Isoperimetric points and Tetrahedrons 2.

Refer to isoperimetric point .

The perimeter $P^{*} = 13 + b + c = 14 + b + d = 15 + c + d \implies b = d + 2 \implies c = d + 1$

Using Heron's formula $A_{APC} = \sqrt{42(d + 8)(d - 5)}, A_{APB} = \sqrt{48(d + 8)(d - 6)}, A_{BPC} = \sqrt{56(d + 8)(d - 7)}$

and

$A_{ABC} = 84 \implies 84 = \sqrt{d + 8}(\sqrt{42(d - 5)} + \sqrt{48(d - 6)} +\sqrt{56(d - 7)})$

Solving for $d$ we obtain: $d = \dfrac{80}{11} \implies \overline{AP} = \dfrac{102}{11}, \overline{CP} = \dfrac{91}{11}$ and $\overline{BP} = \dfrac{80}{11}$ .

Let $\overline{BD}$ be altitude in $\triangle{ABC}:$

$A_{ABC} = 84 = \dfrac{13}{2}(\overline{BD}) \implies \overline{BD} = \dfrac{168}{13} \implies \overline{AD} = \dfrac{70}{13}$ .

Let $A(0,0), \:\ B(\dfrac{70}{13},\dfrac{168}{13}),\:\ C(13,0)$ and $P(x_{0},y_{0})$ .

Using $\triangle{APC} \implies$

$x_{0}^2 + y_{0}^2 = \dfrac{10404}{121}$

$x_{0}^2 - 26x_{0} + 169 = \dfrac{8281}{121}$

$\implies x_{0} = \dfrac{1026}{143}$ and $y_{0} = \dfrac{840}{143}$

Point $P(\dfrac{1026}{143},\dfrac{840}{143})$

The volume of the tetrahedron $V = 28h = 224 \implies h = 8 \implies$ Point $Q(\dfrac{1026}{143},\dfrac{840}{143},8)$ .

$\overline{BA} = -\dfrac{70}{13}\vec{i} -\dfrac{168}{13}\vec{j} + 0\vec{k}$

$\overline{BC} = \dfrac{99}{13}\vec{i} -\dfrac{168}{13}\vec{j} + 0\vec{k}$

$\overline{BQ} = \dfrac{256}{143}\vec{i} -\dfrac{1008}{143}\vec{j} + 8\vec{k}$

$\vec{u} = \overline{BQ} \:\ X \:\ \overline{BA} = \dfrac{1344}{13}\vec{i} -\dfrac{560}{13}\vec{j} + -\dfrac{8736}{143}\vec{k}$

$\vec{v} = \overline{BQ} \:\ X \:\ \overline{BC} = \dfrac{1344}{13}\vec{i} + \dfrac{792}{13}\vec{j} + -\dfrac{4368}{143}\vec{k}$ $\vec{u} \cdot \vec{v} = \dfrac{749952}{121}$ and $|\vec{u}| = \dfrac{112\sqrt{157}}{11}, \:\ |\vec{v}| = \dfrac{24\sqrt{3221}}{11}$ .

Let $\lambda$ be the measure of the angle between the two triangular faces $BQA$ and $BQC$

$\implies \cos(\lambda) = \dfrac{279}{\sqrt{505697}} \implies \lambda \approx \boxed{66.900021}$ .