G4.1 In the figure below, $A B C D$ is a square of side length 2 . A circular arc with centre at $A$ is drawn from $B$ to $D$. A semicircle with centre at $M$, the midpoint of $C D$, is drawn from $C$ to $D$ and sits inside the square. Determine the shortest distance from $P$, the intersection of the two arcs, to side $A D$, that is, the length of $P Z$.
Join $A P, D P, M P$.
Let $F$ be the foot of perpendicular from $P$ to $C D$.
Let $A Z=x$. Then $D Z=2-x=P F, D M=M P=1$
In $\triangle A Z P, A Z^{2}+Z P^{2}=A P^{2}$ (Pythagoras' theorem) $Z P^{2}=4-x^{2} \ldots \ldots(1)$
In $\triangle P M F, M F^{2}+P F^{2}=P M^{2}$ (Pythagoras' theorem)
$M F^{2}=1-(2-x)^{2}=4 x-x^{2}-3 \ldots \ldots$ (2)
$P Z=D F=1+M F$
$4-x^{2}=\left(1+\sqrt{4 x-x^{2}-3}\right)^{2}$
$4-x^{2}=1+2 \sqrt{4 x-x^{2}-3}+4 x-x^{2}-3$
$3-2 x=\sqrt{4 x-x^{2}-3}$
$9-12 x+4 x^{2}=4 x-x^{2}-3$
$5 x^{2}-16 x+12=0$
$(5 x-6)(x-2)=0$
$x=\frac{6}{5}$ or 2 (rejected)
$P Z=\sqrt{4-x^{2}}=\sqrt{4-1.2^{2}}=\sqrt{2.56}=1.6$
Method 2 Let $D$ be the origin, $D C$ be the $x$-axis, $D A$ be the $y$-axis.
Equation of circle $D P C:(x-1)^{2}+y^{2}=1 \Rightarrow x^{2}-2 x+y^{2}=0 \ldots \ldots$. (1)
Equation of circle $B P D: x^{2}+(y-2)^{2}=2^{2} \Rightarrow x^{2}+y^{2}-4 y=0$.
(1) $-(2) \Rightarrow y=\frac{x}{2}$

Sub. (3) into (1): $x^{2}-2 x+\frac{x^{2}}{4}=0 \Rightarrow x=0$ (rejected) or $1.6 \Rightarrow P Z=1.6$