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$\arctan(1/2)$ 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?

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    $\begingroup$ 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. $\endgroup$
    – Blue
    Aug 14, 2023 at 20:21
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    $\begingroup$ I like the approach here a lot. The formula $$\arctan1+\arctan2+\arctan3=\pi$$ is equivalent to yours, and easier to visualize! $\endgroup$ Aug 14, 2023 at 20:57
  • $\begingroup$ @Blue Did you have in mind the same picture I had (see the link)? $\endgroup$ Aug 14, 2023 at 21:00
  • $\begingroup$ @JyrkiLahtonen: That's good, too. See my own link. $\endgroup$
    – Blue
    Aug 14, 2023 at 22:46
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    $\begingroup$ The simplest algebraic reason is $(2+i)(3+i)=5 (1 + i)$. $\endgroup$
    – lhf
    Aug 15, 2023 at 1:11

5 Answers 5

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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.

enter image description here

$\triangle \space ACD$ 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 $\arctan(1/2)+\arctan(1/3)=45°$.

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

enter image description here

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 $\tan^{-1} (\text{rational number})$.

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

cross_angle

$\bf{Added:}$ 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

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

tangent of sums

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.

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    $\begingroup$ Can you include a diagram to emphasize the geometry? $\endgroup$
    – qwr
    Aug 15, 2023 at 2:48
  • $\begingroup$ @qwr: Just added one, thanks for the feedback! $\endgroup$
    – orangeskid
    Aug 15, 2023 at 4:15
  • $\begingroup$ what program did you use? geogebra? it's very hard for me to see $\endgroup$
    – qwr
    Aug 15, 2023 at 4:19
  • $\begingroup$ @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. $\endgroup$
    – orangeskid
    Aug 15, 2023 at 4:23
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    $\begingroup$ yes I enlarge the image, it's just the colors. I can take a stab at making a diagram tomorrow $\endgroup$
    – qwr
    Aug 15, 2023 at 4:28
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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 $w=(a-b,a+b).$

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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:

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

Equilateral right triangle sides sqrt(10)

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