1999 AIME Problem 9

Below is the professionally curated solution for Problem 9 of the 1999 AIME, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AIME solutions, or check the answer key.

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Concepts:complex numberalgebraic manipulation

Difficulty rating: 2270

9.

A function ff is defined on the complex numbers by f(z)=(a+bi)z,f(z) = (a + bi)z, where aa and bb are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin. Given that a+bi=8|a + bi| = 8 and that b2=mn,b^2 = \frac{m}{n}, where mm and nn are relatively prime positive integers, find m+n.m + n.

Solution:

The condition is f(z)z=f(z)|f(z) - z| = |f(z)| for all z,z, that is, (a1+bi)z=(a+bi)z.|(a - 1 + bi)z| = |(a + bi)z|. Dividing by z|z| (for z0z \ne 0) gives a1+bi=a+bi,|a - 1 + bi| = |a + bi|, so (a1)2+b2=a2+b2,(a - 1)^2 + b^2 = a^2 + b^2, which forces a=12.a = \frac{1}{2}.

Since a+bi=8,|a + bi| = 8, we have a2+b2=64,a^2 + b^2 = 64, so b2=6414=2554.b^2 = 64 - \frac{1}{4} = \frac{255}{4}. As gcd(255,4)=1,\gcd(255, 4) = 1, the answer is 255+4=259.255 + 4 = 259.

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