Problem E

2D Geometry 110 in 1!

This is a collection of 110 (in binary) 2D geometry problems.

CircumscribedCircle x1 y1 x2 y2 x3 y3

Find out the circumscribed circle of triangle (x1,y1)-(x2,y2)-(x3,y3). These three points are guaranteed to be non-collinear. The circle is formatted as (x,y,r) where (x,y) is the center of circle, r is the radius.

InscribedCircle x1 y1 x2 y2 x3 y3

Find out the inscribed circle of triangle (x1,y1)-(x2,y2)-(x3,y3). These three points are guaranteed to be non-collinear. The circle is formatted as (x,y,r) where (x,y) is the center of circle, r is the radius.

TangentLineThroughPoint xc yc r xp yp

Find out the list of tangent lines of circle centered (xc,yc) with radius r that pass through point (xp,yp). Each tangent line is formatted as a single real number "angle" (in degrees), the angle of the line (0<=angle<180). Note that the answer should be formatted as a list (see below for details).

CircleThroughAPointAndTangentToALineWithRadius xp yp x1 y1 x2 y2 r

Find out the list of circles passing through point (xp, yp) that is tangent to a line (x1,y1)-(x2,y2) with radius r. Each circle is formatted as (x,y), since the radius is already given. Note that the answer should be formatted as a list. If there is no answer, you should print an empty list.

CircleTangentToTwoLinesWithRadius x1 y1 x2 y2 x3 y3 x4 y4 r

Find out the list of circles tangent to two non-parallel lines (x1,y1)-(x2,y2) and (x3,y3)-(x4,y4), having radius r. Each circle is formatted as (x,y), since the radius is already given. Note that the answer should be formatted as a list. If there is no answer, you should print an empty list.

CircleTangentToTwoDisjointCirclesWithRadius x1 y1 r1 x2 y2 r2 r

Find out the list of circles externally tangent to two disjoint circles (x1,y1,r1) and (x2,y2,r2), having radius r. By "externally" we mean it should not enclose the two given circles. Each circle is formatted as (x,y), since the radius is already given. Note that the answer should be formatted as a list. If there is no answer, you should print an empty list.

For each line described above, the two endpoints will not be equal. When formatting a list of real numbers, the numbers should be sorted in increasing order; when formatting a list of (x,y) pairs, the pairs should be sorted in increasing order of x. In case of tie, smaller y comes first.

Input

There will be at most 1000 sub-problems, one in each line, formatted as above. The coordinates will be integers with absolute value not greater than 1000. The input is terminated by end of file (EOF).

Output

For each input line, print out your answer formatted as stated in the problem description. Each number in the output should be rounded to six digits after the decimal point. Note that the list should be enclosed by square brackets, and tuples should be enclosed by brackets. There should be no space characters in each line of your output.

Sample Input

CircumscribedCircle 0 0 20 1 8 17InscribedCircle 0 0 20 1 8 17TangentLineThroughPoint 200 200 100 40 150TangentLineThroughPoint 200 200 100 200 100TangentLineThroughPoint 200 200 100 270 210CircleThroughAPointAndTangentToALineWithRadius 100 200 75 190 185 65 100CircleThroughAPointAndTangentToALineWithRadius 75 190 75 190 185 65 100CircleThroughAPointAndTangentToALineWithRadius 100 300 100 100 200 100 100CircleThroughAPointAndTangentToALineWithRadius 100 300 100 100 200 100 99CircleTangentToTwoLinesWithRadius 50 80 320 190 85 190 125 40 30CircleTangentToTwoDisjointCirclesWithRadius 120 200 50 210 150 30 25CircleTangentToTwoDisjointCirclesWithRadius 100 100 80 300 250 70 50

Output for the Sample Input

(9.734940,5.801205,11.332389)(9.113006,6.107686,5.644984)[53.977231,160.730818][0.000000][][(112.047575,299.271627),(199.997744,199.328253)][(-0.071352,123.937211),(150.071352,256.062789)][(100.000000,200.000000)][][(72.231286,121.451368),(87.815122,63.011983),(128.242785,144.270867),(143.826621,85.831483)][(157.131525,134.836744),(194.943947,202.899105)][(204.000000,178.000000)]

Rujia Liu's Present 4: A Contest Dedicated to Geometry and CG Lovers
Special Thanks: Di Tang and Yi Chen

Source

Root :: Prominent Problemsetters ::

Root :: Rujia Liu's Presents ::

Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 4. Geometry :: Geometric Computations in 2D ::

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;const double eps=1e-6;int dcmp(double x){if(fabs(x)
           
