Find the radius of the circle cutting sides of equilateral triangle Consider the figure below. The triangle is equilateral. Find the radius of the circle.

My try:
I have dropped perpendiculars from the center of the circle to respective chords naming them $x,y,z$ and named the other line segments as $m,n,p,q,s,t$ as shown below.

By Viviani's theorem we have
$$x+y+z=\frac{\sqrt{3}a}{2}$$ Where $a$ is the side length of the triangle.
By Chords theorem we have:
$$\begin{aligned}
m(m+5) &=t(t+3) \\
q(q+4) &=s(s+3) \\
p(p+4) &=n(n+5) \\
m+n+5 &=p+q+4=s+t+3=a
\end{aligned}$$
Also by Pythagoras theorem if $R$ is the radius of the circle we have:
$$R^{2}-x^{2}=\frac{9}{4}, \quad R^{2}-y^{2}=4 ,\quad R^{2}-z^{2}=\frac{25}{4}$$
I guess this is a very tedious task? Any better ways?
 A: Some may find my methodology a bit too reconstructive on the equilateral triangle but this does not alter the circle whose radius is of interest. It does apparently greatly simplify the math.
In words, shift the equilateral triangle left along the circle so that the sum of length of two chords remains the same. Proceed further to shift that triangle down so that the sum of the three chords remains the same. In other words, reset/re-position the equilateral triangle so that the chord sum is set at 12 (starting at 3, 4, 5) and each now equals 4.
Now, one can relate 4 to the length of the side of the equilateral triangle (which if inscribed has also length 4). This ratio times 120 degrees is the interior angle of the triangle formed from two radius connected to the end of the cord. Use this in the Law of Cosines to derive the radius r as require.
Note: The math in the special case of an inscribed equilateral triangle yields the result the radius would also be 4.
A: 
In our original answer, we presented a geometrical construction to obtain just one of the infinitely many possible configurations for the given triplet of chord lengths of 3, 4, and 5. After reading $\text{@math_for_entertainment}$’s answer, which emphasizes the fact that three chord lengths alone cannot lead to a unique solution, we decided to revise our answer to extend the construction to obtain any of those possible configurations at will just by selecting a parameter.
Without loss of generality, let us assume that $ST\gt MN\gt EF\gt 0$ (or for brevity $\lambda\gt\mu\gt\eta\gt 0$) (see $\mathrm{Fig.\space 1}$). We denote the angle between the radius $OT$ and the longest chord $ST$ by $\theta$, which serves as the aforementioned parameter. Angle $\theta$ must satisfy the following inequality.
$$90^o\gt\theta\ge\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\dfrac{\lambda}{\sqrt{\lambda^2+\mu^2-\lambda\mu}}\right)\tag{1}$$
After choosing a value for $\theta$ within the rage prescribed by (1), draw a line $PQ$ and mark the points $S$ and $T$ so that $ST=\lambda$. Construct two rays at $S$ and $T$ to make $\measuredangle \theta$ with $PQ$ as shown in $\mathrm{Fig.\space 1}$. They do intersect each other at $O$, which is the center of the sought circle. Its radius is equal to the length of $OS$. Now, the center and the radius is known, we can construct the circle $\it{\Omega}$.
After drawing two lines through $O$, each one making an angle of $30^o$ with $ST$, mark points $U$ and $V$ at a convenient location on them as shown in $\mathrm{Fig.\space 1}$. Construct a circle with the radius $r_1 = \dfrac{\mu}{2}$ and its center at $U$. Now, draw a line perpendicular to $OU$ to cut the just drawn circle at $K$ and $L$. Through these two points, we need to draw two lines parallel to $OU$ to intersect the circle $\it\Omega$ at points $M$ and $N$. In order to obtain the vertex $C$ of $\triangle ABC$, draw a line through $M$ and $N$ to meet $PQ$.
The vertex $B$ of $\triangle ABC$ can be located by doing a similar construction after drawing a circle centred at $V$ with radius $r_2 = \dfrac{\eta}{2}$. The vertex $A$ can be found by extending $BE$ and $CM$ to intersect each other.
According to the inequality (1), the smallest value of $\theta$ is given as,
$$\theta_{\large{\text{s}}}= \cos^{-1}\left(\dfrac{\sqrt{3}}{2}\dfrac{\lambda}{\sqrt{\lambda^2+\mu^2-\lambda\mu}}\right).$$
Note that $\theta_{\large{\text{s}}}$ depends only on the lengths of two longest chords. For the given triplet of chord lengths of 3, 4, and 5, $\theta_{\large{\text{s}}}=19.1066053509^o$. $\mathrm{Fig.\space 2}$ illustrates how the configuration looks like if $\theta_{\large{\text{s}}}$ is chosen as the parameter. Note that points $N$ and $T$ coincide with the vertex $C$. Furthermore, if a value less than $\theta_\text{s}$ is selected as the parameter, both $N$ and $T$ are to be found on the extended $AC$ and $BC$ respectively, i.e. exterior of $\triangle ABC$.
A: A circle that fits the given information is not unique, nor is the equilateral triangle.  The attached plots illustrate this.


