2017 AIME I Problem 10

Below is the professionally curated solution for Problem 10 of the 2017 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AIME I solutions, or check the answer key.

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Concepts:complex numberinscribed angleperpendicular bisector

Difficulty rating: 2920

10.

Let z1=18+83i,z_1 = 18 + 83i, z2=18+39i,z_2 = 18 + 39i, and z3=78+99i,z_3 = 78 + 99i, where i=1.i = \sqrt{-1}. Let zz be the unique complex number with the properties that z3z1z2z1zz2zz3\frac{z_3 - z_1}{z_2 - z_1} \cdot \frac{z - z_2}{z - z_3} is a real number and the imaginary part of zz is the greatest possible. Find the real part of z.z.

Solution:

The argument of z3z1z2z1\frac{z_3 - z_1}{z_2 - z_1} is the angle z2z1z3,\angle z_2 z_1 z_3, and the argument of zz2zz3\frac{z - z_2}{z - z_3} is the angle between zz2\overline{zz_2} and zz3.\overline{zz_3}. Their product is real exactly when these angles are equal or supplementary, which by the inscribed angle theorem happens exactly when z1,z_1, z2,z_2, z3,z_3, and zz are concyclic. So zz lies on the circumcircle of z1,z2,z3.z_1, z_2, z_3.

The segment from 18+39i18 + 39i to 18+83i18 + 83i is vertical, so its perpendicular bisector is the horizontal line y=61.y = 61. The segment from z2=18+39iz_2 = 18 + 39i to z3=78+99iz_3 = 78 + 99i has slope 11 and midpoint (48,69),(48, 69), so its perpendicular bisector is y69=(x48).y - 69 = -(x - 48). Setting y=61y = 61 gives x=56,x = 56, so the center is 56+61i.56 + 61i.

The point of the circle with maximal imaginary part is directly above the center, so the real part of zz is 56.56.

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