             0 P在Q的顺时针方向 P*Q < 0 P在Q的逆时针方向 P*Q = 0 PQ共线*/Point Horunit(Point x) {return x/Length(x);}///单位向量Point Verunit(Point x) {return Point(-x.y,x.x)/Length(x);}///单位法向量Point Rotate(Point A,double rad)///逆时针旋转{ return Point(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}double Area2(const Point A,const Point B,const Point C){ return Cross(B-A,C-A);}/// 过两点p1, p2的直线一般方程ax+by+c=0 (x2-x1)(y-y1) = (y2-y1)(x-x1)void getLineGeneralEquation(const Point& p1, const Point& p2, double& a, double&b, double &c){ a = p2.y-p1.y; b = p1.x-p2.x; c = -a*p1.x - b*p1.y;}///P+t*v Q+w*t的焦点Point GetLineIntersection(Point P,Point v,Point Q,Point w){ Point u=P-Q; double t=Cross(w,u)/Cross(v,w); return P+v*t;}///点到直线距离double DistanceToLine(Point P,Point A,Point B){ Point v1=B-A,v2=P-A; return fabs(Cross(v1,v2))/Length(v1);}///点到线段距离double DistanceToSegment(Point P,Point A,Point B){ if(A==B) return Length(P-A); Point v1=B-A,v2=P-A,v3=P-B; if(dcmp(Dot(v1,v2))<0) return Length(v2); else if(dcmp(Dot(v1,v3))>0) return Length(v3); else return fabs(Cross(v1,v2))/Length(v1);}///点到直线投影Point GetLineProjection(Point P,Point A,Point B){ Point v=B-A; return A+v*(Dot(v,P-A)/Dot(v,v));}///推断规范相交bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){ double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1); double c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1); return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;}///一个点是否在直线端点上bool OnSegment(Point p,Point a1,Point a2){ return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;}///多边形有向面积double PolygonArea(Point* p,int n){ double area=0; for(int i=1;i
            
             =0;}///圆const double pi=acos(-1.0);struct Circle{ Point c; double r; Circle(Point _c=0,double _r=0):c(_c),r(_r){} Point point(double a)///依据圆心角算圆上的点 { return Point(c.x+cos(a)*r,c.y+sin(a)*r); }};///a点到b点（逆时针）在圆上的圆弧长度double D(Point a,Point b,Circle C){ double ang1,ang2; Point v1,v2; v1=a-C.c; v2=b-C.c; ang1=atan2(v1.y,v1.x); ang2=atan2(v2.y,v2.x); if(ang2
             
              & sol){ double a=L.v.x,b=L.p.x-C.c.x,c=L.v.y,d=L.p.y-C.c.y; double e=a*a+c*c,f=2*(a*b+c*d),g=b*b+d*d-C.r*C.r; double delta=f*f-4.*e*g; if(dcmp(delta)<0) return 0;//相离 if(dcmp(delta)==0)//相切 { t1=t2=-f/(2.*e); sol.push_back(L.point(t1)); return 1; } //相切 t1=(-f-sqrt(delta))/(2.*e); sol.push_back(L.point(t1)); t2=(-f+sqrt(delta))/(2.*e); sol.push_back(L.point(t2)); return 2;}///圆与圆交点 返回交点个数int getCircleCircleIntersection(Circle C1,Circle C2,vector
              
               & Sol){ double d=Length(C1.c-C2.c); if(dcmp(d)==0) { if(dcmp(C1.r-C2.r)==0) return -1;//重合 return 0; } if(dcmp(C1.r+C2.r-d)<0) return 0; if(dcmp(fabs(C1.r-C2.r)-d)>0) return 0; double a=Angle(C2.c-C1.c); double da=acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*C1.r*d)); Point p1=C1.point(a-da),p2=C1.point(a+da); Sol.push_back(p1); if(p1==p2) return 1; Sol.push_back(p2); return 2;}///P到圆的切线 v[] 储存切线向量int getTangents(Point p,Circle C,Point* v){ Point u=C.c-p; double dist=Length(u); if(dist
               
                rsum*rsum)// two { double ang=acos((A.r-B.r)/sqrt(d2)); a[cnt]=A.point(base+ang); b[cnt]=B.point(pi+base+ang); cnt++; a[cnt]=A.point(base-ang); b[cnt]=B.point(pi+base-ang); cnt++; } return cnt;}///三角形外接圆Circle CircumscribedCircle(Point p1,Point p2,Point p3){ double Bx=p2.x-p1.x,By=p2.y-p1.y; double Cx=p3.x-p1.x,Cy=p3.y-p1.y; double D=2*(Bx*Cy-By*Cx); double cx=(Cy*(Bx*Bx+By*By)-By*(Cx*Cx+Cy*Cy))/D+p1.x; double cy=(Bx*(Cx*Cx+Cy*Cy)-Cx*(Bx*Bx+By*By))/D+p1.y; Point p=Point(cx,cy); return Circle(p,Length(p1-p));}///三角形内切圆Circle InscribedCircle(Point p1,Point p2,Point p3){ double a=Length(p2-p3); double b=Length(p3-p1); double c=Length(p1-p2); Point p=(p1*a+p2*b+p3*c)/(a+b+c); return Circle(p,DistanceToLine(p,p1,p2));}double RtoDegree(double x) {return x/pi*180.;}char op[200];double a[10];Point v[10];double degree[10];vector
                