A: Important note: the side length $a$ of the triangle is supposed to be given and we have to find the radius $R$ of the circle related to $a$. Obviously, if $a$ wasn't given, there would be an infinite number of solutions, as for any circle with a radius large enough, we can build an equilateral triangle satisfying the required relations.
From the Viviani's theorem as noted in the question
$$x+y+z = \frac{\sqrt 3}{2} a$$ and using the equations between $x,y,z$ and $R$, we get the following equation that allows getting $R$ with $a$ as a parameter.
$$\sqrt{R^2 - \frac{9}{4}} + \sqrt{R^2 - \frac{16}{4}} + \sqrt{R^2 - \frac{25}{4}} = \frac{\sqrt 3}{2}a \tag{1}$$
This is what we were looking for... at the caveat of solving this equation!
Additional elements
I use below the notations of the following figure of the question:

Important as the two figures of the question use different notations, at least for $A,B,C$!
Using a comment from Dan to the question, let's set up a Cartesian coordinate system with the origin at the center of the circle and the chord of length equal to $5$ horizontal below the origin.
Based on that, the coordinates of the vertices of the triangle are
$$A= \left(-\frac{2x+z}{\sqrt 3}, -z\right), \, B= \left(\frac{y-x}{\sqrt 3}, x+y\right), \, C= \left(\frac{2y+z}{\sqrt 3}, -z\right).$$
Also note that as $3, 4,5$ is a Pythagorean triple, we have the relation
$$x^2+y^2 - z^2 = R^2$$ and the Viviani's relation as noted in the question
$$x+y+z = \frac{\sqrt 3}{2} a.$$
There may be a way to use those simple relations to get $R$ with $a$ as a parameter in a simpler way than using equation $(1)$ above.
A: If you set up a Cartesian coordinate system with its origin at the center of the circle and $BC$ as a horizontal line beneath it, then the coordinates of the chord endpoints work out to:
$$G = \left( - \frac{5}{2}, - \sqrt{R^2 - \frac{25}{4}} \right)$$
$$T = \left( \frac{5}{2}, - \sqrt{R^2 - \frac{25}{4}} \right)$$
$$N = \left(1 + \frac{\sqrt{3}}{2}\sqrt{R^2 - 4}, \frac{1}{2}\sqrt{R^2 - 4}  - \sqrt{3} \right)$$
$$M = \left(-1 + \frac{\sqrt{3}}{2}\sqrt{R^2 - 4}, \frac{1}{2}\sqrt{R^2 - 4} +  \sqrt{3}\right)$$
$$E = \left(\frac{3}{4} - \frac{\sqrt{3}}{2} \sqrt{R^2 - \frac{9}{4}}, \frac{1}{2}\sqrt{R^2 - \frac{9}{4}} + \frac{3\sqrt{3}}{4} \right)$$
$$F = \left(-\frac{3}{4} - \frac{\sqrt{3}}{2} \sqrt{R^2 - \frac{9}{4}}, \frac{1}{2}\sqrt{R^2 - \frac{9}{4}} - \frac{3\sqrt{3}}{4}\right)$$
From these, we can find equations for the lines that form the three sides of the triangle.
$$\overline{AB} = \overline{MN} : y = -\sqrt{3}x + 2\sqrt{R^2 - 4}$$
$$\overline{AC} = \overline{EF} : y = \sqrt{3}x + 2\sqrt{R^2 - \frac{9}{4}}$$
$$\overline{BC} = \overline{GT} : y = -\sqrt{R^2 - \frac{25}{4}}$$
From the intersections between those three lines, we find the triangle's vertices.
$$A = \overline{AB} \cap \overline{AC} = \left( \frac{1}{\sqrt{3}}\sqrt{R^2 - 4} - \frac{1}{\sqrt{3}}\sqrt{R^2 - \frac{9}{4}}, \sqrt{R^2 - 4} + \sqrt{R^2 - \frac{9}{4}} \right)$$
$$B = \overline{AB} \cap \overline{BC} = \left( \frac{1}{\sqrt{3}}\sqrt{R^2 - \frac{25}{4}} + \frac{2}{\sqrt{3}}\sqrt{R^2 - 4}, -\sqrt{R^2 - \frac{25}{4}} \right)$$
$$C = \overline{AC} \cap \overline{BC} = \left(-\frac{1}{\sqrt{3}}\sqrt{R^2 - \frac{25}{4}} - \frac{2}{\sqrt{3}}\sqrt{R^2 - \frac{9}{4}}, -\sqrt{R^2 - \frac{25}{4}} \right)$$
And by applying the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to any two of these points, the common side length $a$ of the equilateral triangle is:
$$a = \frac{2}{\sqrt{3}}\left( \sqrt{R^2 - \frac{9}{4}} + \sqrt{R^2 - 4} + \sqrt{R^2 - \frac{25}{4}} \right)$$
But this is just what OP already found by Vivani's theorem, and adds no new information.
As indicated by @math for entertainment's answer, there isn't a unique solution by $R$.  We need one more piece of information to find it.
Assuming that information is the side length $a$, then we'll need to solve the above radical equation for $R$.
