给定正多边形的三个点（n > 3），找出给定点可能的正多边形（所有边相同）的最小面积。
例子：
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 .在我们继续之前要注意的一件事是边数必须至少为 4（注意 n > 3 条件）。
在这里，我们要找到一个正多边形可能的最小面积，所以要计算最小可能的面积，我们需要计算所需的 n 值。由于没有给出边长，所以我们首先计算由点组成的三角形的外接半径。它由公式给出
R = abc / 4A
其中 a、b、c 是形成的三角形的边，A 是三角形的面积。在这里，三角形的面积可以通过海伦公式计算。
计算三角形的外接半径后，我们通过公式计算多边形的面积
A = nX ( sin(360/n) xr 2 /2 )
这里 r 表示 n 边形（n 边的正多边形）的外接圆半径。
但是，首先我们必须计算 n 的值。要计算 n 我们首先必须通过余弦公式计算三角形的所有角度
cosA = ( b 2 +c 2 -a 2 ) / 2bc
cosB = ( a 2 +c 2 -b 2 ) / 2ac
cosC = ( a 2 +b 2 -c 2 ) / 2ab
然后，n 由下式给出
n = pi / GCD (A , B, C )
其中 A、B 和 C 是三角形的内角。计算 n 后，我们将此值代入计算多边形面积的公式。
以下是给定方法的实现：
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 divyeshrabadiya07Python3
# 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 RyugaC#
// 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输出：
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