LOJ 1072 - Calm Down

In the first line of input, we will be the number of test cases T. After that we will be given R, the radius of the bigger circle and N, the number of small circles within. R and N are separated by a space for each test case to T times. We have to output the r, the radius of the small circles upto 10 floating points.

If we look at the picture above, we can see that a smaller circle within the bigger circle creates an angle α if we draw two tangent lines starting from the bigger circle's center A on the two sides of the smaller circle touching the bigger circle's parameter line which areAD and AB respectively. We know a circle produces an angle of 2*π. And the smaller circles do not have any space among themselves, so 2*π = N*α from which, we can get α = 2π/N.

Also let's draw AE which goes through the center of the small circle c and here AE = R (radius of the bigger circle) = AD = AB. Drawing AE produces an angle β. Since the smaller circle is split straight through its center, β = α/2 from which we can get β = (2*π/N)/2 = π/N.

Now let's draw one perpendicular line from point C on AB which is CF. If we observe the figure, CF = CE = r (radius of the smaller circle). Now we can easily find out the value of r using the sin(β) ratio.

    sin(β) = CF/(AE-CE) 
           = r/(R-r)
=>  sin(π/N) = r/(R-r)
=>  r = sin(π/N)*(R-r)
      = sin(π/N)*R - sin(π/N)*r
=>  r + sin(π/N)*r = sin(π/N)*R
=>  r*(1+ sin(π/N)) = sin(π/N)*R
=>  r = sin(π/N)*R/(1+ sin(π/N))

So, r = sin(π/N)*R/(1+ sin(π/N))

The above implementation is Accepted.

Solution in C

#include <stdio.h>
#include <math.h>
int main()
{

    int cases;
    scanf("%d", &cases);
    double pi = 2 * acos(0.0);
    for (int i = 1; i <= cases; i++)
    {
        double R;
        int n;
        scanf("%lf %d", &R, &n);
        double r = (R * sin(pi / n * 1.0)) / (1 + sin(pi / n * 1.0));
        printf("Case %d: %.10lf\n", i, r);
    }
    return 0;
}

Tutorial source