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Minimum area of a Polygon with three points given
  • Difficulty Level : Hard
  • Last Updated : 03 Dec, 2020

Given three points of a regular polygon(n > 3), find the minimum area of a regular polygon (all sides same) possible with the points given.
Examples: 

Input : 0.00 0.00
        1.00 1.00
        0.00 1.00
Output : 1.00
By taking point (1.00, 0.00) square is 
formed of side 1.0 so area = 1.00 .

One thing to note in question before we proceed is that the number of sides must be at least 4 (note n > 3 condition)..
Here, we have to find the minimum area possible for a regular polygon, so to calculate the minimum possible area, we need to calculate the required value of n. As the side length is not given, so we first calculate the circumradius of the triangle formed by the points. It is given by the formula 
R = abc / 4A 
Where a, b, c are the sides of the triangle formed and A is the area of the triangle. Here, the area of the triangle can be calculated by Heron’s Formula.
After calculating circumradius of the triangle, we calculate the area of the polygon by the formula 
A = nX ( sin(360/n) xr2 /2 )
Here r represents the circumradius of n-gon (regular polygon of n sides). 
But, first we have to calculate value of n . To calculate n we first have to calculate all the angles of triangle by the cosine formula 
cosA = ( b2+c2-a2 ) / 2bc 
cosB = ( a2+c2-b2 ) / 2ac 
cosC = ( a2+b2-c2 ) / 2ab 
Then, n is given by 
n = pi / GCD (A , B, C ) 
Where A, B and C are the angles of the triangle . After calculating n we substitute this value to the formula for calculating area of polygon .
Below is the implementation of the given approach :
 

C++




// CPP program to find minimum area of polygon of
// number of sides more than three with given three points.
#include <bits/stdc++.h>
using namespace std;
 
// assigning pi value to variable
const double pi = 3.14159265359;
 
// calculating gcd value of two double values .
double gcd(double x, double y)
{
    return fabs(y) < 1e-4 ? x : gcd(y, fmod(x, y));
}
 
// Calculating minimum area of polygon through this function .
double min_area_of_polygon(double Ax, double Ay, double Bx,
                            double By, double Cx, double Cy)
{
    double a, b, c, Radius, Angle_A, Angle_B, Angle_C,
                              semiperimeter, n, area;
 
    // calculating the length of the sides of the triangle
    // formed from given points a, b, c represents the
    // length of different sides of triangle .
    a = sqrt((Bx - Cx) * (Bx - Cx) + (By - Cy) * (By - Cy));
    b = sqrt((Ax - Cx) * (Ax - Cx) + (Ay - Cy) * (Ay - Cy));
    c = sqrt((Ax - Bx) * (Ax - Bx) + (Ay - By) * (Ay - By));
 
    // here we have calculated the semiperimeter of a triangle .
    semiperimeter = (a + b + c) / 2;
 
    // Now from the semiperimeter area of triangle is derived
    // through the heron's formula .
    double area_triangle = sqrt(semiperimeter * (semiperimeter - a)
                                * (semiperimeter - b)
                                * (semiperimeter - c));
 
    // thus circumradius of the triangle is derived from the
    // sides and area of the triangle calculated .
    Radius = (a * b * c) / (4 * area_triangle);
 
    // Now each angle of the triangle is derived from the sides
    // of the triangle .
    Angle_A = acos((b * b + c * c - a * a) / (2 * b * c));
    Angle_B = acos((a * a + c * c - b * b) / (2 * a * c));
    Angle_C = acos((b * b + a * a - c * c) / (2 * b * a));
 
    // Now n is calculated such that area is minimum for
    // the regular n-gon .
    n = pi / gcd(gcd(Angle_A, Angle_B), Angle_C);
 
    // calculating area of regular n-gon through the circumradius
    // of the triangle .
    area = (n * Radius * Radius * sin((2 * pi) / n)) / 2;
 
    return area;
}
 
int main()
{
    // three points are given as input .
    double Ax, Ay, Bx, By, Cx, Cy;
    Ax = 0.00;
    Ay = 0.00;
    Bx = 1.00;
    By = 1.00;
    Cx = 0.00;
    Cy = 1.00;
 
    printf("%.2f", min_area_of_polygon(Ax, Ay, Bx, By, Cx, Cy));
    return 0;
}

Java




// Java program to find minimum
// area of polygon of number of
// sides more than three with
// given three points.
class GFG{
 
