Difference between revisions of "2014 AMC 12B Problems/Problem 13"

Revision as of 21:37, 20 February 2014

Problem

Real numbers $a$ and $b$ are chosen with $1<a<b$ such that no triangles 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 $b$ ?

$\textbf{(A)}\ \frac{3+\sqrt{3}}{2}\qquad\textbf{(B)}\ \frac{5}{2}\qquad\textbf{(C)}\ \frac{3+\sqrt{5}}{2}\qquad\textbf{(D)}}\ \frac{3+\sqrt{6}}{2}\qquad\textbf{(E)}\ 3$ (Error compiling LaTeX. Unknown error_msg)

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 $\frac{2b-1}{b(b-1)} = 1$ Solving for $b$ using the quadratic equation, we get $b^2-3b+1=0 \implies b = \boxed{\textbf{(C)} \ \frac{3+\sqrt{5}}{2}}$

@@ Line 15: / Line 15: @@
 Solving for <math>b</math> using the quadratic equation, we get
 <cmath>b^2-3b+1=0 \implies b = \boxed{\textbf{(C)} \ \frac{3+\sqrt{5}}{2}}</cmath>
-(Solution by kevin38017)

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Difference between revisions of "2014 AMC 12B Problems/Problem 13"

Revision as of 21:37, 20 February 2014

Problem

Solution