import java.io.*;
import java.util.*;
import java.awt.geom.*;

/**
 * Solution to Carpet Stains
 * 
 * Consider a point P in the middle of an equilateral triangle,
 * such that the distance from P to each of the vertices is a, b and c.
 * 
 * We want to find the area of the triangle. Since it's equilateral, 
 * all three sides have the same length. Let's call it x. If we can find x, 
 * then the area is just x*x*sqrt(3)/4.
 * 
 * Well, x forms a triangle with a and b, with b and c, and with a and c.
 * We need to find an x such that the central angles add up to 360 degrees (or 2pi radians).
 * 
 * We'll use a bisection root finding method to find x.
 * 
 * @author vanb
 */
public class carpet_vanb
{
    public Scanner sc;
    public PrintStream ps;
    
    public static final double TWOPI = Math.PI + Math.PI;
    public static final double SQRT3OVER4 = Math.sqrt( 3.0 ) / 4.0;
    public static final double EPSILON = 0.00001;
    
    public float a, b, c;
    
    /**
     * Given the lengths of three sides of a triangle, return one of its angles.
     * The three lengths are a, b and c. This method returns the angle between a and b, opposite c.
     * 
     * Use the Law of Cosines.
     * 
     * c*c = a*a + b*b - 2*a*b*cos(theta)
     * where theta is the angle opposite c.
     * 
     * @param a One leg of a triangle
     * @param b Another leg of a triangle
     * @param c Third leg of a triangle
     * @return The angle between legs a and b, opposite c
     */
    public static double angle( double a, double b, double c )
    {
        double angle = 0.0;
        if( Math.abs(c-a-b)<EPSILON ) angle = Math.PI;
        else if( Math.abs(Math.abs(a-b)-c)<EPSILON ) angle = 0.0; 
        else angle = Math.acos( (a*a+b*b-c*c)/(2.0*a*b) );       
        while( angle<0.0 ) angle += TWOPI;        
        return angle;
    }
    
    public double heron( double a, double b, double c )
    {
        double s = (a+b+c)/2.0;
        return Math.sqrt( s*(s-a)*(s-b)*(s-c) );       
    }
    
    /**
     * This is the function we're trying to find a zero for. 
     * It's the sum of the interior angles, minus 2pi. When that's zero,
     * then the three interior angles add up to 2pi, and that's the x
     * we're looking for.
     * 
     * @param x Side length of the equilateral triangle
     * @return Sum of the interior angles minus 2pi
     */
    public double f( double x )
    {
        return angle( a, b, x ) + angle( a, c, x ) + angle( b, c, x ) - TWOPI;
    }
       
    /**
     * Driver.
     * @throws Exception
     */
    public void doit() throws Exception
    {
        sc = new Scanner( System.in );
        ps = System.out; 
        
        while( sc.hasNext() )
        {
            a = sc.nextFloat();
            b = sc.nextFloat();
            c = sc.nextFloat();
            
            // Differences
            double dab = Math.abs( a-b );
            double dac = Math.abs( a-c );
            double dbc = Math.abs( b-c );
            
            // The lowest and highest the side x can possibly be.
            // If x<|a-b|, then x, a, b cannot form a triangle.
            double lo = Math.max( Math.max( dab, dac ), dbc );
            
            // If x>a+b, then x, a, b cannot form a triangle.
            double hi = Math.min( Math.min( a+b, a+c ), b+c );
            
            // Default, if we can't find an answer
            double area = -1.0;
            
            //System.err.println( lo + " " + hi );
            if( lo<=hi ) //System.err.println(  f(lo) + " " + f(hi) );
            // Check to see if it's even possible.
            if( lo<=hi+EPSILON && f(lo)*f(hi)<=EPSILON )
            {
                // Use bisection method until lo & hi are the same (within epsilon)
                while( hi-lo > 0.000000001 )
                {
                    double mid = (lo+hi)/2.0;
                    if( f(mid)<0.0 ) lo=mid; else hi=mid;
                }
                
                // lo is the x we're looking for.
                // So is hi, since they're close enough (within epsilon)
                area = lo*hi*SQRT3OVER4;
            }
                                   
            ps.printf( "%.3f", area );
            ps.println();
            
// This code was used by the judges to minimize problems with roundoff error
//            
//            String answer = String.format( "%.5f", area );
//            if( answer.endsWith( "50" ) || answer.endsWith( "49" ) ) 
//                System.err.println( "PANIC!! Too close to a cusp!" + answer );
        }
    }
    
    /**
     * @param args
     */
    public static void main( String[] args ) throws Exception
    {
        new carpet_vanb().doit();
    }   
}