                  sol;int main(){ while(scanf("%s",op)!=EOF) { if(strcmp(op,"CircumscribedCircle")==0) { for(int i=0;i<6;i++) scanf("%lf",a+i); Circle C=CircumscribedCircle(Point(a[0],a[1]),Point(a[2],a[3]),Point(a[4],a[5])); printf("(%.6lf,%.6lf,%.6lf)\n",C.c.x,C.c.y,C.r); } else if(strcmp(op,"InscribedCircle")==0) { for(int i=0;i<6;i++) scanf("%lf",a+i); Circle C=InscribedCircle(Point(a[0],a[1]),Point(a[2],a[3]),Point(a[4],a[5])); printf("(%.6lf,%.6lf,%.6lf)\n",C.c.x,C.c.y,C.r); } else if(strcmp(op,"TangentLineThroughPoint")==0) { for(int i=0;i<5;i++) scanf("%lf",a+i); int sz=getTangents(Point(a[3],a[4]),Circle(Point(a[0],a[1]),a[2]),v); for(int i=0;i
                 
                  =0) de-=180.; degree[i]=de; } sort(degree,degree+sz); putchar('[');if(sz==0) putchar(']'); for(int i=0;i

',':']'); putchar(10); } else if(strcmp(op,"CircleThroughAPointAndTangentToALineWithRadius")==0) { for(int i=0;i<7;i++) scanf("%lf",a+i); Point A=Point(a[2],a[3]),B=Point(a[4],a[5]); Circle C(Point(a[0],a[1]),a[6]); Point normal=Verunit(B-A); normal=normal/Length(normal)*a[6]; Point ta=A+normal,tb=B+normal; Line l1=Line(ta,tb-ta); ta=A-normal,tb=B-normal; Line l2=Line(ta,tb-ta); sol.clear(); double t1,t2; int aa=getLineCircleIntersection(l1,C,t1,t2,sol); int bb=getLineCircleIntersection(l2,C,t1,t2,sol); sort(sol.begin(),sol.end()); putchar('['); for(int i=0,sz=sol.size();i<sz;i++) { if(i) putchar(','); printf("(%.6lf,%.6lf)",sol[i].x,sol[i].y); } putchar(']'); putchar(10); } else if(strcmp(op,"CircleTangentToTwoLinesWithRadius")==0) { for(int i=0;i<9;i++) scanf("%lf",a+i); Line LA=Line(Point(a[0],a[1]),Point(a[2],a[3])-Point(a[0],a[1])); Line LB=Line(Point(a[4],a[5]),Point(a[6],a[7])-Point(a[4],a[5])); Line la1=LineTransHor(LA,a[8]),la2=LineTransHor(LA,-a[8]); Line lb1=LineTransHor(LB,a[8]),lb2=LineTransHor(LB,-a[8]); sol.clear(); sol.push_back(GetLineIntersection(la1,lb1)); sol.push_back(GetLineIntersection(la1,lb2)); sol.push_back(GetLineIntersection(la2,lb1)); sol.push_back(GetLineIntersection(la2,lb2)); sort(sol.begin(),sol.end()); putchar('['); for(int i=0,sz=sol.size();i<sz;i++) { if(i) putchar(','); printf("(%.6lf,%.6lf)",sol[i].x,sol[i].y); } putchar(']'); putchar(10); } else if(strcmp(op,"CircleTangentToTwoDisjointCirclesWithRadius")==0) { for(int i=0;i<7;i++) scanf("%lf",a+i); Circle C1=Circle(Point(a[0],a[1]),a[2]+a[6]); Circle C2=Circle(Point(a[3],a[4]),a[5]+a[6]); sol.clear(); getCircleCircleIntersection(C1,C2,sol); sort(sol.begin(),sol.end()); putchar('['); for(int i=0,sz=sol.size();i<sz;i++) { if(i) putchar(','); printf("(%.6lf,%.6lf)",sol[i].x,sol[i].y); } putchar(']'); putchar(10); } } return 0; }

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