Difference between revisions of "2025 AIME I Problems/Problem 8"
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==Solution 1== | ==Solution 1== | ||
| + | The complex number <math>z</math> must satisfy the conditions: | ||
| + | 1. The magnitude between <math>z</math> and <math>(25,20)</math> is <math>5.</math> This can be represented by drawing a circle. | ||
| + | 2. It is equidistant from the points <math>(4+k,0)</math> and <math>(k,3)</math> | ||
| + | |||
| + | For <math>z</math> 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, <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 <math>k=\frac{23}{8}</math> or <math>k=\frac{123}{8},</math> giving a sum of <math>\frac{73}{4} \Longrightarrow \boxed{077}.</math> | ||
Revision as of 18:59, 13 February 2025
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
The complex number must satisfy the conditions:
1. The magnitude between and 
is 
This can be represented by drawing a circle.
2. It is equidistant from the points 
and 
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 
or 
giving a sum of 
