# Problem Link:

**author**: Tasnim Imran Sunny

**tester**: Roman Rubanenko

**editorialist**: Utkarsh Lath

# Difficulty:

Easy

# Pre-requisites:

High School Maths

# Problem:

Given n triangles in cartesian plane, count the number of right triangles

# Explanation:

Given a triangle, you have to tell if it is a right triangle. The author and tester used following property of right triangles.

C

^{2}= A^{2}+ B^{2}

Where C is the length of longest side, A and B are length of other two sides.

```
int dist_square(point p, point q)
return (p.x-q.x)*(p.x-q.x) + (p.y-q.y)*(p.y-q.y)
bool is_right_triangle(point p, point q, point r)
c = dist_square(p, q)
b = dist_square(p, r)
a = dist_square(q, r)
return 2*max(a, b, c) == a+b+c
```

The co-ordinates were kept very small so that there are no integer overflow errors.

# Other Solution

Editorialistâ€™s solution was based on the fact that dot product of two non zero vectors is zero iff they are orthogonal.

```
point operator-(point a, point b)
return point(a.x-b.x, a.y-b.y)
int dot(point p, point q)
return p.x * q.x + p.y * q.y
bool is_right_triangle(point p, point q, point r)
point v1 = p-q
point v2 = q-r
point v3 = r-p
return dot(v1, v2) == 0 or dot(v1, v3) == 0 or dot(v2, v3) == 0
```

# Solutions:

Setterâ€™s solution

Testerâ€™s Solution

Editorialistâ€™s Solution