Area of enclosed overlapping circles within an equilateral triangle

Question

Consider an equilateral triangle of length $\sqrt{6}$ as shown in the below figure. Find the area of the shaded region.

My attempt : Since the side of the triangle is given, hence height = $\frac{3}{\sqrt{2}}$.

Since its an equilateral triangle, so by virtue of symmetry the three circles must meet at the centroid.

Hence inradius = $\frac{1}{\sqrt{2} }$ and Circumradius = $\frac{2}{\sqrt{2} }$.

Please guide me how to proceed from here. Any help will be appreciated.

Answer 1

Note that you can cut each petal in half longways, and then rearrange the half petals by translation to see that the shaded region has the same area as the part of a circle of radius the circumradius of the equilateral triangle minus an inscribed regular hexagon. Since you've computed the circumradius, $r=\sqrt{2}$, the area of the circle is $\pi r^2=2\pi$, and the area of the hexagon is $6r^2\sqrt{3}/4=3\sqrt{3}$. Hence the area of the shaded region is $2\pi - 3\sqrt{3}$.

Answer 2

We divide the shaded portion into six equal parts by three lines, each of them joining the centroid and one vertex.

So area of shaded area(A) = 6 * Area of one segment(S)

$$tan(\phi) = \frac{CD}{DG} = \frac{\sqrt{6}/2}{1/\sqrt{2}} = \frac{1}{\sqrt{3}}$$ So, $$\phi = 60 degrees $$ The orange lines form part of a regular hexagon inscribed in a circle with centre O.

The area of one segment(S) = ( Area of sector OCPGO - Area of triangle OCG) $ = (\pi/6*(\sqrt2)^2 - \sqrt{3}/4*(\sqrt2)^2) $

So, Total area of shaded region = $6S = (2\pi - 3\sqrt{3})$

Answer 3

The center $O_a$ of the circular arc $COB$ with the radius $R_a=|O_aO|=|O_aB|=|O_aC|$ is found as $O_a=DOa\cap EO_a$, $|DO|=|DB|$, $|EO|=|EB|$, $DO_a\perp OB$, $EO_a\perp OC$.

Due to the symmetry, $OO_a$ bisects $\angle COB$, hence $\angle COO_a=60^\circ$.

Also $\angle O_aCO=60^\circ$, $\angle OO_aC=60^\circ$ thus

\begin{align} |O_aC|&=|O_aO|=|O_aC|=|CO|=\tfrac23\cdot\sqrt6\cdot\tfrac{\sqrt3}2 \\ &=\sqrt2 . \end{align}

The area of one of the shaded regions is a doubled difference between a $60^\circ$ circular segment $S_c$ and the area $S_t$ of equilateral $\triangle OO_aC$

\begin{align} S_c&=\tfrac12\,\tfrac\pi3(\sqrt2)^2=\tfrac\pi3 ,\\ S_t&=\tfrac12\,\sqrt2\cdot \sqrt2\cdot\tfrac{\sqrt3}2 =\tfrac{\sqrt3}2 , \end{align}

so the total shaded area is $3\cdot2\cdot(\tfrac\pi3-\tfrac{\sqrt3}2)=2\pi-3\sqrt3.$