Inscribe 3 equal circles in a Reuleaux/curved triangle without overlap How exactly can I inscribe 3 equal circles into an equilateral Reuleaux triangle like the attached image without any overlap? I have the construction of the Reuleaux triangle down, drawn the altitudes, etc. but I haven't been able to find any clear answers of how to inscribe the circles without overlap/where to place the vertices for the central equilateral triangle.

 A: Here is a solution using coordinates geometry.
The figure below features a single inner circle $(C)$ (the other 2 circles can be immediately deduced from it by symmetries and/or rotations).

Let $A(2,0), \ \ B(-1,\sqrt{3}), \ \ C(-1,-\sqrt{3}).$
Let $G(a,0)$ the center of inner circle $(C)$ and $r$ its radius.
Inner circle $(C)$ must be tangent to the line of symmetry passing through $C$ and the origin : this line having polar angle $60°$,
$$GF=GO \cos 30° \ \ \ \iff \ \ \ r=a\frac{\sqrt{3}}{2}\tag{1}$$
Now, the second constraint : circle $(C)$ has to be (internally) tangent to the arc of circle $AB$ of Reuleaux triangle. This means that the centers are aligned with the point of tangency $H$. Therefore,
$$CG=CH-GH \ \ \iff \ \ \sqrt{(a+1)^2+(\sqrt{3})^2}=2 \sqrt{3}-r\tag{2}$$
(using a classical formula for the distance between two points).
Plugging the value of $r$ given by (1) into (2), we get, by squaring, a quadratic equation for $a$:
$$a^2+32a-32=0$$
whose positive root is:
$$a=12\sqrt{2}-16\approx 0.9706 \ \ \ \ \text{giving} \ \ \ \ r=(6\sqrt{2}-8)\sqrt{3}\approx 0.8405$$
A: 
As can be seen in figure you have to draw four circles centered at $O$, $O_1$,$O_2$ and $O_3$. The radius of big circle tangent to three circle can be found using Descartes theorem; if $r$ is the radius of three equal circles then to find the radius of small circle in between and big circle tangent to three circles we must have:
$$\big(3\times \frac 1 r+\frac 1R\big)^2=2\big(3\times \frac 1{r^2}+\frac1 {R^2}\big)\space\space\space\space (1)$$
which gives a quadratic equation for unknown R in terms of known r. It gives two values for R, the large value is for big circle and small one is four little circle.  After finding unknowns follow this:

*

*Draw three circles radius r and big circle radius R tangent to them.


*Take the points of tangencies X, Y and Z as centers of three circles 1, 2 and 3. Draw circle 1, 2 an 3 tangent to circles. These circles intersect at points A, B and C. These are the vertices of your curved triangle.
In  the figure:
$r=35$
$R=75.4$
Radius of curved sides $R_{cs}$ is $R_{cs}=136.82$
Linear distance between vertices is $144$$ units.
Update:

Considering above figure:
In triangle XOZ:
$XZ=2(2r+R_1) \sin 30^o\space\space\space\space (2)$
where $R_1$ is the radius of the small circle.
In triangle OTZ:
$TZ=(2r+R_1)\sin 18.5^o\space\space\space\space (3)$
In triangle XTZ:
$R_{cs}=XT=\sqrt {TZ^2+XZ^2+ TZ\cdot XZ\cdot \cos 100^o}\space\space\space\space (4)$
Now you have $R_{cs}$, start from relation (4). Following system of equations must be solved:
$\begin{cases}R_{cs}=XT=\sqrt {TZ^2+XZ^2+ TZ\cdot XZ\cdot \cos 100^o}\space\space\space\space (4)\\ TZ=(2r+R_1)\sin 18.5^o\space\space\space\space (3)\\XZ=2(2r+R_1) \sin 30^o\space\space\space\space (2)\end{cases}$
which gives $r$ and $R=2r+R_1$. The centers of three circles locate in $120^o$ angular distance and $R_2=r+R_1$ radial distance from center of construction O.
For a construction for inscribing three equal circles in a curved sides triangle we have to have the radius($R_{cs}$) of big circles making the sides inscribing three circles and also  the radius ($R_1$) of little circle inscribed by three circles.
