# Geometric explanation of $\sqrt 2 + \sqrt 3 \approx \pi$

Just curious, is there a geometry picture explanation to show that $\sqrt 2 + \sqrt 3$ is close to $\pi$?

• Well, they are just two numbers that add up to a number close to $\pi$. Would it not be the same as asking for a geometry explanation of why $\frac{22}7$ is a number close to pi? – user130512 Mar 6 '14 at 15:38
• Do you know about mathematical coincidences? – Sawarnik Mar 6 '14 at 15:44
• @Sawarnik, I am not asking for anything related to numerology or something like that. Just hope someone with greater insights will come up with some interesting ideas. Since apparently, all three terms have natural geometry meanings. – ahala Mar 6 '14 at 16:02
• Close up to approximately 1 per mille. – Qmechanic Sep 12 '14 at 18:21

Take the unit circle. Inscribe a square and circumscribe a hexagon. The perimeter of the square is $4\sqrt2$, while the perimeter of the hexagon is $4\sqrt3$. Clearly the circumference of the circle lies somewhere in between:

$$4\sqrt2\lt2\pi\lt4\sqrt3$$

But to the extent that the square and the hexagon are rough approximations to the circle, we might split the difference and say

$$2\pi\approx{1\over2}(4\sqrt2+4\sqrt3)=2(\sqrt2+\sqrt3)$$

• this is the most clear answer. – ahala Sep 12 '14 at 19:01
• @ahala, thanks. Note, this "proof" doesn't say why the approximation $\sqrt2+\sqrt3=3.14626\ldots$ is as good as it is, nor why it's slightly too large. It's be nice to have a simple explanation for those features. – Barry Cipra Sep 12 '14 at 19:19 This is what I'm trying to show on a diagram indicating explicitly all quantities and the approximation is quite rough. Here is a circle with center $O$. Quantities $OA=OB=OC=OD=AB=1$ and $OC\perp OA, OD\parallel OA$. Line segments $AC$ and $BD$ have an intersection $E$. We can easily deduce the following quantity. $$AC^2=OA^2+OC^2=2\qquad BD^2=AD^2-AB^2=3$$ Hence we have $AC=\sqrt2,BD=\sqrt3$ and the length of half perimeter $ABCD$ is $\pi$. Also note that \begin{align}\sqrt2+\sqrt3&=AC+BD\\&=(AE+EC)+(BE+ED)\\&=AE+DE+(BE+CE)\\&\approx AB+CD+BC\\&\approx\mathrm{arc}AB+\mathrm{arc}CD+\mathrm{arc}BC\\&\approx\mathrm{arc}ABCD=\pi\end{align}

• Very neat. I love simple geometric arguments like this. +1 – Cameron Williams Mar 6 '14 at 17:27
• (-1) $AE + BE + CE + DE \approx AB + BC + CD$ is just a random invented approximation to get your argument to work. In fact, working out the details, before the first approximation you have $\sqrt{2} + \sqrt{3} \approx 3.15$, after the first approximation you have $1 + \sqrt{2} + \sqrt{2}/(1 + \sqrt{3}) \approx 2.93$, and then by replacing the straight lines by arcs you get $\pi \approx 3.14$. So this is nonsense; it is certainly not a motivation why $|\sqrt{2} + \sqrt{3} - \pi| \approx 0.0047$ is so small. – TMM Mar 10 '14 at 10:12
• @TMM I mentioned that this is a quite rough approximation showing geometrically two quantities are close, instead of so close. For approximation process, it's close from 3.15 to 2.93, from 2.93 to 3.14. So it's close from 3.15 to 3.14. This indeed makes sense. It will be nonsense if directly compute it analytically. And of course it is useful since at least this motivated another visualization right behind here. – Shuchang Mar 10 '14 at 12:46
• Like I said, (1) there is no motivation why $AE + BE + CE + DE \approx AB + BC + CD$ (and indeed, they are not so close), and (2) the approximations are both very far off, considering the small distance between $\sqrt 2 + \sqrt 3$ and $\pi$. The fact that two large errors cancel each other out is just a coincidence and a fabricated trick to make your flawed argument work. – TMM Mar 10 '14 at 13:31
• Similarly I could argue that $\sqrt 2, \sqrt 3 \approx \pi/2$ and conclude that $\sqrt 2 + \sqrt 3 \approx \pi$, saying that these are "quite rough approximations." And the result is correct, so my argumentation must be correct, right?! – TMM Mar 10 '14 at 13:36 Starting with unit circle $\bigcirc O$, one easily constructs $A$, $B$, $C$, $D$ with $|\overline{AB}| = \sqrt{2}$ and $|\overline{CD}| = \sqrt{3}$. Quadrisecting $\overline{AB}$ and $\overline{CD}$ one draws $\bigcirc A$ and $\bigcirc{C}$ to provide chords of length $\sqrt{2}/4$ and $\sqrt{3}/4$. Chains of congruent circles lead us to $B^\prime$ and $D^\prime$, such that we have polygonal lengths $$|\widehat{AB^\prime}| = |\overline{AB}| = \sqrt{2} \qquad |\widehat{CD^\prime}| = |\overline{CD}| = \sqrt{3}$$

