Inner 6-7-8 triangle

If the central angles are $\alpha,\beta,\gamma$ , then $a = \sqrt{85 - 84\cos\alpha}$ , $b = \sqrt{113-112\cos\beta}$ and $c = \sqrt{100-96\cos\gamma}$ , and so we want to maximise $a + b + c \; = \; \sqrt{85 - 84\cos\alpha} + \sqrt{113-112\cos\beta} + \sqrt{100-96\cos\gamma}$ subject to the restraint $\alpha + \beta + \gamma = \pi$ . Lagrange multipliers tell us that $\frac{42\sin\alpha}{a} \; = \; \frac{56\sin\beta}{b} \; = \; \frac{48\sin\gamma}{c} \; =\; k$ for some contant $k$ , and hence $\begin{aligned} k^2 & = \; \frac{42^2}{a^2}\left[1 - \left(\frac{85 - a^2}{84}\right)^2\right] \; = \; \frac{56^2}{b^2}\left[1 - \left(\frac{113-b^2}{112}\right)^2\right] \; = \; \frac{48^2}{c^2}\left[1 - \left(\frac{100 - c^2}{96}\right)^2\right] \\ & = \; \frac{(a^2-1)(169-a^2)}{4a^2} \; = \; \frac{(b^2-1)(225-b^2)}{4b^2} \; = \; \frac{(c^2-4)(196-c^2)}{4c^2} \\ 226 - 4k^2 & = \; a^2 + \frac{13^2}{a^2} + 56 \; = \; b^2 + \frac{15^2}{b^2} \; = \; c^2 + \frac{28^2}{c^2} + 26 \end{aligned}$ making the answer $13 + 56 + 15 + 28 + 26 = \boxed{138}$ .

Let $A$ , $B$ , $C$ , $P$ be points in the plane such that $x = AP$ , $y = BP$ , $z = CP$ are fixed distances. Then by the arguments in the following links, when the perimeter of triangle $ABC$ is maximized, $P$ is the incenter of triangle $ABC$ .

https://math.stackexchange.com/questions/1684501/finding-max-perimeter-of-triangle-of-three-circulating-points

https://www.cut-the-knot.org/Curriculum/Geometry/TripolarOptimization.shtml

Let $r$ be the inradius of triangle $ABC$ . Then $r$ is the height of triangle $PBC$ with respect to base $BC$ , so by Heron's formula, $\begin{aligned} r^2 &= \left( \frac{2 [PBC]}{BC} \right)^2 \\ &= \frac{4 [PBC]^2}{a^2} \\ &= \frac{((y + z)^2 - a^2)(a^2 - (y - z)^2)}{4a^2} \\ &= \frac{-a^4 + 2(y^2 + z^2) a^2 - (y^2 - z^2)^2}{4a^2}. \end{aligned}$ Then $-4r^2 = a^2 - 2(y^2 + z^2) + \frac{(y^2 - z^2)^2}{a^2}.$

Likewise, $\begin{aligned} -4r^2 &= b^2 - 2(x^2 + z^2) + \frac{(x^2 - z^2)^2}{b^2}, \\ -4r^2 &= c^2 - 2(x^2 + y^2) + \frac{(x^2 - y^2)^2}{c^2}, \end{aligned}$ so $a^2 - 2(y^2 + z^2) + \frac{(y^2 - z^2)^2}{a^2} = b^2 - 2(x^2 + z^2) + \frac{(x^2 - z^2)^2}{b^2} = c^2 - 2(x^2 + y^2) + \frac{(x^2 - y^2)^2}{c^2}.$

Plugging in $x = 8$ , $y = 6$ , $z = 7$ gives us $a^2 - 170 + \frac{13^2}{a^2} = b^2 - 226 + \frac{15^2}{b^2} = c^2 - 200 + \frac{28^2}{c^2},$ which leads to $a^2 + \frac{13^2}{a^2} + 56 = b^2 + \frac{15^2}{b^2} = c^2 + \frac{28^2}{c^2} + 26.$

Let $\theta_1, \theta_2$ and $\theta_3$ denote the interior angles that split the original triangle into 3 triangles.

