Complex Number Analysis
Let $ z_1 = 18 + 83i $, $ z_2 = 18 + 39i, $ and $ z_3 = 78 + 99i, $ where $ i^2 = -1 $. Let $ z $ be the unique complex number with the properties that $ \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 $ z $ is the greatest possible. Find the real part of $ z $.
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Solution
Let $ z = a + bi, $ where $ a $ and $ b $ are real numbers. Then
\begin{align*}
\frac{z_3 - z_1}{z_2 - z_1} \cdot \frac{z - z_2}{z - z_3} &= \frac{60 + 16i}{-44i} \cdot \frac{(a - 18) + (b - 39)i}{(a - 78) + (b - 99)i} \\
&= \frac{-4 + 15i}{11} \cdot \frac{[(a - 18) + (b - 39)i][(a - 78) - (b - 99)i]}{(a - 78)^2 + (b - 99)^2}.\end{align*}This expression is real if and only if the imaginary part is 0. In other words,
\[(-4 + 15i)[(a - 18) + (b - 39)i][(a - 78) - (b - 99)i]\]has imaginary part 0. In turn this is equivalent to
\[(-4)(-(a - 18)(b - 99) + (a - 78)(b - 39)) + 15((a - 18)(a - 78) + (b - 39)(b - 99)) = 0.\]This simplifies to $ a^2 - 112a + b^2 - 122b + 4929 = 0 $. Completing the square, we get
\[(a - 56)^2 + (b - 61)^2 = 1928,\]so
\[(a - 56)^2 = 1928 - (b - 61)^2.\]When $ b $ is maximized, the right-hand side is 0, and $ a = \boxed{56} $.
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