Late for my Date!

Since we are dealing with time rather than discrete numbers, we use geometry. On the coordinate plane, let the $x$ -axis represent the time at which Alex arrives, the $y$ -axis represent the time at which his date arrives, and the origin represent $6\!:\!30$ for both axes. Also, one unit equals one minute.

From the condition that Alex will wait no more than $12$ minutes, we get the inequality $y \leq x + 12.$ From the condition that his date will wait no more than $6$ minutes, we get the inequality $y \geq x - 6.$ We also have $0 \leq y \leq 30,$ since Alex's date arrives between $6\!:\!30$ and $7\!:\!00.$ Similarly, $15 \leq x \leq 45,$ since Alex arrives between $6\!:\!45$ and $7\!:\!15.$

The graph of these inequalities is shown below. The square region marked by the black points represents the total possible ordered pairs, and the dark blue region within this region represent our desired cases.

Now, we just find the area of the blue region to the total square region. Let $A = (15, 30).$ The line $y = x + 12$ intersects $y = 30$ at $B(18, 30)$ and $x = 15$ at $C(15, 27).$ The line $y = x - 6$ intersects $y = 30$ at $D(36, 30)$ and $x = 15$ at $E(15, 9).$ The area of the blue region is then

$\begin{aligned} [BDEC] &= [ADE] - [ABC] \\ &= \frac{(AD)(DE)}{2} - \frac{(AB)(AC)}{2} \\ &= \frac{21^2}{2} - \frac{3^2}{2} \\ &= \frac{18(24)}{2} \\ &= 216. \end{aligned}$

The area of the square region is just $30^2 = 900.$ Thus, our probability is $\frac{216}{900} = \frac{6}{25}$ , so $m + n = 6 + 25 = \boxed{31}.$

(The moral of this story: Never have a math contest on the same day as Valentine's Day!)

UPDATE: Unfortunately, SMT has been canceled. It has been replaced by a new contest called ASMT, which is in the same format as the SMT.