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## 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)

Categories

## Can we Prove that ……..

The length of any side of a triangle is not more than half of its perimeter

### Key Concepts

Triangle Inequality

Perimeter

Geometry

Answer: Yes we can definitely prove that by Triangle Inequality

Mathematical Circles – Chapter 6 – Inequalities Problem 3

Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg

## Try with Hints

We can start this sum by using this picture below

The length of the three sides of this triangle are a,b and c. So if we apply triangle inequality which implies that the length of one side of a triangle is less than the sum of the lengths of the two sides of that triangle. In reference to the theorem

b + c > a

So can you try to do the rest of the sum ????????

According to the question we have to find the perimeter at first

Perimeter is the sum of the length of all sides of the triangle = a + b + c

And the length of each side is a or b or c.

We have to prove : a + b + c > length of any one side

This can be one of the most important hint for this problem. Try to do the rest of the sum …………………………..

Here is the rest of the sum :

As stated above if we use triangle inequality :

b + c > a

Lets add a to both the sides

a + b + c > a + a

a + b + c > 2 a

The left hand side of the above inequality is the perimeter of this triangle.

perimeter > 2 a

So , $\frac {perimeter}{2} > a$

$\frac {perimeter}{2}$ = semi perimeter

Hence this is proved that the length of one side of a triangle is less than half of its perimeter.