Find the maximum side

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I am taking r=Inradius and $R=Circumradius$

$AO=R$ because $AO$ is the circumradius and $DI=r$ because it is the inradius.

In the right triangle $ADI$ we have

$sin(A/2)=\frac { DI }{ AI } =\frac { r }{ AI } \quad \Rightarrow AI=rcosec(A/2)\quad (i)$

By euler's theorom distance between incentre and circumcentre is

${ OI }^{ 2 }={ R }^{ 2 }-2Rr\quad \quad \quad (ii)$

Given ${ AIO }^{ 0 }\le { 90 }^{ 0 }$ we have $cos(AIO)\ge 0$

Applying cosine rule in triangle $AIO$ we get $\frac { { (A }I^{ 2 }+{ OI }^{ 2 }-{ A0 }^{ 2 }) }{ 2.AO.OI } =cos(AIO)\ge 0$

Hence we have ${ A }I^{ 2 }+{ IO }^{ 2 }\ge { AO }^{ 2 }$

Putting values we get ${ r }^{ 2 }{ cosec }^{ 2 }(A/2)+({ R }^{ 2 }-2Rr)\ge { R }^{ 2 }\\ \Rightarrow r{ cosec }^{ 2 }(A/2)\ge 2R\quad \quad \Rightarrow r\ge 2R{ sin }^{ 2 }(A/2)$

Using $r=(s-a)tan(A/2)\quad and\quad 2R=a/sin(A)$ we get

$\Rightarrow (s-a)tan(A/2)sin(A)\ge a{ sin }^{ 2 }(A/2)\\ \Rightarrow 2(s-a)\ge a\quad \quad \Rightarrow (a+b+c)-2a\ge a\\ \Rightarrow \frac { b+c }{ 2 } \ge a$

Putting the values $b=102,c=100$ we get $a\le 101$

So $\boxed { { a }_{ max } } =101$