Find $\angle BAC$

The photo of explanation Let $D'$ and $D''$ be the reflection of $D$ about $AB$ and $AC$ respectively. Then, $\min {\left( C_{\triangle DEF}\right)} = \min {\left( DE+EF+FD \right)} = \min {\left( D'E+EF+FD'' \right)} = D'D'' = 5$ . Observe that $\triangle AD'D''$ is a equilateral triangle, therefore $\angle D'AD"=60\degree$ . Also, by the reflection, $\angle BAD=\angle BAD'$ and $\angle CAD=\angle CAD''$ . Therefore, $\angle D'AD"=\angle BAD+\angle BAD'+\angle CAD+\angle CAD''=2\left( \angle BAD+\angle CAD \right) = 2\angle BAC = 2\alpha\degree =60\degree \rightarrow \alpha=\boxed{30}$

Consider $\triangle ABC$ with side lengths $a$ , $b$ , and $c$ . With a common base $b$ , using AM-GM inequality , we have $a+c \ge 2\sqrt{ac}$ . That is the shortest $a+c$ occurs when $a=c$ or for a common base, an isosceles triangle has the shortest perimeter. Therefore for a common base $EF$ , the shortest perimeter occurs when $\triangle DEF$ is isosceles or $AD$ bisects $\angle BAC$ and $EF \perp AD$ . Let $EF$ cuts $AD$ at $P$ , $AP=x$ , and $EP = y$ . Then the perimeter of $\triangle DEF$ is given by $p = 2y + 2\sqrt{y^2 + (5-x)^2}$ . Let $\angle BAC = \theta$ and $t = \tan \frac \theta 2$ ; then $y = x \tan \frac \theta 2 = tx$ and:

$\begin{aligned} \frac p2 & = tx + \sqrt{t^2x^2 + (5-x)^2} \\ \frac 12 \cdot \frac {dp}{dx} & = t + \frac {t^2x + x - 5}{\sqrt{t^2x^2 + (5-x)^2}} \end{aligned}$

Extrema occur when $\dfrac {dp}{dx} = 0$ , we have:

$\begin{aligned} t \sqrt{t^2x^2 + (5-x)^2} & = -(t^2x + x - 5) & \small \color{#3D99F6} \text{Squaring both sides} \\ t^2(1+t^2) x^2 - 10t^2 x + 25t^2 & = (1+2t^2+t^4)x^2 - 10(1+t^2)x + 25 \\ (1+t^2)x^2 - 10x + 25(1-t^2) & = 0 \\ (x-5)^2 + t^2(x^2 - 25) & = 0 \\ (x-5)(x-5 + t^2(x+5)) & = 0 \end{aligned}$

$\implies \begin{cases} x = 5 & \implies p = 20 t & \small \color{#3D99F6} \text{when }p \text{ is maximum.} \\ x = \dfrac {5(1-t^2)}{1+t^2} & \implies p = \dfrac {20t}{1+t^2} & \small \color{#3D99F6} \text{when }p \text{ is minimum.} \end{cases}$

Therefore, $\dfrac {20t}{1+t^2} = 10 \sin \theta = 5 \implies \sin \theta = \frac 12 \implies \theta = \boxed{30}^\circ$ .