Four centers problem

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

Set the triangle to coordinate B (0,0) C (0,8) A (15,0) See H=B O is midpoint of AC so (15/2, 4) Note coordinate of I is (r,r) where r is in radius A=rs 60=20r r=3 so I =(3,3) G is a little tricky lets choose the B median which has equation y=8x/15 Now we see midpoint of AC is (0,4) take with A (15,0) gives equation for A median as y=-4x.15+4 solve for two intersections of median we get (5, 8/3) Now shoe lace to get 21/4 -> 25