// Assigning pi value to variable
static double pi = 3.14159265359;
       
public static double fmod(double a,
                          double b)
{
  int result = (int) Math.floor(a / b);
  return a - result * b;
}
     
// calculating gcd value of
// two double values .
public static double gcd(double x,
                         double y)
{
  return Math.abs(y) < 1e-4 ? x :
         gcd(y, fmod(x, y));
}
       
// Calculating minimum area of polygon through this function .
public static double min_area_of_polygon(double Ax, double Ay,
                                         double Bx, double By,
                                         double Cx, double Cy)
{
  double a, b, c, Radius, Angle_A, Angle_B, Angle_C, 
         semiperimeter, n, area;
 
  // Calculating the length of the sides
  // of the triangle formed from given
  /// points a, b, c represents the 
  // length of different sides of triangle .
  a = Math.sqrt((Bx - Cx) * (Bx - Cx) +
                (By - Cy) * (By - Cy));
  b = Math.sqrt((Ax - Cx) * (Ax - Cx) +
                (Ay - Cy) * (Ay - Cy));
  c = Math.sqrt((Ax - Bx) * (Ax - Bx) +
                (Ay - By) * (Ay - By));
 
  // Here we have calculated the
  // semiperimeter of a triangle .
  semiperimeter = (a + b + c) / 2;
 
  // Now from the semiperimeter area
  // of triangle is derived
  // through the heron's formula .
  double area_triangle = Math.sqrt(semiperimeter *
                                  (semiperimeter - a) *
                                  (semiperimeter - b) *
                                  (semiperimeter - c));
 
  // Thus circumradius of the triangle
  // is derived from the sides and
  // area of the triangle calculated .
  Radius = (a * b * c) / (4 * area_triangle);
 
  // Now each angle of the triangle
  // is derived from the sides
  // of the triangle .
  Angle_A = Math.acos((b * b + c * c - a * a) /
                      (2 * b * c));
  Angle_B = Math.acos((a * a + c * c - b * b) /
                      (2 * a * c));
  Angle_C = Math.acos((b * b + a * a - c * c) /
                      (2 * b * a));
 
  // Now n is calculated such that
  // area is minimum for the regular n-gon .
  n = pi / gcd(gcd(Angle_A, Angle_B), Angle_C);
 
  // calculating area of regular n-gon
  // through the circumradius of the triangle .
  area = (n * Radius * Radius *
          Math.sin((2 * pi) / n)) / 2;
 
  return area;
}
   
// Driver code
public static void main(String[] args)
{
  // Three points are given as input .
  double Ax, Ay, Bx, By, Cx, Cy;
  Ax = 0.00;
  Ay = 0.00;
  Bx = 1.00;
  By = 1.00;
  Cx = 0.00;
  Cy = 1.00;
  System.out.println(String.format("%.2f",
                     min_area_of_polygon(Ax, Ay,
                                         Bx, By,
                                         Cx, Cy)));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3




# Python3 program to find minimum area of
# polygon of number of sides more than three
# with given three points.
 
# from math lib import every function
from math import *
 
# assigning pi value to variable
pi = 3.14159265359
 
# calculating gcd value of two double values .
def gcd(x, y) :
 
    if abs(y) < 1e-4 :
        return x
    else :
        return gcd(y, fmod(x, y))
 
 
# Calculating minimum area of polygon
# through this function .
def min_area_of_polygon(Ax, Ay, Bx,
                        By, Cx, Cy) :
 
    # calculating the length of the sides of
    # the triangle formed from given points
    # a, b, c represents the length of different
    # sides of triangle
    a = sqrt((Bx - Cx) * (Bx - Cx) +
             (By - Cy) * (By - Cy))
    b = sqrt((Ax - Cx) * (Ax - Cx) +
             (Ay - Cy) * (Ay - Cy))
    c = sqrt((Ax - Bx) * (Ax - Bx) +
             (Ay - By) * (Ay - By))
 
    # here we have calculated the semiperimeter
    # of a triangle .
    semiperimeter = (a + b + c) / 2
 
    # Now from the semiperimeter area of triangle
    # is derived through the heron's formula
    area_triangle = sqrt(semiperimeter *
                        (semiperimeter - a) *
                        (semiperimeter - b) *
                        (semiperimeter - c))
 
