Geometry in the EMIC

Let the midpoint of $AB$ be $N$ and $l = PA = PB =PM$ . When $l$ is minimum when $PM$ is perpendicular to $CD$ and $MPN$ is a straight line. Then $PN = 24 - l$ . By Pythagoras theorem, we have:

$AN^2 + PN^2 = AP^2$

$\Rightarrow 12^2 + (24 - l)^2 = l^2$

$\quad 144 + 576 - 48l + l^2 = l^2$

$\Rightarrow l = \dfrac{720}{48} = \boxed{15}$

$Let\quad ABCD\quad in\quad X-Y\quad plane\quad such\quad that\\ A(-12,-12)\quad \& \quad B(12,-12)\quad \& \quad C(12,12)\\ \& \quad D(-12,12).\quad Since\quad PA=PB=PC\\ so\quad 'P'\quad must\quad lie\quad on\quad Y-axis\quad (By\quad Symmetry)\\ let\quad P(0,\alpha ).\quad Now\quad PM=PA\\ \Longrightarrow { \quad (\alpha -12 })^{ 2 }=\quad { (\alpha +12) }^{ 2 }+{ (12 })^{ 2 }\\ \Longrightarrow \quad \alpha =-3\\ \Longrightarrow PM=15$ .

If $M$ is the midpoint of $CD$ , and $PA=PB=PM$ , then $P$ is the circumcentre of $\triangle ABM$ . Now, $\angle AMB = 180-2\tan^{-1} (2)$ . Thus, $\sin\angle AMB = 0.8$

By sine rule, $\frac{AB}{\sin\angle AMB}=2R$ or $\frac{24}{0.8}=2PM$

Thus, $PM=\boxed{15}$