Skip to content
Related Articles

Related Articles

Find three vertices in an N-sided regular polygon such that the angle subtended at a vertex by two other vertices is closest to A

View Discussion
Improve Article
Save Article
  • Last Updated : 09 Jul, 2021
View Discussion
Improve Article
Save Article

Given an N-sided regular convex polygon and an angle A in degrees, the task is to find any 3 vertices i, j, and k, such that the angle subtended at vertex j by vertex i and k is closest to A

Note: Each vertex is numbered from 1 to N in an anticlockwise direction starting from any vertex.

Examples:

Input: N = 3, A = 15
Output: 2 1 3
Explanation:
The given polygon is a triangle, therefore the minimum angle that can be subtended on any vertex is equal to 30 which is closest to A and includes all the vertices of the polygon.

Input: N = 4, A = 90
Output: 2 1 4

Approach: The given problem can be solved based on the following observations:

  • The vertices of a regular polygon lie on a circle.
  • The central angle of n sided regular polygon is 360/N.
  • The angle subtended on the center of the circle by vertices separated by X edges is equals to (360*X)/N.
  • By the theorem, the angle subtended on the circum-circle by an arc is half of the angle subtended at the center by the same arc. Therefore, the angle subtended on a vertex by an arc formed by X edges would be equal to (180*X)/N.
  • As the polygon is regular therefore the same angle can be subtended by taking any 2 vertices of edge difference equal to X.

Follow the steps below to solve the problem:

  • Initialize two variables, say ans as 0 and minElement as INT_MAX to store the count of edges which subtends the angle closest to A and to store the closest angle respectively.
  • Iterate over the range [1, N – 2] using the variable i and perform the following steps:
    • Find the angle subtended on circum-circle by an arc formed by i edges using the above formula and store it in a variable, angle.
    • Check if the absolute value of (A – angle) is less than the absolute value of (minElement – A), then update the value of i to ans and the value of angle to minElement.
  • After completing the above steps, print the vertices {2, 1, 2 + ans} as the resultant vertices.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
#define ll long long
#define MAXN 1000000
 
// Function to find three vertices that
// subtends an angle closest to A
void closestsAngle(int N, int A)
{
    // Stores the closest angle to A
    double mi = INT_MAX;
 
    // Stores the count of edge which
    // subtend an angle of A
    int ans = 0;
 
    // Iterate in the range [1, N-2]
    for (int i = 1; i < N - 1; i++) {
 
        // Stores the angle subtended
        double angle = 180.0 * i / N;
 
        // If absolute(angle-A) is less
        // than absolute(mi-A)
        if (fabs(angle - A) < fabs(mi - A)) {
 
            // Update angle to mi, and
            // also update i to ans
            mi = angle;
            ans = i;
        }
    }
 
    // Print the vertices
    cout << 2 << ' ' << 1 << ' '
         << 2 + ans;
}
 
// Driver Code
int main()
{
    int N = 3, A = 15;
    closestsAngle(N, A);
 
    return 0;
}

Java




// Java program for the above approach
 
import java.io.*;
 
class GFG {
 
    // Function to find three vertices that
    // subtends an angle closest to A
    static void closestsAngle(int N, int A)
    {
        // Stores the closest angle to A
        double mi = Integer.MAX_VALUE;
 
        // Stores the count of edge which
        // subtend an angle of A
        int ans = 0;
 
        // Iterate in the range [1, N-2]
        for (int i = 1; i < N - 1; i++) {
 
            // Stores the angle subtended
            double angle = 180.0 * i / N;
 
            // If absolute(angle-A) is less
            // than absolute(mi-A)
            if (Math.abs(angle - A) < Math.abs(mi - A)) {
 
                // Update angle to mi, and
                // also update i to ans
                mi = angle;
                ans = i;
            }
        }
 
        // Print the vertices
        System.out.print(2 + " " + 1 + " " + (2 + ans));
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 3, A = 15;
        closestsAngle(N, A);
    }
}
 
// This code is contributed by subhammahato348.

Python3




# Python program for the above approach
# Function to find three vertices that
# subtends an angle closest to A
import math
import sys
def closestsAngle(N, A):
 
    # Stores the closest angle to A
    mi = sys.maxsize
 
    # Stores the count of edge which
    # subtend an angle of A
    ans = 0
 
    # Iterate in the range [1, N-2]
    for i in range(N ):
 
        # Stores the angle subtended
        angle = 180.0 * i / N
 
        # If absolute(angle-A) is less
        # than absolute(mi-A)
        if (math.fabs(angle - A) < math.fabs(mi - A)): 
 
            # Update angle to mi, and
            # also update i to ans
            mi = angle
            i+=1
            ans = i
     
    # Pr the vertices
    print(2 , 1 , 2 + ans)
 
# Driver Code
if __name__ == '__main__':
      
    N = 3
    A = 15
    closestsAngle(N, A)
 
# This code is contributed by Shivani

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to find three vertices that
// subtends an angle closest to A
static void closestsAngle(int N, int A)
{
     
    // Stores the closest angle to A
    double mi = Int32.MaxValue;
 
    // Stores the count of edge which
    // subtend an angle of A
    int ans = 0;
 
    // Iterate in the range [1, N-2]
    for(int i = 1; i < N - 1; i++)
    {
         
        // Stores the angle subtended
        double angle = 180.0 * i / N;
 
        // If absolute(angle-A) is less
        // than absolute(mi-A)
        if (Math.Abs(angle - A) <
            Math.Abs(mi - A))
        {
             
            // Update angle to mi, and
            // also update i to ans
            mi = angle;
            ans = i;
        }
    }
 
    // Print the vertices
    Console.Write(2 + " " + 1 +
                      " " + (2 + ans));
}
 
// Driver Code
public static void Main()
{
    int N = 3, A = 15;
    closestsAngle(N, A);
}
}
 
// This code is contributed by subhammahato348

Javascript




<script>
  
        // JavaScript program for the above approach
 
 
        // Function to find three vertices that
        // subtends an angle closest to A
        function closestsAngle(N, A) {
            // Stores the closest angle to A
            let mi = Number.MAX_VALUE;
 
            // Stores the count of edge which
            // subtend an angle of A
            let ans = 0;
 
            // Iterate in the range [1, N-2]
            for (let i = 1; i < N - 1; i++) {
 
                // Stores the angle subtended
                let angle = 180.0 * i / N;
 
                // If absolute(angle-A) is less
                // than absolute(mi-A)
                if (Math.abs(angle - A) < Math.abs(mi - A)) {
 
                    // Update angle to mi, and
                    // also update i to ans
                    mi = angle;
                    ans = i;
                }
            }
 
            // Print the vertices
            document.write(2 + ' ' + 1 + ' '
                + parseInt(2 + ans));
        }
 
        // Driver Code
 
        let N = 3, A = 15;
        closestsAngle(N, A);
 
 
      // This code is contributed by Potta Lokesh
       
</script>

Output: 

2 1 3

 

Time Complexity: O(N)
Auxiliary Space: O(1)


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!