    # thus circumradius of the triangle is derived
    # from the sides and area of the triangle calculated
    Radius = (a * b * c) / (4 * area_triangle)
 
    # Now each angle of the triangle is derived
    # from the sides of the triangle
    Angle_A = acos((b * b + c * c - a * a) / (2 * b * c))
    Angle_B = acos((a * a + c * c - b * b) / (2 * a * c))
    Angle_C = acos((b * b + a * a - c * c) / (2 * b * a))
 
    # Now n is calculated such that area is
    # minimum for the regular n-gon
    n = pi / gcd(gcd(Angle_A, Angle_B), Angle_C)
 
    # calculating area of regular n-gon through
    # the circumradius of the triangle
    area = (n * Radius * Radius *
            sin((2 * pi) / n)) / 2
 
    return area
 
# Driver Code
if __name__ == "__main__" :
 
    # three points are given as input .
    Ax = 0.00
    Ay = 0.00
    Bx = 1.00
    By = 1.00
    Cx = 0.00
    Cy = 1.00
 
    print(round(min_area_of_polygon(Ax, Ay, Bx,
                                    By, Cx, Cy), 1))
 
# This code is contributed by Ryuga

C#




// C# program to find minimum
// area of polygon of number of
// sides more than three with
// given three points. 
using System;
using System.Collections.Generic;
 
class GFG
{
     
    // Assigning pi value to variable
    static double pi = 3.14159265359;
            
    static double fmod(double a, double b)
    {
      int result = (int) Math.Floor(a / b);
      return a - result * b;
    }
          
    // calculating gcd value of
    // two double values .
    static double gcd(double x, double y)
    {
      return Math.Abs(y) < 1e-4 ? x : gcd(y, fmod(x, y));
    }
            
    // Calculating minimum area of polygon through this function .
    static double min_area_of_polygon(double Ax, double Ay,
                                             double Bx, double By,
                                             double Cx, double Cy)
    {
      double a, b, c, Radius, Angle_A, Angle_B, Angle_C, 
             semiperimeter, n, area;
      
      // Calculating the length of the sides
      // of the triangle formed from given
      /// points a, b, c represents the 
      // length of different sides of triangle .
      a = Math.Sqrt((Bx - Cx) * (Bx - Cx) +
                    (By - Cy) * (By - Cy));
      b = Math.Sqrt((Ax - Cx) * (Ax - Cx) +
                    (Ay - Cy) * (Ay - Cy));
      c = Math.Sqrt((Ax - Bx) * (Ax - Bx) +
                    (Ay - By) * (Ay - By));
      
      // Here we have calculated the
      // semiperimeter of a triangle .
      semiperimeter = (a + b + c) / 2;
      
      // Now from the semiperimeter area
      // of triangle is derived
      // through the heron's formula .
      double area_triangle = Math.Sqrt(semiperimeter *
                                      (semiperimeter - a) *
                                      (semiperimeter - b) *
                                      (semiperimeter - c));
      
      // Thus circumradius of the triangle
      // is derived from the sides and
      // area of the triangle calculated .
      Radius = (a * b * c) / (4 * area_triangle);
      
      // Now each angle of the triangle
      // is derived from the sides
      // of the triangle .
      Angle_A = Math.Acos((b * b + c * c - a * a) /
                          (2 * b * c));
      Angle_B = Math.Acos((a * a + c * c - b * b) /
                          (2 * a * c));
      Angle_C = Math.Acos((b * b + a * a - c * c) /
                          (2 * b * a));
      
      // Now n is calculated such that
      // area is minimum for the regular n-gon .
      n = pi / gcd(gcd(Angle_A, Angle_B), Angle_C);
      
      // calculating area of regular n-gon
      // through the circumradius of the triangle .
      area = (n * Radius * Radius *
              Math.Sin((2 * pi) / n)) / 2;
      
      return area;
    }  
   
  // Driver code
  static void Main()
  {
      // Three points are given as input .
      double Ax, Ay, Bx, By, Cx, Cy;
      Ax = 0.00;
      Ay = 0.00;
      Bx = 1.00;
      By = 1.00;
      Cx = 0.00;
      Cy = 1.00;
      Console.WriteLine(String.Format("{0:0.00}", min_area_of_polygon(Ax, Ay,
                                             Bx, By,
                                             Cx, Cy)));
  }
}
 
// This code is contributed by divyeshrabadiya07

Output: 
 

1.00



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