tetrahedron

题意：

求解一个四面体的内切球。

解法：

首先假设内切球球心为$(x0,x1,x2)$，可以用$r = \frac{3V}{S_1+S_2+S_3+S_4}$得出半径，

这样对于四个平面列出三个方程，解得

$x_n = \sum_{i=0}^3{Ai_{x_n} \cdot S_i } / (S_1 + S_2 + S_3 + S_4)$

这样，即可得出内切球。

时间复杂度$O(1)$。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>

 #define LD double
 #define sqr(x) ((x)*(x))
 #define eps 1e-13

 using namespace std;

 struct node
 {
     LD x,y,z;
     void scan()
     {
         scanf("%lf%lf%lf",&x,&y,&z);
     }
     void print()
     {
         printf("%.4lf %.4lf %.4lf\n",x,y,z);
     }
     LD length()
     {
         return sqrt(sqr(x)+sqr(y)+sqr(z));
     }
     node operator+(const node &tmp)
     {
         return (node){x+tmp.x,y+tmp.y,z+tmp.z};
     }
     node operator-(const node &tmp)
     {
         return (node){x-tmp.x,y-tmp.y,z-tmp.z};
     }
     node operator/(LD tmp)
     {
         return (node){x/tmp,y/tmp,z/tmp};
     }
     node operator*(LD tmp)
     {
         return (node){x*tmp,y*tmp,z*tmp};
     }
 };

 node cross(node a,node b)
 {
     node ans;
     ans.x = a.y*b.z - b.y*a.z;
     ans.y = b.x*a.z - a.x*b.z;
     ans.z = a.x*b.y - b.x*a.y;
     return ans;
 }

 LD dist(node a,node b)
 {
     return (b-a).length();
 }

 LD dot(node a,node b)
 {
     return a.x*b.x + a.y*b.y + a.z*b.z;
 }

 LD get_angle(node a,node b)
 {
     LD tmp = dot(a,b)/a.length()/b.length();
     return acos(tmp);
 }

 node get_node(node A,node B,node C)
 {
     LD Lth = (B-A).length() + (C-A).length() + (C-B).length();
     cout << sqr(Lth-) << endl;
     LD r = fabs(cross(B-A,C-A).length()) / Lth;
     cout << r*r << endl;
     node v1 = C-A;
     node v2 = B-A;
     node v = (v1+v2)/(v1+v2).length();
     LD d = (C-A).length()/;
     LD L = sqrt(sqr(d)+sqr(r));
     v = v*L;
     return A+v;
 }

 int main()
 {
     node A,B,C,D;
     while(~scanf("%lf%lf%lf",&A.x,&A.y,&A.z))
     {
         B.scan();
         C.scan();
         D.scan();
         if(fabs(dot(cross(B-A,C-A),D-A)) < eps)
         {
             puts("O O O O");
             continue;
         }
         LD S1 = fabs(cross(B-D,C-D).length())/;
         LD S2 = fabs(cross(D-A,C-A).length())/;
         LD S3 = fabs(cross(B-A,D-A).length())/;
         LD S4 = fabs(cross(B-A,C-A).length())/;
         LD Ve = fabs(dot(cross(B-A,C-A),D-A))/;
         LD R = *Ve / ((S1+S2+S3+S4));
         node ans;
         ans.x = (S1*A.x + S2*B.x + S3*C.x + S4*D.x)/(S1+S2+S3+S4);
         ans.y = (S1*A.y + S2*B.y + S3*C.y + S4*D.y)/(S1+S2+S3+S4);
         ans.z = (S1*A.z + S2*B.z + S3*C.z + S4*D.z)/(S1+S2+S3+S4);
         printf("%.4lf %.4lf %.4lf %.4lf\n",ans.x,ans.y,ans.z,R);
     }
     return ;
 }
 /*
 0 0 0 2 0 0 0 0 2 0 2 0
 0 0 0 2 0 0 3 0 0 4 0 0
 */