By cosine rule , $\cos\theta_1 = \frac{85-a^2}{84} \hspace{5em} \cos\theta_2 = \frac{113-b^2}{112} \hspace{5em} \cos\theta_3 = \frac{100-c^2}{96}$

We want to maximize $P := a + b + c = \sqrt{85 - 84 \cos \theta_1} + \sqrt{113 - 112 \cos \theta_2} + \sqrt{100 - 96 \cos\theta_3}$

Note that $\cos(\theta_3) = \cos[2\pi - (\theta_1 + \theta_2)] = \cos(\theta_1 + \theta_2) = \cos(\theta_1) \cos(\theta_2) - \sin(\theta_1) \sin(\theta_2).$

Thus, we can express $P$ as a two variable function (in terms of $\theta_1$ and $\theta_2$ ). When $P$ is maximized, its partial derivatives are both equal to $0 ,$ $\dfrac{\partial P}{\partial \theta_1} = 0 , \dfrac{\partial P}{\partial \theta_2} = 0 .$

$\dfrac{\partial P}{\partial \theta_1} = 0 \quad\implies \quad \dfrac{48 \sin(\theta_1 + \theta_2)}{\sqrt{100 - 96 \cos(\theta_1 + \theta_2)}} + \dfrac{42\sin\theta_1}{\sqrt{85 - 84\cos \theta_1}} = 0 \tag1$

$\dfrac{\partial P}{\partial \theta_2} = 0 \quad\implies \quad \dfrac{48 \sin(\theta_1 + \theta_2)}{\sqrt{100 - 96 \cos(\theta_1 + \theta_2)}} + \dfrac{56\sin\theta_2}{\sqrt{113 - 112\cos \theta_2}} = 0 \tag2$

Because $(1) = (2)$ :

$\dfrac{42\sin\theta_1}{\sqrt{85 - 84\cos \theta_1}} = \dfrac{56\sin\theta_2}{\sqrt{113 - 112\cos \theta_2}}$

Square both sides of the equation, then utilize the identity $\sin^2 (Z) = 1 - \cos^2(Z) ,$

$\dfrac9{a^2} \left[1 - \left(\dfrac{85-a^2}{84} \right)^2 \right ] = \dfrac{16}{b^2} \left[1 - \left(\dfrac{113-b^2}{112} \right)^2 \right ]$

Upon simplifying, $a^2 + \dfrac{169}{a^2} + 56 = b^2 + \dfrac{225}{b^2} \tag3.$

Likewise we can also express $P$ as a two variable function (in terms of $\theta_1$ and $\theta_3$ ), and by using $\cos(\theta_2) = \cos(\theta_1 + \theta_3) = \cos(\theta_1) \cos(\theta_3) - \sin(\theta_1) \sin(\theta_3) .$

Using the similar reasoning as above, we gather that

$\dfrac{\partial P}{\partial \theta_1} = 0 \quad\implies \quad \dfrac{56\sin(\theta_1 + \theta_3)}{\sqrt{113 - 112 \cos(\theta_1 + \theta_3)}} + \dfrac{42\sin\theta_1}{\sqrt{85 - 84\cos \theta_1}} = 0 \tag4$

$\dfrac{\partial P}{\partial \theta_3} = 0 \quad\implies \quad \dfrac{56 \sin(\theta_1 + \theta_3)}{\sqrt{113 - 112 \cos(\theta_1 + \theta_3)}} + \dfrac{48\sin\theta_3}{\sqrt{100 - 96\cos \theta_3}} = 0 \tag5$

Again, because $(4) = (5):$

$\dfrac{42\sin\theta_1}{\sqrt{85 - 84\cos \theta_1}} = \dfrac{48\sin\theta_3}{\sqrt{100 - 96\cos \theta_3}} \quad \implies \quad \dfrac{49}{a^2} \left[1 - \left(\dfrac{85-a^2}{84} \right)^2 \right ] = \dfrac{64}{b^2} \left[1 - \left(\dfrac{100-b^2}{96} \right)^2 \right ]$

Simplify once more $a^2 + \dfrac{169}{a^2} + 30 = c^2 + \dfrac{784}{c^2} \tag6$

Combining $(3)$ and $(6)$ yields

$a^2 + \dfrac{169}{a^2} + 56 = b^2 + \dfrac{225}{b^2} = c^2 + \dfrac{784}{c^2} + 26 .$

Hence, $P = \sqrt{169} = 13, Q = 56, R = \sqrt{225} = 15, S = \sqrt{784} = 28, T = 26 .$ The answer is $13 + 56 + 15 + 28 + 26 = \boxed{138} .$