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
Check the Answer
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)