Consider $\Delta ABC$ such that $AB = 13, BC = 14, AC = 15$

Let $AD \perp BC, BE \perp AC, CF \perp AB$ and let $H$ be its orthocenter

Let $R(ABC)$ denote circumradius of $\Delta ABC$

Let $r_0$ be inradius and $r_1, r_2, r_3$ be the exradii of $\Delta ABC$

Then find the value of

$R(ABC) + R(HAB) + R(HBC) + R(HAC) + \sum_{i=0}^3 r_i$

The answer is 73.

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$\text{Let P , Q , R be the reflections of H in BC , AC , AB respectively.}$

$m\angle BAC = a \quad, \quad m\angle ABC = b \quad , \quad m\angle ACB = c \\ a^| = 90-a \quad,\quad b^| = 90-b\quad,\quad c^| = 90-c$

By angle chasing we can clearly get that quadrilaterals ABPC , ABCQ , ARBC are cyclic since their opposite angles are supplementary.

$\implies R(ABC)=R(BPC)=R(AQC)=R(ARB)$

Since $\Delta BHC \cong \Delta BPC \quad\Delta AHC\cong \Delta AQC\quad\Delta AHB \cong \Delta ARB \\ \implies \boxed{R(ABC)=R(HAB)=R(HBC)=R(HAC)}$

Area of $\Delta ABC = 84$ (Using Heron's formula).Also ,

$\Delta ABC = \dfrac{\text{product of all sides}}{4R} \Rightarrow 4R = \dfrac{2730}{84} = 32.5$ where 'R' is circumradius of $\Delta ABC$

Area of $\Delta ABC = r_0.s \Rightarrow r_0 = \dfrac{84}{21} = 4$ where 's' is the semiperimeter of the triangle.

We also have the relation $r_1+r_2+r_3=r_0+4R$

$\displaystyle \sum_{i=0}^3 = 4R +2r_0= 32.5 + 8 = 40.5$

So , $\large R(ABC)+R(HAB)+R(HBC)+R(HAC)+\displaystyle \sum_{i=0}^3 r_i$

$\Rightarrow 4R + 4R + 2r_0 = 32.5+40.5 = \huge\boxed{73}$

@Azhaghu Roopesh M