Welcome 2016! Part 30

$Let\quad AD=x\quad ,\quad so\quad DB=11/x\\ \\ Now,\quad since\quad area\quad of\quad triangle\quad \triangle =rs,\quad where\quad r\quad is\quad the\quad inradius\quad and\quad s\quad is\quad the\quad semi-perimeter.\\ \\ Here,\quad s\quad =\quad (AB+BC+CA)/2\\ \\ \Rightarrow s=\{ (x+\frac { 11 }{ x } )+(\frac { 11 }{ x } +r)+(x+r)\} /2\\ \\ \Rightarrow s=x+\frac { 11 }{ x } +r\\ \\ \therefore \quad \triangle =rx+\frac { 11r }{ x } +{ r }^{ 2 }\quad \longmapsto equation\quad 1\\ \\ Now,applying\quad Pythagoras\quad theorem\quad \\ \\ { (r+x) }^{ 2 }+{ (r+\frac { 11 }{ x } ) }^{ 2 }{ =(\frac { 11 }{ x } +x) }^{ 2 }\\ \\ \Rightarrow 2rx+\frac { 22r }{ x } +2{ r }^{ 2 }=22\\ \\ \Rightarrow rx+\frac { 11r }{ x } +{ r }^{ 2 }=11\\ \\ Hence\quad from\quad equation\quad 1,\quad area\quad of\quad triangle\quad \triangle =11\quad square\quad units.\\ \\$

Let P be the center of the inscribed circle. Draw in $AP$ and $PB$ as well as the perpendiculars from the center of the circle to the sides. We know $[ABC]=rs=r(r+x+y)$ where $xy=11$ . We can also find the area in a different way by just doing base multiplied by height divided by 2. This gives us $\frac{(r+x)(r+y)}{2}$ as the area. Setting the two equal gives $r^2+rx+ry-xy=0$ . This can be rewritten as $(r+x)(r+y)=2xy$ or $\frac{(r+x)(r+y)}{2}=xy$ . Note the LHS is just the area of the triangle while the RHS is 11, thus, our answer is 11.