Question

Real numbers $a$ and $b$ are chosen with $1 \lt a \lt b$ such that no triangle with positive area has side lengths $1, a,$ and $b$ or $\tfrac1b, \tfrac1a,$ and $1.$ What is the smallest possible value of $b?$

$\dfrac{3+\sqrt{3}}{2}$

$\dfrac{5}{2}$

$\dfrac{3+\sqrt{5}}{2}$

$\dfrac{3+\sqrt{6}}{2}$

$3$

Answer: C

Concepts:triangle inequality quadratic bounding to limit cases

Difficulty rating: 1870

Solution:

Since $b$ is the largest of $1, a, b,$ no such triangle exists exactly when $b \ge a+1.$ Since $1$ is the largest of $\tfrac1b, \tfrac1a, 1,$ no such triangle exists exactly when $1 \ge \tfrac1a + \tfrac1b,$ that is $a \le \tfrac{b}{b-1}.$

Both conditions hold with $b$ smallest when $a+1 = b$ and $a = \tfrac{b}{b-1}$ meet, giving $b - 1 = \tfrac{b}{b-1},$ or $b^2 - 3b + 1 = 0.$

The root larger than $1$ is $b = \dfrac{3+\sqrt5}{2}.$

Thus, the correct answer is C.

2014 AMC 12B Problem 13

Answer: C

Difficulty rating: 1870

Solution:

Problem 13 in Other Years