Find a segment value of Isosceles Triangle $M$ and $N$ are points from equal sides $CA$ and $CB$ from isosceles triangle $ABC$ as what $CM = CN$. Determine the segment $CM$, knowing that $AB = 2k$ and that $2P$ and $2p$ are the perimeters respectively from triangle $ABC$ and the trapeze AMNB.
Answer: $\Large CM = \frac{(P-k)(P-p)}{P-2k}$


Given isosceles $\triangle ABC$,
\begin{align}
AB+BC+AC&=2P, \tag{1}\label{1}\\
AB+BN+MN+AM&=2p. \tag{2}\label{2}
\end{align}
Subtraction $\eqref{1}-\eqref{2}$ gives
\begin{align}
2x-y&=2P-2p. \tag{3}\label{3}
\end{align}
$\triangle MNC\propto\triangle ABC$, hence
\begin{align}
\frac{y}{2k}&=
\frac{x}{AC}=\frac{x}{BC},\\
\frac{y}{2k}&=
\frac{y+x+x}{2k+AC+BC}
=\frac{y+2x}{2P},\\
y&=\frac{2x k}{P-k}.
\tag{4}\label{4}
\end{align}
Finally, a combination of $\eqref{3}$ and $\eqref{4}$ gives the result 
\begin{align}
x&=
\frac{(P-p)(P-k)}{P-2k}.
\end{align}
