# Minimum area of a Polygon with three points given

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 ` `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 `

Output:

```1.00

```

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