Ratul's Triangle Revisited

We will work with the formulas $r = \dfrac{1}{s}\sqrt{s(s - a)(s - b)(s - c)}$ for the inradius and $R = \dfrac{abc}{4sr}$ for the circumradius .

From the second equation we see that $3 = \dfrac{abc}{4 \times 7 \times 1} \Longrightarrow abc = 84$ .

From the first equation we see that

$1 = \dfrac{1}{7}\sqrt{7(7 - a)(7 - b)(7 - c)} \Longrightarrow 7 = (7 - a)(7 - b)(7 - c) \Longrightarrow$

$7 = (49 - 7(a + b) + ab)(7 - c) = 343 - 49(a + b + c) + 7(ab + ac + bc) - abc$ .

Substituting in $a + b + c = 2s = 14$ and $abc = 84$ and simplifying yields that

$7 = 343 - 49 \times 14 + 7(ab + ac + bc) - 84 \Longrightarrow ab + ac + bc = 62$ .

So $a^{2} + b^{2} + c^{2} = (a + b + c)^{2} - 2(ab + ac + bc) = 14^{2} - 2 \times 62 = \boxed{72}$ .

Edit: Note that, as Michael Mendrin points out in his solution, such a triangle does exist and has side lengths $4 - \sqrt{2}, 4 + \sqrt{2}, 6$ .

In general, $a^2 + b^2 + c^2 = 2s^2 - 2r^2 - 8Rr$ .

Inradius $\color{#3D99F6}{r=\dfrac{A}{s}}$ , Circumradius $R=\dfrac{abc}{4A} \longrightarrow \color{#20A900}{abc=4AR}$

$\begin{aligned} A&=\sqrt{s(s-a)(s-b)(s-c)} \\ A^{2}&=s(s-a)(s-b)(s-c) \\ A \times \color{#3D99F6}{\dfrac{A}{s}}&=(s-a)(s-b)(s-c) \\ A\color{#3D99F6}{r}&=(s-a)(s-b)(s-c) \\ Ar&=s^{3}-(a+b+c)s^{2}+(ab+ac+bc)s-\color{#20A900}{abc} \\Ar&=s^{3}-(2s)s^{2}+(ab+ac+bc)s-\color{#20A900}{4AR} \\ -(ab+ac+bc)s&=s^{3}-2s^{3}-Ar-4AR \\ ab+ac+bc&=\dfrac{-s^{3}-Ar-4AR}{-s} \\ ab+ac+bc&=s^{2}+\color{#3D99F6}{\dfrac{A}{s}} \color{#333333}{\times r +4 \times} \color{#3D99F6}{\dfrac{A}{s}} \color{#333333}{\times R} \\ ab+ac+bc&=s^{2}+\color{#3D99F6}{r} \color{#333333}{\times r +4} \times \color{#3D99F6}{r} \color{#333333}{\times R} \\ ab+ac+bc&=s^{2}+r^{2}+4Rr \end{aligned}$

Then,

$\begin{aligned} a^{2}+b^{2}+c^{2}&=(a+b+c)^{2}-2(ab+ac+bc) \\ a^{2}+b^{2}+c^{2}&= (2s)^{2}-2(s^{2}+r^{2}+4Rr) \\ a^{2}+b^{2}+c^{2}&=4s^{2}-2s^{2}-2r^{2}-8Rr \\ a^{2}+b^{2}+c^{2}&=2s^{2}-2r^{2}-8Rr \end{aligned}$

So, $a^{2}+b^{2}+c^{2}=2 \times 7^{2} -2 \times 1^{2} -8 \times 3 \times 1=72$

It's a right triangle of sides $\left(4+\sqrt{2}, \; 4-\sqrt{2}, \; 6\right)$ , which has a semi-perimeter of $7$ , an inradius of $1$ , and a circumradius of $3$ . So, the sum of the squares of the sides is the same as twice the square of the hypotenuse, or $72$