Difference between revisions of "2025 AIME I Problems/Problem 8"
(→Solution 1) |
(→Solution 1) |
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Line 22: | Line 22: | ||
label("$(k,3)$", (3,3), N); | label("$(k,3)$", (3,3), N); | ||
draw(rightanglemark((3,3),(5,3/2),(21,23), 20)); | 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)); | draw((25,20) -- (21,23)); | ||
</asy> | </asy> |
Revision as of 21:18, 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 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
or
giving
or
, having a sum of
~nevergonnagiveup
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.