2018 AIME II Problem 6

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Concepts:polynomialfactoringquadraticgeometric probability

Difficulty rating: 2510

6.

A real number aa is chosen randomly and uniformly from the interval [20,18].[-20, 18]. The probability that the roots of the polynomial x4+2ax3+(2a2)x2+(4a+3)x2x^4 + 2ax^3 + (2a - 2)x^2 + (-4a + 3)x - 2 are all real can be written in the form mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Solution:

Group the terms by whether they involve a:a: (x42x2+3x2)+2a(x3+x22x)=(x1)(x+2)(x2x+1)+2ax(x1)(x+2),(x^4 - 2x^2 + 3x - 2) + 2a(x^3 + x^2 - 2x) = (x - 1)(x + 2)(x^2 - x + 1) + 2ax(x - 1)(x + 2), so the polynomial factors as (x1)(x+2)(x2+(2a1)x+1).(x - 1)(x + 2)\left(x^2 + (2a - 1)x + 1\right).

All four roots are real exactly when the quadratic factor has real roots, i.e. when (2a1)240,(2a - 1)^2 - 4 \ge 0, which means a12a \le -\frac{1}{2} or a32.a \ge \frac{3}{2}. The excluded interval (12,32)\left(-\frac{1}{2}, \frac{3}{2}\right) has length 22 inside [20,18],[-20, 18], which has length 38,38, so the probability is 3638=1819.\frac{36}{38} = \frac{18}{19}. The requested sum is 18+19=37.18 + 19 = 37.

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