#include <bits/stdc++.h>
using namespace std;
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template<typename T> bool mmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool mmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
template<typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
#define FI first
#define SE second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))

using pd = pair<double, double>;
double eps = 1e-10;

double inner(pd a, pd b)
{
    return a.first * b.first + a.second * b.second;
}
double outer(pd a, pd b)
{
    return a.first * b.second - a.second * b.first;
}
double norm2(pd a)
{
    return inner(a, a);
}
pd normalize(pd p)
{
    double no = sqrt(norm2(p));
    p.first /= no;
    p.second /= no;
    return p;
}
vector<pd> P(12);
double ret;

pd equ(int i)
{
    pd u = normalize(P[i + 1] - P[i]) , v = normalize(P[i + 2] - P[i]);
    return normalize(u + v);
}

// void ren(pd P, pd u, pd Q, pd v, double &a, double &b)
// {
//     // P + au == Q + bv を解く
//     double det = v.first * u.second - v.second * u.first;
//     if (abs(det) < eps) exit(8);
//     pd dPQ = Q - P;
//     a = (dPQ.first * (-v.second) + dPQ.second * v.first) / det;
//     b = (dPQ.first * (-u.second) + dPQ.second * u.first) / det;
//     if (a < 0 or b < 0) exit(8);
// }

pd pdmul(pd p, double a, pd v)
{
    p.first += a * v.first;
    p.second += a * v.second;
    return p;
}

double Rmax;
void solve(int i)
{
    // double a, b;
    pd u = equ(i), v = equ(i + 1);
    pd w = normalize(P[i + 1] - P[i]);
    double s1 = abs(outer(u, w)), s2 = abs(outer(v, w));
    // ren(P[i], u, P[i + 1], v, a, b);
    double lo = 0.0, hi = Rmax;
    while (hi - lo > 1e-11) // or (hi - lo) / hi > 1e-11)
    {
        double R = (hi + lo) / 2;
        if (norm2(pdmul(P[i], R / s1, u) - pdmul(P[i + 1], R / s2, v)) > 4 * R * R) lo = R;
        else hi = R;
    }
    mmax(ret, lo);
}
int main()
{
    REP(i, 3) cin >> P[i].first >> P[i].second;
    REP(i, 9) P[i + 3] = P[i];
    double men2 = abs(outer(P[2] - P[0], P[1] - P[0]));
    double len = 0;
    REP(i, 3) len += sqrt(norm2(P[i + 1] - P[i]));
    Rmax = men2 / len;

    ret = -1;
    REP(i, 3) solve(i);
    cout << ret << endl;
}


Measured memory_bytes: 256