Triangle Inequality

Triangle Inequality Theorem is the theorem used to find the values of the side of a triangle. You can’t just give a triangle with 3 sides without basis because it sometimes can’t form a triangle. I have seen questions for this theorem in PMO, MMC as well as Brilliant.org. The theorem states that the sum of two sides of the triangle is greater than the third side.

I woke up this morning thinking that I can prove this theorem in my own way by construction and a little common sense that everyone can understand.

Proof by Construction:

Given a two sides of triangle with fixed length of a and b, where b>c. We need to find third side a.

Below is an interactive multimedia using geogebra to show this inequality. You can drag the points to follow the construction proof.

Segments b and c have constant length. By constructing circles with radius \overline{AB} and \overline{AC} we will be able to see the paths of points B and C. The circle is dotted since that is just a guide. Think of it as imaginary circle.

The maximum value of a is attained by moving point B counterclockwise as B approaches collinearity with A and C

a=b+c

This equality attained if the three points are collinear making a straight line not a triangle. Thus, to have a triangle a<b+c

The minimum value of is attained by moving point clockwise as B approaches collinearity with A and C. As we move B closer to the value of a is getting smaller and approaching b-a, but can’t be equal to b-a because it will not make a triangle but a straight line. 

Thus, we can conclude the following inequality to get the third side.

\boxed{b-c<a<b+c}

Problem 1: Two sides of the triangle is 5 and 9. What is the range of the third side of the triangle?

Solution:

9-5<third side<9+5

4<third side<14

Problem 2: You are given 2 sticks with length 15 and 17. If you are given a third side with integral length. How many triangles can you make?

Solution:

17-15<third side<17+15

2<third side<32

Since the third side has an integer length, we can have an integral length from 3 to 31. That makes 29 triangles.

Problem 3: Two sides of the triangle are 9 and 10. What is the largest area of triangle that can be made using these two sides?

Solution:

Since we are only given two sides, you might thinking to get the maximum integer length the third side which is 18. That is a misconception. A triangle with sides 9,10,18 maybe big but it’s not a triangle with maximum area.

Using the area of triangle,

A=\displaystyle\frac{absin\theta}{2}

a and b are two sides and \theta is the included angle. Here, a and b are constant. To maximize the area, we just need to maximize sin\theta. Since -1\le sin\theta\le 1. The maximum value is 1 when \theta=90°

Maximum area of triangle

A=\displaystyle\frac{9\cdot 10\cdot 1}{2}

A=45 sq. units

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