AIME I

Geometrically, $|a-b|$ represents the distance between complex numbers $a$ and $b$ in the complex plane. Thus $|(25+20i)-z| = 5$ means that the distance from $25+20i$ to $z$ is $5$. The set of solutions for $z$ is then a circle with radius $5$ and center $25+20i$. Similarly, $|z-(4+k)| = |z-(k+3i)|$ means that the distance from $z$ to $4+k$ equals the distance from $z$ to $k+3i$. Geometrically, the set of points equidistant from two fixed points $A$ and $B$ is the perpendicular bisector of $\overline{AB}$. Thus the solution set for this equation is the line that is the perpendicular bisector of the segment connecting $4+k$ and $k+3i$.

Any intersection of the circle from the first equation and the line from the second equation is a solution to the system of equations. For the system to have exactly one complex solution, the line and the circle must be tangent. There are two such lines, $\ell_1$ and $\ell_2$, as shown below.

Switching to Cartesian coordinates, for any $k$, the slope of the line between $(k, 3)$ and $(4+k, 0)$ equals $-\frac{3}{4}$, and hence the slopes of lines $\ell_1$ and $\ell_2$ are both equal to $\frac{4}{3}$. Because line $\ell_1$ passes through $(2+k_1, \frac{3}{2})$, the equation of this line is
$$y - \frac{3}{2} = \frac{4}{3}(x - (2 + k_1)),$$
and the $x$-coordinate of its $x$-intercept is $k_1 + \frac{7}{8}$. Similarly, the $x$-coordinate of the $x$-intercept of $\ell_2$ is $k_2 + \frac{7}{8}$.

The line $\ell_3$, parallel to lines $\ell_1$ and $\ell_2$, whose $x$-intercept is the midpoint of the $x$-intercepts of lines $\ell_1$ and $\ell_2$, passes through the center of the circle, which is at $(25, 20)$. Thus the equation of line $\ell_3$ is
$$y - 20 = \frac{4}{3}(x - 25),$$
and the $x$-coordinate of its $x$-intercept is $10$.

Therefore
$$\frac{k_1 + k_2}{2} + \frac{7}{8} = 10,$$
from which $k_1 + k_2 = \frac{73}{4}$. The requested sum is $73 + 4 = 77$.