The Three Square Geometry Problem - Numberphile

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Recommends yberman · 2015-01 · Mathematics Stack Exchange · on “An example of a problem which is difficult but is made easier when a diagram is drawn”

I hate just posting a link (I do not have enough reputation to comment), but Zvezdelina Stankova provided a nice problem which is only approachable if you draw the right picture. See [this video] .

Recommends Arthur · 2016-06 · Mathematics Stack Exchange · on “Derive a relation between angles A,B and C”

If what they're really after is that $C = A + B$, then it is proven, for instance, in this video [this video] , by Numberphile (note that what they prove there is that $A + B + C = 90^\circ$, but if you know that $C = 45^\circ$, then we also have $A + B = 45^\circ$, so $C = A+B$ follows). The proof works by extending the figure and adding two more right triangles with angle $B$. One then easily sees that the angles $B$ and $C$ appear next to the angle $A$, forming a right angle, $A+B+C=90^\circ$ [LINK]

Recommends Blue · 2023-08 · Mathematics Stack Exchange · on “Is it possible to explain geometrically why $\arctan (1/2) +\arctan (1/3) = 45$ degrees?”

This is the gist of the well-known "Three Squares Problem" popularized by Martin Gardner. Numberphile has a YouTube video [this video] about it; my Trigonography site has a picture proof [LINK] . I'm *pretty sure* I've seen it mentioned here, but I can't seem to find any instances after a cursory search.

Cites David · 2014-12 · Mathematics Stack Exchange · on “Is it possible to find the limit of this?”

After recently watching a Numberphile video about a square problem [this video] I started thinking about what would happen to the sum of all angles if you had n amount of squares. After a bit of testing, I noticed that: $$In\space degrees\space and\space assuming\space each\space square\space is\space 1 \times 1\space units$$ $$\left(\sum\limits_{n=1}^{10^{x+1}} arctan\left(\frac{1}{n}\right)\right) - \left(\sum\limits_{n=1}^{10^x} arctan\left(\frac{1}{n}\right)\right) \approx 131.93 $$ So, in words, the sum of the angles of 100 squares stacked next to each other minus the sum of the angles of 10 squares stacked next to each other is approximately equal to 131.93. This is true for any $10^{x+1}$

Cites arunoruto · 2015-06 · Mathematics Stack Exchange · on “Evaluating $\lim\limits_{n \rightarrow \infty}\sum\limits_{k=1}^n \arctan\left(\frac{1}{\sqrt{k^2+1}}\right)$”

So I was watching an "old" video from Numberphile, "The Three Square Geometry Problem" [this video] . Here is a screenshot: [LINK] It's pretty easy to see that the sum of three angles is 90°, but now I am curious: What if we keep going on? What if we had more than 3 squares and with that more than 3 angles? What would the sum be? Basically I want to find the value of: $$\lim\limits_{n \rightarrow \infty}\sum\limits_{k=1}^n \arctan\left(\frac{1}{\sqrt{k^2+1}}\right)$$ If there is an answer I would like to know it. Thank you. P.S. $\arctan$ stands for $\tan^{-1}$

Cites Blue · 2016-07 · Mathematics Stack Exchange · on “Interesting Trigonometry Identity”

@SirJony: People may have seen it mentioned by Numberphile [this video] a couple of years ago. As I note in my Trigonography entry [LINK] , the puzzle was featured by Martin Gardner in 1996, and dates back to at least 1971.

Cites Blue · 2017-11 · Mathematics Stack Exchange · on “Elementary trigonometric proof problem using multiple angles in the sine rule”

I had thought that "sine rule" meant "Law of Sines", but perhaps not. Consider this derivation of the relation at issue: $$\begin{align} \sin(180^\circ-\angle ADB-\angle AEB) &= \sin(\angle ADB+\angle AEB) \tag{1}\\[4pt] &= \sin \angle ADB \cos \angle AEB + \cos \angle ADB \sin \angle AEB \tag{2}\\[4pt] &= \frac{|\overline{AB}|}{|\overline{AD}|}\frac{|\overline{AB}|}{|\overline{AE}|}+\frac{|\overline{BD}|}{|\overline{AD}|}\frac{|\overline{BE}|}{|\overline{AE}|} \tag{3}\\[4pt] &= \frac{1}{\sqrt{5}}\frac{3}{\sqrt{10}} + \frac{2}{\sqrt{5}} \frac{1}{\sqrt{10}} \tag{4}\\[4pt] &= \frac{5}{\sqrt{5}}\frac{1}{\sqrt{10}} \tag{5}\\[4pt] &= \frac{1}{\sqrt{2}} \tag{6} \end{align}$$ These steps are consis

Cites Thang Pang Ern · 2021-12 · Mathematics Stack Exchange · on “A Conjecture about Complex Numbers”

I thought of this question on Complex Numbers (perhaps related to Number Theory) a couple of months ago. It states that $$\prod\limits_{r=1}^{n}{\left( r+i \right)}$$ is purely imaginary if and only if $n=3$ . I believe that this is true but not sure how to prove it rigorously. My friend verified it for up to 9-digit numbers using a code but still haven't found any contradiction. Motivation I came across the Three Square Geometry Problem on Numberphile [this video] years ago and found it pretty fascinating that one could use Complex Numbers to solve this problem. In particular, using arguments, it is easy to show that $$\arg \left(( 1+i)(2+i)(3+i) \right)=\arg(1+i)+\arg(2+i)+\arg(3+i)=\frac{\pi}{