Minimize the Perimeter

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 $C$ .

Substituting in area of triangle equations $T = \frac{1}{2} h_c c$ and $T = \frac{1}{2} ab \sin C$ , we get $P_{\text{min}} = \cfrac{8T^2}{abc}$ .

By Heron's Theorem, the area of $\triangle ABC$ is $T = 84$ . Therefore, $P_{\text{min}} = \cfrac{8T^2}{abc} = \cfrac{8 \cdot 84^2}{13 \cdot 14 \cdot 15} = \cfrac{1344}{65}$ , so $p = 1344$ , $q = 65$ , and $p + q = \boxed{1409}$ .

See triangle $(0,0),(15,0)$ , ( $\frac{42}{5}\,\frac{56}{5}$ ). Find the coordinates of bases the heights $D,E,F$ and find perimeter $DE+EF+FD$

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