2014 AMC 12B Problem 13

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

With

1 \lt a \lt b,

no triangle

1, a, b

means

b \ge a+1

No triangle

\tfrac1b, \tfrac1a, 1

means

1 \ge \tfrac1a + \tfrac1b,

i.e.

a \le \tfrac{b}{b-1}

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.

Problem 13 in other years