Number of rectangles in a circle of radius R

Given a circular sheet of radius R and the task is to find the total number of rectangles with integral length and width that can be cut from the circular sheet, one at a time.

Examples:

Input: R = 2
Output: 8
8 rectangles can be cut from a circular sheet of radius 2.
These are: 1×1, 1×2, 2×1, 2×2, 1×3, 3×1, 2×3, 3×2.



Input: R = 1
Output: 1
Only one rectangle with dimensions 1 X 1 is possible.

Approach
Consider the following diagram,
Circle_largest_rect

Its easy to see, that ABCD is the largest rectangle that can be formed in the given circle with radius R and centre O, having dimensions a X b

Drop a perpendicular AO such that, ∠AOD = ∠AOB = 90°

Consider the following diagram for further analysis,
rectange_circle_square_proof

Consider triangles AOD and AOB,
In these triangles,
AO = AO (Common Side)
∠AOD = ∠AOB = 90° 
OD = OB = R
Thus, by SAS congruence ▵AOD ≅ ▵AOB
∴ AD = AB by CPCT(i.e Corresponding Parts on Congruent Triangles)
or, a = b
=> The rectangle ABCD is a square

The diameter BD is the maximum diagonal the rectangle can have to be able to be cut from the Circular Sheet.

Thus, all the combinations of a and b can be checked to form all possible rectangles, and if the diagonal of any such rectangle is less than or equal to the length of the diagonal of the largest rectangle formed (i.e 2 * R, where R is the Radius of the circle as explained above)

Now, the maximum length of a and b will always be strictly less than the diameter of the circle so all possible values of a and b will lie in the closed interval [1, (2 * R – 1)].

Below is the implementation of the above approach:

C++

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// C++ program to find the number of rectangles
// that can be cut from a circle of Radius R
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the total possible
// rectangles that can be cut from the circle
int countRectangles(int radius)
{
  
    int rectangles = 0;
  
    // Diameter = 2 * Radius
    int diameter = 2 * radius;
  
    // Square of diameter which is the square
    // of the maximum length diagnal
    int diameterSquare = diameter * diameter;
  
    // generate all combinations of a and b
    // in the range (1, (2 * Radius - 1))(Both inclusive)
    for (int a = 1; a < 2 * radius; a++) {
        for (int b = 1; b < 2 * radius; b++) {
  
            // Calculate the Diagnal length of
            // this rectange
            int diagnalLengthSquare = (a * a + b * b);
  
            // If this rectangle's Diagonal Length
            // is less than the Diameter, it is a
            // valid rectangle, thus increment counter
            if (diagnalLengthSquare <= diameterSquare) {
                rectangles++;
            }
        }
    }
  
    return rectangles;
}
  
// Driver Code
int main()
{
  
    // Radius of the circle
    int radius = 2;
  
    int totalRectangles;
    totalRectangles = countRectangles(radius);
    cout << totalRectangles << " rectangles can be"
         << "cut from a circle of Radius " << radius;
    return 0;
}

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// Java program to find the  


// number of rectangles that  


// can be cut from a circle 


// of Radius R 


import java.io.*; 


  


class GFG 


{ 


      


// Function to return  


// the total possible 


// rectangles that can  


// be cut from the circle 


static int countRectangles(int radius) 


{ 


    int rectangles = 0; 


  


    // Diameter = 2 * Radius 


    int diameter = 2 * radius; 


  


    // Square of diameter  


    // which is the square 


    // of the maximum length  


    // diagnal 


    int diameterSquare = diameter * 


                         diameter; 


  


    // generate all combinations  


    // of a and b in the range  


    // (1, (2 * Radius - 1)) 


    // (Both inclusive) 


    for (int a = 1;  


             a < 2 * radius; a++) 


    { 


        for (int b = 1; 


                 b < 2 * radius; b++) 


        { 


  


            // Calculate the  


            // Diagnal length of 


            // this rectange 


            int diagnalLengthSquare = (a * a +  


                                       b * b); 


  


            // If this rectangle's Diagonal 


            // Length is less than the Diameter,  


            // it is a valid rectangle, thus  


            // increment counter 


            if (diagnalLengthSquare <= diameterSquare) 


            { 


                rectangles++; 


            } 


        } 


    } 


  


    return rectangles; 


} 


  


// Driver Code 


public static void main (String[] args)  


{ 


      


// Radius of the circle 


int radius = 2; 


  


int totalRectangles; 


totalRectangles = countRectangles(radius); 


System.out.println(totalRectangles +  


             " rectangles can be " +  


            "cut from a circle of" +  


               " Radius " + radius); 


} 


} 


  


// This code is contributed 


// by anuj_67. 



















