JEE Novice - (31)

clearly orthocenter is $(0,0$

magnitude of inradius, which is radius of incircle of $\Delta OAB$ and incenter denoted by $I$

$PI=QI=r=(s-o) tan(\frac{O}{2})$

where $o=\sqrt{13},~b=2,~a=3$ note that $o$ is the length of side facing $O$ , $b$ is the length of side facing $B$ and $a$ is the length of side facing $A$

angle $AOB=90^{o}$

$r=\frac{5-\sqrt{13}}{2}$

We have to find $OI$

Clearly by pythagoras theorem

$OI=\sqrt{PI^{2}+OP^{2}}=\sqrt{PI^{2}+QI^{2}}$

$OI=\sqrt{2} PI=\sqrt{2} r$

$OI=\sqrt{2}(\frac{5-\sqrt{13}}{2})$

On rationalising and simplifying

$\huge{\boxed{OI=\frac{6\sqrt{2}}{5+\sqrt{13}}}}$

A very nice solution from Tanishq. An upvote for that.

I, here, have the standard JEE approach.

Let the vertices of the triangle formed be $A(0,0) , B(0,3) , C(2,0)$ and the sides opposite to them be $a = \sqrt{13} , b = 2 , c = 3$ respectively.

The coordinates of the incentre of a triangle whose vertex coordinates are $A(x_1,y_1) , B(x_2,y_2) , C(x_3,y_3)$ is

$(\dfrac{ax_1 + bx_2 + cx_3}{a+b+c} , \dfrac{ay_1 + by_2 + cy_3}{a+b+c} )$

Using the above formula we get,

$I = (\dfrac{6}{5+\sqrt{13}}, \dfrac{6}{5+\sqrt{13}})$

Orthocentre of a righta-triangle is the vertex opposite to the Hypotenuse.

Hence the distance is(using distance formula)

$\dfrac{6\sqrt{2}}{5 + \sqrt{13}}$

$3x+2y=6,~~\implies~\dfrac X 2+\dfrac y 3=1\\ \therefore~The~vertices~are~(0,3),~(0,0),~(2,0).\\ Sides~3-2-\sqrt{13}. \implies~inradius, r=\dfrac{2*3}{2+3+\sqrt{13}}.\\ So~r \left (\dfrac{2*3}{2+3+\sqrt{13}},\dfrac{2*3}{2+3+\sqrt{13}} \right ).\\ Orthocenter ~is~the~vertex~with~90^o,~that~is~(0,0).\\ So~the~distance~=\sqrt{ \left ( \dfrac{2*3}{2+3+\sqrt{13}}-0 \right )^2+\left (\dfrac{2*3}{2+3+\sqrt{13}}-0 \right )^2}\\ So~the~distance~=\dfrac{6\sqrt2}{5+\sqrt{13}}=\dfrac{a\sqrt b}{c+\sqrt{d}}.\\ a+b+c+d=6+2+5+13=\Large~~\color{#D61F06}{26}.$