2014 AMC 12B Problems/Problem 13

Problem

Real numbers $a$ and $b$ are chosen with $1<a<b$ 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 $b^2 - 3b + 1 = 0$ . Solving for $b$ we get $\frac{3 + \sqrt{5}}{2}$ . So the answer is C. ~coolmath_2018

Solution 4

$1+a \leq b, \quad \frac{1}{b} + \frac{1}{a} \leq 1$

$\implies a \leq b-1, \quad \frac{a+b}{ab} \leq 1$

$\implies a+b \leq ab \implies b \leq ab-a \implies b \leq(b-1)a \implies a \geq \frac{b}{b-1}$

$\therefore \frac{b}{b-1} \leq a \leq b-1$

$\frac{b}{b-1} \leq b-1 \implies b \leq b^2-2b+1 \implies b^2-3b+1 \geq 0$

$b= \frac{3 \pm \sqrt{9-4}}{2}= \frac{3 \pm \sqrt{5}}{2}$

$b=\frac{3 + \sqrt{5}}{2}$

select C

~ Ji Yang

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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