A Triangle Problem 2!

From the figure , let's take $\displaystyle \cos(\alpha) = \dfrac{a^2+b^2-c^2}{2ab}$

From this $2ab\cos(\alpha)-a^2 =b^2-c^2$ and $2ab\cos(\alpha)-b^2= a^2-c^2$

Now $\displaystyle \dfrac{a^3+b^3+c^3}{a+b+c} =c^2 \implies a(a^2-c^2) +b(b^2-c^2) = 0$ $\implies a(2ab\cos(\alpha)-b^2 )+b(2ab\cos(\alpha)-a^2) = 0 \implies 2(a+b)\cos(\alpha)= (a+b)$ $\implies \cos(\alpha)= \dfrac{1}{2} \implies \alpha =\dfrac{π}{3}$

Now , generally $\displaystyle\int_0^{\alpha} \dfrac{f(x)}{f(\alpha-x)+f(x)} dx= \dfrac{\alpha}{2}$

So here answer is $\boxed{\dfrac{\alpha}{2} = \dfrac{π}{6} }$

To prove: $\displaystyle\int_{a}^{b} \dfrac{f(x)}{f(a + b - x) + f(x)} dx = \dfrac{b - a}{2}$

Let $I = \displaystyle\int_{a}^{b} \dfrac{f(x)}{f(a + b - x) + f(x)} dx$ and $u = a + b - x$

$\implies I = \displaystyle\int_{a}^{b} \dfrac{f(a + b - u)}{f(u) + f(a + b - u)} du$

Since $u$ is a dummy variable we can replace $u$ by $x$ to obtain:

$I = \displaystyle\int_{a}^{b} \dfrac{f(a + b - x)}{f(x) + f(a + b - x)} dx$

Adding $I = \displaystyle\int_{a}^{b} \dfrac{f(x)}{f(a + b - x) + f(x)} dx$ and $I = \displaystyle\int_{a}^{b} \dfrac{f(a + b - x)}{f(x) + f(a + b - x)} dx$

$\implies 2I = \displaystyle\int_{a}^{b} dx = b - a \implies \boxed{I = \dfrac{b - a}{2}}$ $\implies \displaystyle\int_{0}^{\alpha} \dfrac{f(x)}{f(\alpha - x) + f(x)} dx = \dfrac{\alpha}{2}$ .

Let $\dfrac{a^3 + b^3 + c^3}{a + b + c} = c^2$

Using the law of cosines $\implies c^2 = a^2 + b^2 - 2ab\cos(\alpha) \implies$

$a^3 + b^3 + c^3 = c^2(a + b + c) = (a + b + c)(a^2 + b^2 - 2ab\cos(\alpha)) =$

$a^3 + b^3 + ab^2 + ba^2 + ca^2 + cb^2 - 2(a^2b + ab^2 + abc)\cos(\alpha)) \implies$

$c^3 - c(a^2 + b^2) = ab^2 + ba^2 - 2(a^2b + ab^2 + abc)\cos(\alpha)) \implies$

$c(c^2 - (a^2 + b^2)) = ab^2 + ba^2 - 2(a^2b + ab^2 + abc)\cos(\alpha) \implies$

$c(-2ab\cos(\alpha)) = -2abc\cos(\alpha) = ab^2 + ba^2 - 2(a^2b + ab^2 + abc)\cos(\alpha) \implies$

$ab^2 + ba^2 - 2(a^2b + ab^2)\cos(\alpha) = 0 \implies (ab^2 + ba^2)(1 - 2\cos(\alpha)) = 0 \implies$

$ab(a + b)(1 - 2\cos(\alpha)) = 0$ and $a > 0$ and $b > 0 \implies 1 - 2\cos(\alpha) = 0 \implies$

$\cos(\alpha) = \dfrac{1}{2} \implies \alpha = \dfrac{\pi}{3} \implies$

$\displaystyle\int_{0}^{\alpha} \dfrac{f(x)}{f(\alpha - x) + f(x)} dx = \dfrac{\pi}{6} \approx \boxed{0.5235987755982989}$ .