Question

Given $f(z) = z^2 - 19z,$ there are complex numbers $z$ with the property that $z,$ $f(z),$ and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z).$ There are positive integers $m$ and $n$ such that one such value of $z$ is $m + \sqrt{n} + 11i.$ Find $m + n.$

答案

Solution:

Since $f(w) - w = w(w - 20),$ the two legs at $f(z)$ are $f(z) - z = z(z - 20), \qquad f(f(z)) - f(z) = f(z)\bigl(f(z) - 20\bigr) = z(z - 19)(z - 20)(z + 1),$ using $f(z) = z(z - 19)$ and $f(z) - 20 = (z - 20)(z + 1).$ They are perpendicular exactly when their quotient $(z - 19)(z + 1)$ is purely imaginary and nonzero.

Write $z = x + 11i.$ The real part of $(z - 19)(z + 1) = z^2 - 18z - 19$ is $x^2 - 121 - 18x - 19,$ so we need $x^2 - 18x - 140 = 0,$ giving $x = 9 \pm \sqrt{221}.$ The form $m + \sqrt{n} + 11i$ with $m, n$ positive integers requires $x = 9 + \sqrt{221}.$

Hence $m + n = 9 + 221 = 230.$

2019 AIME I Problem 12

Solution:

Problem 12 in Other Years