Common Tangent Problem.

$\lim_{n \rightarrow \infty} \sum_{j = 1}^{n} (\dfrac{j}{n})^n x^{n - j} = \sum_{j = 0}^{n - 1} (1 - \dfrac{j}{n})^{n} x^{j} = \sum_{n = 0}^{\infty} (\dfrac{x}{e})^n = \dfrac{e}{e - x}$ on $|x| < e$

$\implies m(x) = \lim_{n \rightarrow \infty} \dfrac{\sum_{j = 1}^{n} x^{j}}{\sum_{j = 1}^{n} (\dfrac{j}{n})^n x^{n - j}} = (\dfrac{x}{1 - x})(\dfrac{e - x}{e}) = \dfrac{x(e - x)}{e(1 - x)}$ on $|x| < 1$ .

$\lim_{x \rightarrow 0} p(x) = \lim_{x \rightarrow 0} m(x) = 0 \implies c = 0$ .

$m(e - 2) = \dfrac{2(e - 2)}{e(3 - e)} = p(e - 2) = (e - 2)^2 a + (e - 2) b \implies$

$\boxed{(e - 2)a + b = \dfrac{2}{e(3 - e)}}$

$\dfrac{dm}{dx} = \dfrac{x^2 - 2x + e}{e(1 - x)^2}$ and $\dfrac{dp}{dx} = 2ax + b$

$\dfrac{d}{dx}(m(x))|_{x = e - 2} = \dfrac{e^2 - 5e + 8}{e(3 - e)^2} = \dfrac{d}{dx}(p(x))|_{x = e - 2} = 2(e - 2)a + b \implies$

$\boxed{2(e - 2)a + b = \dfrac{e^2 - 5e + 8}{e(3 - e)^2}}$

and

$\boxed{(e - 2)a + b = \dfrac{2}{e(3 - e)}}$

Solving the system above we obtain:

$a = \dfrac{e - 1}{e(3 - e)^2}$ and $b = -\dfrac{e^2 - e - 4}{e(3 - e)^2}$

$\implies a + b + c = \dfrac{e + 1}{e(3 - e)} \approx \boxed{4.85548885}$ .

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Note: Using point $A: (e - 2, \dfrac{2(e - 2)}{e(3 - e)})$ the equation of the common tangent line is $y = (\dfrac{e^2 - 5e + 8}{e(3 - e)^2})x - \dfrac{(e - 2)^2 (e - 1)}{e(3 - e)^2}$ .

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Checking:

$\dfrac{dp}{dx} =\dfrac{1}{e(3 - e)^2}(2(e - 1)x - (e^2 - e - 4)) \implies \dfrac{d}{dx}(p(x))|_{x = e - 2} = \dfrac{e^2 - 5e + 8}{e(3 - e)^2} = \dfrac{d}{dx}(m(x))|_{e - 2}$ and $p(e - 2) = \dfrac{2(e - 2)}{e(3 - e)} = m(e - 2)$ .