A geometric puzzle: Area enclosed by three circular arcs

[bob012345]

TL;DR

Find the area enclosed by the three circular arcs.

This figure is bounded by three circular arcs of equal radius R. The arcs are tangent where they meet. Find the enclosed area in terms of R.

 

[kuruman]

Good one!

Spoiler

The area of the equilateral triangle of side $2R$ is $A_1=R^2\sqrt{3}.$
Each 60° "pizza" piece is 1/6 of the area of a circle so the 3 pieces together have area $A_2=\frac{1}{2}\pi R^2.$
The answer is the difference $\Delta A=R^2\left(\sqrt{3}-\frac{1}{2}\pi\right)=0.161 R^2.$

 

[Roberto Pavani]

Exagonal (L=R) area minus half circle (R=R) area?
Exagon vertex are 3 circle centers plus 3 figure vertex

 

[bob012345]

That was perhaps a bit too easy, even though I tried to obscure it a bit. So, as an additional exercise, what is the radius $R$ of the large circle in terms of the radius $r$ of the smaller circles?

 

[Roberto Pavani]

$h$ = height from big circle center to inner triangle
$r$ = inner radius
$3 r h = A$, (Triangle area)
$A = r^2 sqrt(3)$
$h = r / sqrt(3)$
$sqrt(r^2+h^2) = sqrt(r^2+3 r^2) = r\, sqrt(3)$
$R = r (1+sqrt(3))$ (R is big radius)

 

[bob012345]

$R = r (1+sqrt(3))$ (R is big radius)

Here is a sanity check. From the picture in post #4, $R$ is just a little bit larger than $2r$.

 

[Roberto Pavani]

You are right... my mistake... a calculation error:

$h$ = height from big circle center to inner triangle
$r$ = inner radius
$3 r h = A$, (Triangle area)
$A = r^2 sqrt(3)$
$h = r / sqrt(3)$

$h^2 = \frac{r^2} {3}$

## sqrt(r^2+h^2) = sqrt(r^2+3 r^2) = r\, sqrt(3) ## <--- Mistake

$\sqrt {r^2+h^2} = \sqrt {r^2+ \frac {r^2}{3} } = \frac {2 r}{\sqrt 3}$

$R = r (1+sqrt(3))$ (R is big radius) <-- mistake, uses the wrong value

$R = r + \frac {2 r}{\sqrt 3} = r (1 + \frac 2 {\sqrt 3} )$