# 2014 AMC 12B Problems/Problem 13

## Problem

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

$\textbf{(A)}\ \frac{3+\sqrt{3}}{2}\qquad\textbf{(B)}\ \frac{5}{2}\qquad\textbf{(C)}\ \frac{3+\sqrt{5}}{2}\qquad\textbf{(D)}\ \frac{3+\sqrt{6}}{2}\qquad\textbf{(E)}\ 3$

## Solution 1

Notice that $1>\frac{1}{a}>\frac{1}{b}$. Using the triangle inequality, we find $$a+1 > b \implies a>b-1$$ $$\frac{1}{a}+\frac{1}{b} > 1$$ In order for us the find the lowest possible value for $b$, we try to create two degenerate triangles where the sum of the smallest two sides equals the largest side. Thus we get $$a=b-1$$ and $$\frac{1}{a} + \frac{1}{b}=1$$ Substituting, we get $$\frac{1}{b-1}+\frac{1}{b}=\frac{b+b-1}{b(b-1)}=1$$ $$\frac{2b-1}{b(b-1)} = 1$$ $$2b-1=b^2-b$$ Solving for $b$ using the quadratic equation, we get $$b^2-3b+1=0 \implies b = \boxed{\textbf{(C)} \ \frac{3+\sqrt{5}}{2}}$$

## Solution 2 (similar to Solution 1)

We can form degenerate triangles from the given information. Since $b>a>1$, we know that $a,b,1$ must satisfy $a + 1 \le b$ to be degenerate in the first scenario.

Similarly, since $\frac{1}{n} + 1 > 1$ for all positive integers $n$, the only condition that is guaranteed to form a degenerate triangle from the second instance is $\frac{1}{a} + \frac{1}{b} \le 1$.

Solving the second inequality for $a$ in terms of $b$ (for more-easily apparent inequality manipulation) yields $b\ge \frac{a}{a-1}$. Notice that $b> a+1 \ge \frac{a}{a-1}$ for almost all positive real numbers $a>1$ (see note at the end).

Thus, it suffices to solve the inequality $a+1 \ge \frac{a}{a-1}$ for $a$ whose only positive solution is $a \ge \frac{1 + \sqrt{5}}{2}$.

Substituting back into the first inequality yields $\boxed{b \ge \frac{3 + \sqrt{5}}{2}}$ and thus the answer is $\text{C}$.

Note: this solution assumes that $a$ is not less than $1.618$, the points at which the last inequality.

~baldeagle123

## Solution 3

By the triangle inequality we have $a+b \ge 1$, $a + 1 \ge b$, $b+1 \ge a$ and $b(1+a) \ge a$, $a(1+b) \ge b$, $a+b \ge ab$. Clearly a triangle can't have negative area so, our triangle must have $0$ area which means equality must be obtained for all the inequalities above. Since $a = 1+b$ we know that $(1+b)^2 = b$ which simplifies to $b^2 - 3b + 1$. Solving for $b$ we get $\frac{3 + \sqrt{5}}{2}$. So the answer is C. ~coolmath_2018