Now $\overline{B^\prime D^\prime}$ looks very close to being perpendicular to $\overline{AC}$. (Is it?) This gives us the grand approximation $$\sqrt{2} + \sqrt{3} = |\widehat{AB^\prime}| + |\widehat{CD^\prime}| \approx |\stackrel{\frown}{AB^\prime}| + |\stackrel{\frown}{CD^\prime}| \approx |\stackrel{\frown}{AC}| = \pi$$

(Of course, there are multiple approximations going on here. The polygonal lengths approximate the arc lengths with different roughness, and the combined arcs only approximate a semicircle (although they do that quite well!).)

• Very nice! thanks. can I ask what do you use to draw the diagram? – ahala Mar 6 '14 at 18:30
• I'm quite pleased with how neatly it all worked out. :) BTW, I use GeoGebra for my figures. – Blue Mar 6 '14 at 18:31
• (+1) This is probably the closest you can get to showing $\sqrt 2 + \sqrt 3 \approx \pi$. I suppose if you split the red and blue lines in more pieces, the pink line will look more and more perpendicular to $AC$. (But it will never be perpendicular, since $\sqrt{2} + \sqrt{3} > \pi$.) Nice work! – TMM Mar 10 '14 at 10:23

I can only see a relation with trigonometric functions, i.e.something like $\sqrt{3}=2-\tan (\pi/12)$ and $$\sqrt{2}=\frac{4\cos(\pi/12)}{3-\tan(\pi/12)}.$$ If we consider the continued fraction of $\sqrt{2}$, $\sqrt{3}$ and $4/\pi$ we see some similarities, too. However, there are also arguments indicating that both numbers are only accidentally close. Since $\pi$ is not algebraic, an expression $\pi=\sqrt{a}+\sqrt{b}$ is impossible. Indeed, the number $\sqrt{2}+\sqrt{3}$ is integral over $\mathbb{Z}$, which means that there is a monic polynomial $f\in \mathbb{Z}[x]$ with $f(\sqrt{2}+\sqrt{3})=0$. This cannot exist for $\pi$ of course.

By the way, much closer are $\pi^4+\pi^5$ and $e^6$.

Divide this approximation by four:

\displaystyle \begin{align*}\pi &\approx \sqrt{2}+\sqrt{3} \\ &\Leftrightarrow \\ \frac{\pi}{4} &\approx \frac{\sqrt{2}+\sqrt{3}}{4}\\ &\Leftrightarrow \\ \frac{\pi}{4} &\approx \frac{\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{3}}{2} = \frac{\sin{(\frac{\pi}{4})}+\sin{(\frac{\pi}{3})}}{2} , \end{align*}

So this approximation tells that $\frac{\pi}{4}$ is roughly the average of $\sin{(\frac{\pi}{4})}$ and $\sin{(\frac{\pi}{3})}$, which is visualized below: If the green arc segment with length $\frac{\pi}{4}$ is made straight, its red tip follows a cycloid (https://en.wikipedia.org/wiki/Cycloid), like the valve of a running bicycle wheel, and ends up between $\frac12 \sqrt{2}$ and $\frac12 \sqrt{3}$.

I can show geometrical (kinda: Dante Alighieri meets Plato): $$3+\frac{\sqrt{2}}{10}<\pi<\sqrt{2}+\sqrt{3}$$ by using the Pythagorean theorem and the intercept theorem : 