In a triangle $ABC$, the heights $AF$, $BH$ and $CE$ are drawn, with $O$ being the orthocenter. Calculate the radius of the circle circumscribed around the triangle $ABC$, If: $AB^2 + BC^2 - AC^2 = 10$ and $AO^2 + CO^2 - OB^2 = 15$.
(Answer:$2.5$)
I managed to make the drawing to scale and tried to relate the triangles formed but I had no success.
$BC^2=BE^2+CE^2\tag1$ $AC^2=AE^2+CE^2\tag2$ ${\small{(1)-(2)}}:BC^2-AC^2=BE^2-AE^2$
$AH^2+BH^2=AB^2\tag3$ $BH^2+CH^2=BC^2\tag4$ ${\small{(3)-(4)}}:AB^2-AC^2=AH^2-CH^2$
$BF^2+AF^2=AB^2\tag5$ $CF^2+AF^2=AC^2\tag6$ ${\small{(5)-(6)}}:AB^2-AC^2=BF^2-CF^2$
Just use the property of the relationship between the distance from the circumcenter($d$) to the base of the triangle and from the orthocenter($d_1$) to the vertex of the triangle: $d_1=2d$.
$\triangle AHJ: R^2= (\frac{p}{2})^2+(\frac{b}{2})^2\implies 4R^2 = p^2+b^2\tag1$
$\triangle CHG: R^2=(\frac{n}{2})^2+(\frac{a}{2})^2 \implies 4R^2=n^2+a^2\tag2$
$\triangle AIH: R^2= (\frac{c}{2})^2+(\frac{m}{2})^2 \implies 4R^2=m^2+c^2 \tag3$
$a^2+c^2-b^2=10\tag4$
$n^2+m^2-p^2=15\tag5$
$(4)+(5): \underbrace{a^2+n^2}_{4R^2}+\underbrace{c^2+m^2}_{4R^2}-(\underbrace{b^2+p^2}_{4R^2}) = 25$ $\implies R=\sqrt{\frac{25}{4}}=\boxed{\frac{5}{2}}$