The sides of triangles

You would have often seen the the equation which states that any given side in a triangle is shorter than the sum of other two sides. Have you ever wondered why ?
The answer is simple:
Consider a straight line AC with B being a point on AC. Now, to make this a triangle let us bend AB and BC around B and join AC. Clearly AC will now be shorter than what it was earlier ie. , AC is shorter than AB+BC. So basically the triangle whose one side is equal to the other two will have and area = 0.

Lets try some problems to learn this better.
Problem: What is number of distinct triangles with integral valued sides and perimeter as 14 ?
a) 6 b) 5 c) 4 d) 3 [ CAT 2000 ]
This is a beautiful CAT problem testing both logical thinking and conceptual knowledge. Let us see the approach to this problem :
If a,b,c are sides of such a triangle then given that a+b+c =14.
The first thing to infer here is that any of the sides cannot be 7 or more as the sum of the other two sides should be larger than the third side. Therefore the largest side of such a triangle is 6. Let the smallest side be 1. So the other two sides add up to 13.
Cleary this is not possible as no side is 7 or more.
Let the smallest side be 2.
We can now have a triangle with sides (2,6,6). No other triangle is possible with smallest side 2.
Let the smallest side be 3.
We can now have a triangle with sides (3,5,6). No other triangle is possible with smallest side 3.
Let the smallest siden now have be 4.
We can now have a triangle with sides (4,5,5) & (4,4,6). No other triangle is possible with smallest side 4.

No other triangles are possible with smallest sides 5 or 6. Therefore a total of four triangles are possible with integer sides and perimeter 14.

Problem: There are 10 points in a plane. Except for 4 points , which are collinear, no 3 points are in a straight line. The number of triangles that can be formed with vertices as these points are
a) 120 b) 80 c) 119 d) 116
This is a problem of permutations involving simple concept of triangles. The idea of giving this problem here is to emphasise the fact that no topic can be viewed in isolation in CAT and concepts from different topics are often combined to create problrms that appear in CAT.
The appoach here is to identify that any 3 points can form a triangle.
Therefore 10 points will give 10C3 = 10x9x8 ∕ 6 = 120 triangles.

Now, the 4 points can form 4C3 = 4 triangles.
But 4 collinear points cannot form any triangle.So we have counted 4 triangles extra. Therefore the number of triangles are 120 – 4 = 116 triangles.