, 2022

Let $\alpha = \angle ABC$, $AB = a$, $BC = b$, $CD = c$, $DA = d$ and $AC = e$. Then $\angle ADC = 180^\circ - \alpha$, $2c = d + e$, $2a = b + e + c$ and it follows from the cosine theorem for $\triangle ADC$ that
$$d^2 + c^2 + \frac{dc}{2} = e^2 = (2c - d)^2 \iff 2c = 3d.$$
Since $2c = d + e$ we obtain $e = 2d$ and $2a = b + e + c$ implies $2a = b + \frac{7e}{4}$. The cosine theorem for $\triangle ABC$ gives
$$a^2 + b^2 - \frac{ab}{2} = e^2 = \left(\frac{4}{7}(2a - b)\right)^2 \iff 33b^2 + \frac{79ab}{2} - 15a^2 = 0.$$
Since the roots of the equation $33x^2 + \frac{79x}{2} - 15 = 0$ are $x_1 = \frac{10}{33}$ and $x_2 < 0$ we have $\frac{b}{a} = \frac{10}{33} \iff b = \frac{10a}{33}$.
Now $2a = b + \frac{7e}{4}$ implies $a = \frac{33e}{32} = \frac{33d}{16}$ and $b = \frac{5d}{8}$. Finally $a = \frac{33d}{16}$, $b = \frac{5d}{8}$, $c = \frac{3d}{2}$, $e = 2d$. The least value of $d$ for which $a$ is an integer is $d = 16$. Therefore $AB = 33$, $BC = 10$, $CD = 24$, $DA = 16$, $AC = 32$ and the least value of the perimeter equals $83$.