# Triangle Inequality Problem - AMC 12B, 2014 - Problem 13

Try this beautiful problem from AMC 12 based on Triangle inequality problem.

## Problem - Triangle Inequality

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?

• $\frac {3+\sqrt 3}{2}$
• $\frac {5}{2}$
• $\frac {3+\sqrt 5}{2}$
• $\frac {3+\sqrt 6}{2}$

### Key Concepts

Triangle Inequality

Inequality

Geometry

Answer: $\frac {3+\sqrt 5}{2}$

American Mathematics Competition - 12B ,2014, Problem Number - 13

Secrets in Inequalities.

## Try with Hints

Here is the first hint where you can start this sum:

It is given $1 >\frac {1}{a} > \frac {1}{b }$ . Use Triangle Inequality here :

a+1>b

a>b-1

$\frac {1}{a} + \frac {1}{b} >1$

If we want to find the lowest possible value of b , we create 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

Now try to do the rest of the sum.......................

We already know $\frac {1}{a} + \frac {1}{b} = 1$

After substituting we will get :

$\frac {1}{b - 1} + \frac {1}{b} = \frac {b+b-1}{b(b-1)} = 1$

$\frac {2b - 1}{b(b-1)} = 1$

Now do the rest of the calculation .............................

Here is the rest of steps to check your problem :

$2b - 1 = b^2 - b$

Now Solving for b using the quadratic equation, we get

$b^2 - 3b + 1 = 0$

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

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