The answer is 25.

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$ABC~ is~ a ~right~ \Delta.~ Let~ M ~be ~the~ midpoint~ of~ AC.\\ \therefore~ on~ co-ordinate~ plane,~ let~~ B(0,0),~ A(0,15),~ C(8,0),~ M(4,7.5).\\ We ~have~ HGO~ ~ (in~ that~ order)~~ is~ the ~Euler~ Line. \\ \therefore ~quadrilateral~ IHGO ~is ~the ~degenerated~\Delta~~ IHO.\\ We~ know~ for~ a ~right~\Delta ,~~ H=B,~~O=M,~~~ so~~ HO=BM=AC/2=17/2.\\ Line~ BM ~is,~~ 15X-8Y+0=0.\\ In~ radius,~ r=\dfrac{Area} s=\dfrac{1/2 * 15 * 8}{1/2(15+8+17)}=3.~~ So~~ I(3,3).\\ \therefore~ the~ \bot ~ distance~ from~ I~ to ~line~ BM,~~ p =\dfrac{15 * 3 - 8 * 3 + 0}{\sqrt{15^2+8^2}} = \dfrac {21}{17}. \\ \therefore~ the~ Area~ IHGO=Area~ IHO=1/2 *p* HO =1/2 * 21/17*17/2 =21/4=a/b.\\ \therefore~ a+b=21+4=\Large~~~~~ \color{#D61F06}{25}.$