2025 AIME I Problem 8

Below is the professionally curated solution for Problem 8 of the 2025 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AIME I solutions, or check the answer key.

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Concepts:complex numberperpendicular bisectortangent line

Difficulty rating: 2560

8.

Let kk be a real number such that the system 25+20iz=5|25 + 20i - z| = 5 z4k=z3ik|z - 4 - k| = |z - 3i - k| has exactly one complex solution z.z. The sum of all possible values of kk can be written as mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n. Here i=1.i = \sqrt{-1}.

Solution:

The first equation says zz lies on the circle of radius 55 centered at (25,20).(25, 20). The second says zz is equidistant from P1=(k+4,0)P_1 = (k + 4, 0) and P2=(k,3),P_2 = (k, 3), i.e. it lies on the perpendicular bisector of P1P2.\overline{P_1 P_2}. The system has exactly one solution precisely when this line is tangent to the circle.

The midpoint is (k+2,32)\left(k + 2, \frac{3}{2}\right) and P1P2\overline{P_1 P_2} has slope 34,-\frac{3}{4}, so the bisector has slope 43:\frac{4}{3}: in standard form 8x6y(8k+7)=0.8x - 6y - (8k + 7) = 0. Tangency requires 8256208k782+62=738k10=5,\frac{|8 \cdot 25 - 6 \cdot 20 - 8k - 7|}{\sqrt{8^2 + 6^2}} = \frac{|73 - 8k|}{10} = 5, so 8k=73±50,8k = 73 \pm 50, giving k=1238k = \frac{123}{8} or k=238.k = \frac{23}{8}.

The sum is 1468=734,\frac{146}{8} = \frac{73}{4}, so m+n=73+4=77.m + n = 73 + 4 = 77.

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