Family of Triangles and their Circumradii

Here's a graphic of two isosceles triangles in the family of triangles with the greatest and least circumradadii (dark blue circles). All triangles in this family have the same inradius of $2$ (red circle), which is the property of any triangle in which its area and its perimeter are the same.

Let $(a_1=b_1,c_1)$ and $(a_2=b_2,c_2)$ be the sides of the two isosceles triangles, with corresponding base angles $\theta_1, \theta_2$ . Then we have the following (true for both)

$c=4Cot(\dfrac{1}{2}\theta)$
$a=\dfrac{1}{2}cSec(\theta)$
$P=a+b+c$
$R=\dfrac{abc}{4\cdot2\cdot\dfrac{1}{2}P}$

Then we solve the simultaneous equations by numerical means

$P_1=P_2$
$2R_1=R_2$

with the solution $\theta_1=85.39347…$ and $\theta_2=15.93082…$ , which, when worked out, gives us the area of the triangles $58.315…$

We need $2s = \Delta = \sqrt{s(s-a)(s-b)(s-c)}$ , and hence $4s = (s-a)(s-b)(s-c)$ . Writing $u=s-a$ , $v=s-b$ , $w=s-c$ , we deduce that $u,v,w > 0$ are such that $s = u+v+w$ , $a = v+w$ , $b=u+w$ , $c = u+v$ , and we require $4(u+v+w) \; = \; uvw$ Since $\tfrac12bc\sin A = \Delta$ , we deduce that $\begin{aligned} 2R & = \frac{abc}{2\Delta} \; = \; \frac{(u+v)(u+w)(v+w)}{2\Delta} \\ & = \frac{u^2v + u^2w + uv^2 + v^2w + uw^2 + vw^2 + 2uvw}{2\Delta} \; = \; \frac{(u+v+w)(uv + uw + vw) - uvw}{2\Delta} \\ & = \frac{(u+v+w)(uv + uw + vw - 4)}{2\Delta} \; = \; \frac{uv + uw + vw - 4}{4} \end{aligned}$ If we put $p = 8R$ , then $u,v,w$ are positive real roots of the equation $f(X) \; = \; X^3 - sX^2 + (p+4)X - 4s \; = \; 0$ and so we deduce that $s^2 \ge 3(p+4)$ , $f(x)$ has turning points $\tfrac13\big[s \pm \sqrt{s^2 - 3(p+4)}\big]$ , with $f\Big(\tfrac12\big(s \pm \sqrt{s^2 - 3(p+4)}\big)\Big) \; =\; \tfrac{1}{27}s(9p-72-2s^2) \mp \tfrac{2}{27}\big(s^2 - 3(p+4)\big)^{\frac32}$ and so the fact that $f(X)$ has three positive real roots can be stated by the condition $\begin{aligned} \big| s(9p-72-2s^2) \big| & \le 2\big(s^2 - 3(p+4)\big)^{\frac32} \\ s^2(2s^2 - 9p + 72)^2 & \le 4\big(s^2 - 3(p+4)\big)^3 \\ 4p^3 - (s^2 - 48)p^2 + (192 - 80s^2)p + 16(s^2+4)^2 & \le 0 \end{aligned}$ If we put $s = 2q$ , this inequality becomes $p^3 - (q^2 - 12)p^2 + (48 - 80q^2)p + 64(q^2 + 1)^2 \; \le \; 0$ One of the roots of the cubic equation $g(X)\; = \; X^3 - (q^2 - 12)X^2 + (48 - 80q^2)X + 64(q^2 + 1) \; = \; 0$ must be negative. We want the other roots to be of the form $2t$ , $t$ for positive $t$ . Thus we want the roots of the cubic to be $2t$ , $t$ , $q^2 - 12 - 3t$ . Matching the other coefficients of the cubic, we deduce that $2t^2 + 3t(q^2 - 12 - 3t) \; =\; 48 - 80q^2 \hspace{2cm} 2t^2(q^2 - 12 - 3t) \; = \; -64(q^2 + 1)^2$ and hence $7t^2 - 3(q^2 - 12)t + (48 - 80q^2) \; = \; 0 \hspace{2cm} 6t^3 - 2(q^2 - 12)t^2 - 64(q^2 + 1)^2 \; = \; 0$ Solving these equations simultaneously, we deduce that $t \; = \; \frac{16 (44q^4 + 161q^2 + 13)}{3 (q^4 + 256q^2 - 24)}$ Since $t$ is a root of the cubic $g(X)$ , we deduce that $q$ must satisfy the polynomial equation $9q^{10} - 1954q^8 + 9673q^6 - 197876 q^4 - 1144q^2 - 16 \; = \; 0$ Thus the problem finally reduces to that of solving a quintic equation in $q^2$ . Solving numerically, we find that we have exactly one positive root $q$ , namely $14.578787$ . This leads to the desired area of $58.3152$ . This is $\boxed{58.315}$ to $3$ decimal places.