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Python3














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# Python3 program to find  


# the number of rectangles 


# that can be cut from a  


# circle of Radius R Function 


# to return the total possible 


# rectangles that can be cut 


# from the circle 


def countRectangles(radius): 


      


    rectangles = 0


      


    # Diameter = 2 * Radius 


    diameter = 2 * radius 


      


    # Square of diameter which  


    # is the square of the  


    # maximum length diagnal 


    diameterSquare = diameter * diameter 


      


    # generate all combinations  


    # of a and b in the range  


    # (1, (2 * Radius - 1))(Both inclusive) 


    for a in range(1, 2 * radius):  


        for b in range(1, 2 * radius): 


              


            # Calculate the Diagnal  


            # length of this rectange 


            diagnalLengthSquare = (a * a + 


                                   b * b) 


              


            # If this rectangle's Diagonal  


            # Length is less than the  


            # Diameter, it is a valid 


            # rectangle, thus increment counter 


            if (diagnalLengthSquare <= diameterSquare) : 


                rectangles += 1


      


    return rectangles 


  


# Driver Code 


  


# Radius of the circle 


radius = 2


totalRectangles = countRectangles(radius) 


print(totalRectangles , "rectangles can be" ,  


      "cut from a circle of Radius" , radius) 


  


# This code is contributed by Smita 



















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C#














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// C# program to find the  


// number of rectangles that  


// can be cut from a circle 


// of Radius R 


using System; 


  


class GFG 


{ 


      


// Function to return  


// the total possible 


// rectangles that can  


// be cut from the circle 


static int countRectangles(int radius) 


{ 


    int rectangles = 0; 


  


    // Diameter = 2 * Radius 


    int diameter = 2 * radius; 


  


    // Square of diameter  


    // which is the square 


    // of the maximum length  


    // diagnal 


    int diameterSquare = diameter * 


                         diameter; 


  


    // generate all combinations  


    // of a and b in the range  


    // (1, (2 * Radius - 1)) 


    // (Both inclusive) 


    for (int a = 1;  


             a < 2 * radius; a++) 


    { 


        for (int b = 1; 


                 b < 2 * radius; b++) 


        { 


  


            // Calculate the  


            // Diagnal length of 


            // this rectange 


            int diagnalLengthSquare = (a * a +  


                                        b * b); 


  


            // If this rectangle's  


            // Diagonal Length is  


            // less than the Diameter,  


            // it is a valid rectangle,  


            // thus increment counter 


            if (diagnalLengthSquare <=  


                             diameterSquare) 


            { 


                rectangles++; 


            } 


        } 


    } 


  


    return rectangles; 


} 


  


// Driver Code 


public static void Main ()  


{ 


      


// Radius of the circle 


int radius = 2; 


  


int totalRectangles; 


totalRectangles = countRectangles(radius); 


Console.WriteLine(totalRectangles +  


            " rectangles can be " +  


           "cut from a circle of" +  


              " Radius " + radius); 


} 


} 


  


// This code is contributed 


// by anuj_67. 



















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<?php 


// PHP program to find the  


// number of rectangles that 


// can be cut from a circle 


// of Radius R 


  


// Function to return the  


// total possible rectangles 


// that can be cut from the circle 


function countRectangles($radius) 


{ 


  


    $rectangles = 0; 


  


    // Diameter = 2 * $Radius 


    $diameter = 2 * $radius; 


  


    // Square of diameter which  


    // is the square of the  


    // maximum length diagnal 


    $diameterSquare = $diameter * 


                      $diameter; 


  


    // generate all combinations 


    // of a and b in the range  


    // (1, (2 * Radius - 1))(Both inclusive) 


    for ($a = 1;  


         $a < 2 * $radius; $a++)  


    { 


        for ($b = 1;  


             $b < 2 * $radius; $b++)  


        { 


  


            // Calculate the Diagnal  


            // length of this rectange 


            $diagnalLengthSquare = ($a * $a + 


                                    $b * $b); 


  


            // If this rectangle's Diagonal  


            // Length is less than the 


            // Diameter, it is a valid  


            // rectangle, thus increment counter 


            if ($diagnalLengthSquare <= $diameterSquare)  


            { 


                $rectangles++; 


            } 


        } 


    } 


  


    return $rectangles; 


} 


  


// Driver Code 


  


// Radius of the circle 


$radius = 2; 


  


$totalRectangles; 


$totalRectangles = countRectangles($radius); 


echo $totalRectangles , " rectangles can be " , 


      "cut from a circle of Radius " , $radius; 


  


// This code is contributed 


// by anuj_67. 


?> 



















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Output:


8 rectangles can be cut from a circle of Radius 2





Time Complexity: O(R2), where R is the Radius of the Circle





    

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