5,12,13),$$ $$(7,24,25), \ldots\}$$ will give new identities involving $$\tan^{-1} (\text{rational number})$$.

• For every Pythagorean triple $(a,b,c)$ with $c$ as hypotenuse: $\arctan(\frac{c-a+b}{c+a+b})+\arctan(\frac{c+a-b}{c+a+b})=45°$. $(3,4,5)\implies\arctan(1/2)+\arctan(1/3)$, $(5,12,13)\implies\arctan(1/5)+\arctan(2/3)$. Aug 15, 2023 at 20:45

Just a comment on the beautiful answer of @Oscar Lanzi:, if we add four strips of width $$1$$ and length $$x$$ to a square of side $$1$$, we get again a cross with an inscribed square, and the equality

$$\arctan \frac{1}{2x + 1} + \arctan\frac{x}{x+1} = \frac{\pi}{4}$$

$$\bf{Added:}$$ In the picture below, the yellow segments have length $$1$$, the light blue have length $$x$$.

$$\bf{Added:}$$ With the same method, with some care, we can prove the addition formula for the tangent. Indeed, consider two rectangles with sides, $$a$$, $$b$$ , and two with sides $$t a$$, $$t b$$. (in the picture below, sides $$2,5$$ and $$4$$, $$10$$) . Arrange them so we from their diagonals we get a rectangle with the ratio of the sides $$t$$. Then we get the formula

$$\arctan \frac{a}{b} + \arctan\frac{t b - a}{t a + b} = \arctan t$$

In our case

$$\arctan\frac{2}{5} + \arctan\frac{8}{9} = \arctan 2$$

Note: If we start from an arbitrary symmetric cross we might not get a rectangle, but only a parallelogram, so we modified the approach a bit. In general, to get a cross that produces a rotated rectangle, start with a container rectangle and intersect its sides with a central circle.

• Can you include a diagram to emphasize the geometry?
– qwr
Aug 15, 2023 at 2:48
• @qwr: Just added one, thanks for the feedback! Aug 15, 2023 at 4:15
• what program did you use? geogebra? it's very hard for me to see
– qwr
Aug 15, 2023 at 4:19
• @qwr: yes, it is geogebra, I should probably share the file somewhere. Can you click on the image, view, and then enlarge? I reduced the size due to bandwidth concerns of some people. Aug 15, 2023 at 4:23
• yes I enlarge the image, it's just the colors. I can take a stab at making a diagram tomorrow
– qwr
Aug 15, 2023 at 4:28

With $$0 < a < b,$$

$$\arctan \frac{a}{b} + \arctan \frac{-a+b}{a+b} = \frac{\pi}{4}$$

as vectors, in order to add the angles, we need to ask the angle between vectors $$v=(-1,2)$$ and $$w=(1,3).$$ This is what I was doing in answering your previous question.

$$\cos \theta = \frac{v \cdot w}{|v| \; |w|} = \frac{5}{ \sqrt 5 \; \sqrt{10}} = \frac{5}{\sqrt{50}}$$

so $$\cos \theta = \frac{5}{5 \sqrt 2} = \frac{1}{ \sqrt 2}$$

We get something similar with $$v=(-1,5)$$ and $$w=(2,3).$$

Or with $$v=(2,5)$$ and $$w=(-3,7).$$

Or with any integers $$0 < a < b,$$ vectors $$v=(a,b)$$ and $$w=(a-b,a+b).$$

Just to add to the variety of many fine geometrical answers, express the measures of the following three angles in terms of arctangent and compare the results:

$$\angle CAD, \angle DAB, \angle CAB$$