/* Solution to Quadrilateral Area problem by Mees.
 *
 * Expected answer: ACCEPTED.
 *
 * Does a ternary search over one of the diagonals to find the optimal diagonal
 * length.
 */

#include <algorithm>
#include <cmath>
#include <iostream>
#include <stack>

using namespace std;

int main() {
    // Four sides.
    double a, b, c, d;
    cin >> a >> b >> c >> d;

    // Bounds on the possible sizes of the diagonal.
    double low = max(abs(b-a),abs(c-d));
    double high = min(a+b, c+d);

    // A ternary search on the size of the diagonal, looking for maximal area.
    // This is a concave optimization. We use Heron's formula to find the area.
    double mid1, mid2, area1, area2, sper1, sper2;
    do { 
        mid1 = (2.0*low + high)/3.0;
        mid2 = (low + 2.0*high)/3.0;

        sper1 = (a + b + mid1)/2;
        sper2 = (c + d + mid1)/2;
        area1 = sqrt(sper1*(sper1-a)*(sper1-b)*(sper1-mid1))
            + sqrt(sper2*(sper2-c)*(sper2-d)*(sper2-mid1));

        sper1 = (a + b + mid2)/2;
        sper2 = (c + d + mid2)/2;
        area2 = sqrt(sper1*(sper1-a)*(sper1-b)*(sper1-mid2))
            + sqrt(sper2*(sper2-c)*(sper2-d)*(sper2-mid2));

        if(area1 > area2) 
            high = mid2;
        else 
            low = mid1;
    } while(high - low > 0.0000001);

    printf("%.8lf\n", area1);
    return 0;
}