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Difference between revisions of "2025 AIME I Problems/Problem 8"

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For <math>z</math> to have one solution, the perpendicular bisector of the segment connecting the two points must be tangent to the circle.  
 
For <math>z</math> to have one solution, the perpendicular bisector of the segment connecting the two points must be tangent to the circle.  
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This bisector must pass the midpoint, <math>(2+k,\frac{3}{2}),</math> and have slope <math>\frac{4}{3}.</math> The segment connecting the point of tangency to the center of the circle has slope <math>\frac{-3}{4},</math> meaning the points of tangency can be <math>(29,17)</math> or <math>(21,23).</math> Solving the equation for the slope of the perpendicular bisector gives <cmath>\frac{\frac{3}{2}-23}{k+2-21}=\frac{4}{3}</cmath> or <cmath>\frac{\frac{3}{2}-17}{k+2-29}=\frac{4}{3},</cmath> giving <math>k=\frac{23}{8} \text{or} \frac{123}{8}</math>, having a sum of <math>\frac{73}{4} \Longrightarrow \boxed{077}.</math>
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This bisector must pass the midpoint, <math>(2+k,\frac{3}{2}),</math> and have slope <math>\frac{4}{3}.</math> The segment connecting the point of tangency to the center of the circle has slope <math>\frac{-3}{4},</math> meaning the points of tangency can be <math>(29,17)</math> or <math>(21,23).</math> Solving the equation for the slope of the perpendicular bisector gives <cmath>\frac{\frac{3}{2}-23}{k+2-21}=\frac{4}{3}</cmath> or <cmath>\frac{\frac{3}{2}-17}{k+2-29}=\frac{4}{3},</cmath> giving <math>k=\frac{23}{8}</math> or <math>\frac{123}{8}</math>, having a sum of <math>\frac{73}{4} \Longrightarrow \boxed{077}.</math>
  
 
~nevergonnagiveup
 
~nevergonnagiveup

Revision as of 19:16, 13 February 2025

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

The complex number $z$ must satisfy the following conditions:

$1.$ 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

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