How to check if 3 sides form a triangle in C++

Question

I am trying to check if 3 sides form a triangle in C++, but the answer for all possible numbers I tried it says wrong...

#include <iostream>
#include <cmath>

using namespace std;

int main()
{
    int a, b, c;

    cin >> a >> b >> c;

    if (pow(a,2) == pow(b,2) * pow(c,2) || pow(b,2) == pow(a,2) * pow(c,2) || pow(c,2) == pow(a,2) * pow(b,2))
        cout << "The sides form a triangle" << endl;
    else
        cout << "The sides do not form a triangle." << endl;
    return 0;
}

Answer 1

An efficient approach will be to sort the given sides. This will be efficient if you are given an entire array and you are asked whether the given array elements form a triangle or not. This can be applied for n number of given sides. However, this can also be applied for 3 sides. Suppose the given array is b. In your case array b is of length h=3.

sort(b,b+h);
for (int j=0;j<(h-2);j++){
    if (b[j]+b[j+1]>b[j+2])
    {
return true;
    }
}
else {
return false;
}

Answer 2

Let's say that a, b, c is the sides of the triangle. Therefore, it must be satisfy this criteria :

1. a + b > c
2. a + c > b
3. b + c > a

All the criteria must be true. If one of them are false, then a, b, c will not create the triangle.

#include <iostream>
#include <cmath>

using namespace std;

int main()
{
    int a, b, c;
    cin >> a >> b >> c;
    // check whether a, b, c can form a triangle
    if (a+b > c && a+c > b && b+c > a)
        cout << "The sides form a triangle" << endl;
    else
        cout << "The sides do not form a triangle." << endl;
    return 0;
}

Answer 3

Triangle conditions to check for,

(a + b > c),
(b + c > a),
(c + a > b)

Answer 4

Assuming you are only testing for right angled triangles then the logic to use is z^2 = x^2 + y+2 So there's a mistake in the logic:

 if (pow(a,2) == pow(b,2) * pow(c,2) || pow(b,2) == pow(a,2) * pow(c,2) || pow(c,2) == pow(a,2) * pow(b,2))

This should be:

 if (pow(a,2) == pow(b,2) + pow(c,2) || pow(b,2) == pow(a,2) + pow(c,2) || pow(c,2) == pow(a,2) + pow(b,2))

But even with this change the result might be might wrong due to testing equality on floating point numbers. Make a specific function to test 2 floating point numbers are close enough given some tolerance you decide on then use that for comparisons.

If you do not want to limit your approach to only right angled triangles then you might wish to read up on the triangle inequality. In summary the triangle inequality just states that the length of any edge in a triangle must be smaller than the sum of the other 2 edges.

Answer 5

For a normal triangle

1. sum of any two sides is greater than third side (or)
2. difference of any two sides is less than third side

hint :  a+b > c || ...

For a right angled triangle

1) sum of the squares of two sides equals the square of the longest side

Hint:

Find the longest side of three sides, that is find longest number in the three..
square the remaining two nums, add them and equate it to square of longest number

Answer 6

Actually, given any three sides, you only need to check one condition: that the longest side, say c, is less than the sum of the two shorter sides, say a and b. That is,

if c < a+b {
   return true;
} else return false;

This is the essence of the triangle inequality theorem. The other conditions will be trivially true when it is a triangle and irrelevant if this one condition is not. The key, of course, will be to sort the three sides to find the longest side by using a simple sorting algorithm. The assumption in the code is that the sides have already been sorted so that c is the longest.