2025 AIME I Problems/Problem 8
Problem
Let 
be a real number such that the system
\begin{align*}
&|25 + 20i - z| = 5 \\
&|z - 4 - k| = |z - 3i - k|
\end{align*}
has exactly one complex solution 
. The sum of all possible values of can be written as 
, where 
and 
are relatively prime positive integers. Find 
. Here 
.
Solution 1
![[asy] size(300); draw((0, 0) -- (0, 20), EndArrow(10)); label("$y$", (0, 20), NW); dot((25,20)); draw((0, 0) -- (25, 0), EndArrow(10)); label("$x$", (25, 0), SE); draw(circle((25,20),5)); label(scale(0.7)*"$(25,20)$", (25,20), S); draw((7,0) -- (3,3), blue); draw((5,3/2) -- (21,23), dashed); label("$(4+k,0)$", (7,0), S); label("$(k,3)$", (3,3), N); draw(rightanglemark((3,3),(5,3/2),(21,23), 20)); draw(rightanglemark((25,20),(21,23),(5,3/2), 20)); draw((25,20) -- (21,23)); [/asy]](http://latex.artofproblemsolving.com/5/5/b/55b4293d356e1004283f0668760ab19ef31280a5.png)
The complex number must satisfy the following conditions on the complex plane:
The magnitude between and 
is 
This can be represented by drawing a circle with center and radius
It is equidistant from the points 
and 
Hence it must lie on the perpendicular bisector of the line connecting these points.
For to have one solution, the perpendicular bisector of the segment connecting the two points must be tangent to the circle.
This bisector must pass the midpoint, 
and have slope 
The segment connecting the point of tangency to the center of the circle has slope 
meaning the points of tangency can be 
or 
Solving the equation for the slope of the perpendicular bisector gives ![\[\frac{\frac{3}{2}-23}{k+2-21}=\frac{4}{3}\]](http://latex.artofproblemsolving.com/7/9/d/79daae48af26f2a59fbee0a959646a1f97a2a5be.png)
or ![\[\frac{\frac{3}{2}-17}{k+2-29}=\frac{4}{3},\]](http://latex.artofproblemsolving.com/6/b/9/6b991387e76aa0f7ef73984385688104ec4ad1ac.png)
giving 
or 
, having a sum of 
~nevergonnagiveup
Solution 2 (Systematic + Algebra)
We first look at each equation, and we convert each to algebra (note that the absolute value sign of 
means the magnitude). Let's convert z to 
.
Note that the first equation becomes:
Note that this is the equation of a circle centered at 
with radius 
.
And the second equation becomes:
You can see that the many similar terms that cancel out, simplfying, you get:
Now we must isolate B
This equation can be seen as a line with a 
slope, and a y-intercept of 
.
Note that the question only wants one solution, so we want two tangent lines, one above the circle, and one below the circle. You can see Solution 1 for the picture.
Because the slope is , the circle must have a slope coming out of center of its reciprocal, 
. So the points on the circle where this line with a must intersect must be 
and 
. We can easily use point-slope form to find the equations of these lines.
and
Now we must match the y-intercepts to the equations with in it.
Solving the equations:
we get that 
and 
Adding them up and simplifying, we get a sum of
~Marcus :)
See also
| 2025 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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