Answer: 14

Solution:

Let chords $BC$ and $AC$ meet $\overline{MN}$ at $P$ and $Q,$ and set $x = \frac{CM}{CN}.$ Since $\angle MBN = 90^\circ$ (angle in a semicircle) and $MB = \frac{3}{5},$ we get $BN = \frac{4}{5};$ also $AM = AN = \frac{\sqrt{2}}{2}.$ Because $P$ lies on both $MN$ and $BC,$ the ratio $\frac{MP}{PN}$ equals the ratio of the distances from $M$ and $N$ to line $BC,$ i.e. $\frac{[BMC]}{[BNC]}.$ In cyclic quadrilateral $MBNC$ the angles $BMC$ and $BNC$ are supplementary, so their sines are equal and $$\begin{aligned} \frac{MP}{PN} &= \frac{BM \cdot MC}{BN \cdot NC} = \frac{3x}{4}, \\ \frac{MQ}{QN} &= \frac{AM \cdot MC}{AN \cdot NC} = x. \end{aligned}$$

Since $MN = 1,$ these give $MP = \frac{3x}{3x + 4}$ and $MQ = \frac{x}{x + 1},$ so $$\begin{aligned} &d = MQ - MP \\ &= \frac{x}{x + 1} - \frac{3x}{3x + 4} \\ &= \frac{x}{3x^2 + 7x + 4} \\ &= \frac{1}{3x + \frac{4}{x} + 7}. \end{aligned}$$

As $C$ ranges over the far semicircle, $x$ takes every positive value. By AM-GM, $3x + \frac{4}{x} \ge 2\sqrt{12} = 4\sqrt{3},$ with equality at $x = \frac{2}{\sqrt{3}}.$ Hence the largest value of $d$ is $$\frac{1}{7 + 4\sqrt{3}} = 7 - 4\sqrt{3},$$ since $(7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 1.$ Then $r + s + t = 7 + 4 + 3 = 14.$