# 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

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 $$b^2-3b+1=0 \implies b = \boxed{\textbf{(C)} \ \frac{3+\sqrt{5}}{2}}$$

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