jmc

Let $Q$ be the tangency point on $\overline{AC}$, and $R$ on $\overline{BC}$. By the Two Tangent Theorem, $AP=AQ=23$, $BP=BR=27$, and $CQ=CR=x$. Using $rs=A$, where $s=\frac{27\cdot 2+23\cdot 2+x\cdot 2}{2}=50+x$, we get $(21)(50+x)=A$. By Heron's formula, $A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{(50+x)(x)(23)(27)}$. Equating and squaring both sides, \begin{eqnarray} [21(50+x)]^2 &=& (50+x)(x)(621)\\ 441(50+x) &=& 621x\\ 180x = 441 \cdot 50 &\Longrightarrow & x = \frac{245}{2} \end{eqnarray} We want the perimeter, which is $2s=2\left(50+\frac{245}{2}\right)=\boxed{345}$.