//数学题！设两楼间的距离为s，分别由m和n构成; x和y两侧的对应楼高分别为a和b
//由相似三角形可知: c/a = n/s ① c/b = m/s ② m + n = s ③ 可求出：c=a*b/(a+b) 由此根据a、b、c可二分出距离s

#include <iostream>
#include <math.h>
#include <algorithm>

using namespace std;

double x,y,c;

bool check(double mid){
    double a = sqrt(x*x - mid*mid);
    double b = sqrt(y*y - mid*mid);
    return a*b/(a+b) > c;
}

int main(){

    cin >> x >> y >> c;
    double l = 0, r = min(x,y);
    while(r - l > 1e-5){
        double mid = (l+r)/2;
        if(check(mid)){
            l = mid;
        }else{
            r = mid;
        }
    }

    printf("%.3f",l);

    return 0;
}