Inspirception

Reflect $D$ and $E$ in $AB$ (note I have labelled the vertices of the transcribed circle more conventionally) to obtain points $D_2$ and $E_1$ . Reflect $E$ and $F$ in $BC$ to obtain $E_2$ and $F_1$ . Finally reflect $F$ and $D$ in $AC$ to obtain $F_2$ and $D_1$ . Note that distances $D_1E + EF + FD_2 \; = \; E_1F + FD + DE_2 \; = \; F_1D + DE + EF_2 \; = \; DE + EF + FD \; = \; P$ are all equal, where $P$ is the perimeter of the triangle $DEF$ , and so we will have the shortest perimeter $P$ for the inscribed triangle $DEF$ if we can choose points $D,E,F$ such that the points $D_1,E,F,D_2$ are collinear, as well as the points $E_1,F,D,E_2$ and $F_1,D,E,F_2$ . This happens when $D,E,F$ are the feet of the perpendiculars from the vertices $A,B,C$ to the opposite sides, so that $DEF$ is the orthic triangle of $ABC$ .

Note that $AF = b\cos A$ and $AE = c\cos A$ . This implies that $AEF$ and $ABC$ are similar, and hence that $EF = a\cos A$ . Similarly $FD = b\cos B$ and $DE = c\cos C$ . Thus we deduce that $P \; = \; a\cos A + b\cos B + c\cos C \; = \; 2R(\sin A \cos A + \sin B \cos B + \sin C \cos C) \; = \; R(\sin2A + \sin2B + \sin2C)$ where $R$ is the radius of the circumcircle of $ABC$ . On the other hand, the area of the triangle $ABC$ is $\Delta \; = \; \tfrac12R^2(\sin2A + \sin2B +\sin2C) \; = \; \tfrac12PR$ In this case $R=25$ and $P=56$ , so that $\Delta = \boxed{700}$ .

Reflect $\triangle ABC$ in $CB$ and then reflect $\triangle A'BC$ in $CA'$ .

Then the perimeter of $\triangle DEF$ is equivalent to $FE + ED' + D'F''$ , which is at a minimum when $E$ and $D'$ are on $FF''$ .

Since reflections preserve lengths and angles, $\triangle ABC \cong \triangle A'BC \cong \triangle A'B'C$ , so $CF = CF''$ and $\angle FCF'' = 2\angle ACB$ .

By the law of cosines on $\triangle FCF''$ , $FF'' = \sqrt{CF^2 + CF^2 - 2 \cdot CF \cdot CF \cos 2 \angle ACB} = 2 \cdot CF \cdot \sin ACB$ .

Since $2 \cdot \sin ACB$ is a positive constant, $FF''$ reaches a minimum when $CF$ is a minimum, which is when $CF$ is the altitude of $\triangle ABC$ . Therefore, the minimum perimeter is $P_{\text{min}} = 2 h_c \sin C$ , where $h_c$ is the altitude of $\triangle ABC$ from $C$ .

Using triangle area equations $T = \frac{1}{2} h_c c$ and $T = \frac{1}{2} ab \sin C$ , we get $P_{\text{min}} = \cfrac{8T^2}{abc}$ , and using the circumradius equation $R = \cfrac{abc}{4T}$ , we get $T = \frac{1}{2}RP_{\text{min}}$ .

In this question, $R = 25$ and $P_{\text{min}} = 56$ , so $T = \frac{1}{2} \cdot 25 \cdot 56 = \boxed{700}$ .

If DEF has the smallest possible perimeter then we know that it's the orthic triangle in which case, $\frac{r}{R}$ = $\frac{Perimeter of △DEF}{Perimeter of △ABC}$ or r x Perimeter of △ABC = 56 x 25 or Semi-perimeter of △ABC x r =28 x 25 = 700 s.u.

I couldn't solve this without cheating. Like this: For a given reference triangle ( $\triangle ABC$ ), its orthic triangle ( $\triangle DEF$ ) produces the minimum perimeter. I placed three points on a Geogebra circumcircle until its orthic triangle was close to 56. It was surprisingly easy.

Anyway, I just wanted to say this is a great problem! Thank you, David.