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2025 AIME I Problems/Problem 8

Problem

Let $k$ 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 $z$. The sum of all possible values of $k$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. Here $i = \sqrt{-1}$.

Solution 1 (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 $A + Bi$.


Note that the first equation becomes: $(25 - A)^2 + (20 - B)^2 = 25$

Note that this is the equation of a circle centered at $(25, 20)$ with radius $5$.


And the second equation becomes: $(A-4-k)^2 + B^2 = (A - k)^2 + (B-3)^2$

You can see that the many similar terms that cancel out, simplfying, you get:

$-8(A - k) + 16 + 6B = 9$

Now we must isolate B

$B= \frac{4}{3}(A-k) - \frac{7}{6}$

$B = \frac{4}{3}A - \frac{4}{3}k - \frac{7}{6}$

This equation can be seen as a line with a $\frac{4}{3}$ slope, and a y-intercept of $\frac{4}{3}k - \frac{7}{6}$.

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 $\frac{4}{3}$, the circle must have a slope coming out of center of its reciprocal, $-\frac{3}{4}$. So the points on the circle where this line with a $\frac{4}{3}$ must intersect must be $(21, 23)$ and $(29, 17)$. We can easily use point-slope form to find the equations of these lines. $y - 23 = \frac{4}{3}(x - 21)$

and

$y - 17 = \frac{4}{3}(x - 29)$

Now we must match the y-intercepts to the equations with $k$ in it. Solving the equations:

$\frac{4}{3}(-21) + 23 = - \frac{4}{3}k - \frac{7}{6}$

$\frac{4}{3}(-29) + 17 = - \frac{4}{3}k - \frac{7}{6}$

we get that $k = \frac{23}{8}$ and $k = \frac{123}{8}$ Adding them up and simplifying, we get a sum of $\frac{73}{4} \Longrightarrow \boxed{077}.$

~Marcus :) (feel free to correct my Latex)


Solution 2

[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] The complex number $z$ must satisfy the following conditions on the complex plane:

$1.$ The magnitude between $z$ and $(25,20)$ is $5.$ This can be represented by drawing a circle with center $(25,20)$ and radius $5.$

$2.$ It is equidistant from the points $(4+k,0)$ and $(k,3).$ Hence it must lie on the perpendicular bisector of the line connecting these points.


For $z$ 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, $(2+k,\frac{3}{2}),$ and have slope $\frac{4}{3}.$ The segment connecting the point of tangency to the center of the circle has slope $\frac{-3}{4},$ meaning the points of tangency can be $(29,17)$ or $(21,23).$ Solving the equation for the slope of the perpendicular bisector gives \[\frac{\frac{3}{2}-23}{k+2-21}=\frac{4}{3}\] or \[\frac{\frac{3}{2}-17}{k+2-29}=\frac{4}{3},\] giving $k=\frac{23}{8}$ or $\frac{123}{8}$, having a sum of $\frac{73}{4} \Longrightarrow \boxed{077}.$

~nevergonnagiveup


There's actually an easier way to do it using this method by utilizing the distance between point and line formula after building off of what is shown above. First we find the standard form of the perpendicular bisector, which can be found using the point-slope form: $y-b = m(x-a)$, where $a$ and $b$ are the $x$ and $y$ coordinates of a point on the line. By plugging in $(2+k, \frac{3}{2})$, we get $y-\frac{3}{2} = \frac{4}{3}(x-2-k)$, we can eventually find the standard form as $8x-6y-7-8k=0$. Now we use the distance between point and line formula on the center of the circle at $(25, 20)$ and the perpendicular bisector. We get $d = \frac{|8\cdot 25-6\cdot 20-7-8k|}{\sqrt{6^2+8^2}}$. Plugging in $d$ = 5 we can simplify this to $50 = |200-120-7-8k|$. We can finally solve for the absolute value equality and figure out $k = \frac{23}{8}$ or $k = \frac{123}{8}$. Adding them together, we get $\frac{146}{8} = \frac{73}{4}$, hence the answer which we desire is $\Longrightarrow \boxed{077}.$

~Mathycoder (edited by MathKing555)

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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