uva11178 Morley's Theorem

Morleys theorem states that that the lines trisecting the angles of an
arbitrary plane triangle meet at the vertices of an equi- lateral
triangle. For example in the gure below the tri-sectors of angles A,
B and C has intersected and created an equilateral triangle DEF. Of
course the theorem has various gen- eralizations, in particular if all
of the tri- sectors are intersected one obtains four other equilateral
triangles. But in the original theorem only tri-sectors nearest to BC
are allowed to intersect to get point D, tri-sectors nearest to CA are
allowed to intersect point E and tri-sectors nearest to AB are inter-
sected to get point F. Trisector like BD and CE are not allowed to
intersect. So ultimately we get only one equilateral triangle DEF. Now
your task is to nd the Cartesian coordinates of D, E and F given the
coordinates of A, B, and C. Input First line of the input le contains
an integer N (0 < N< 5001) which denotes the number of test cases to
follow. Each of the next lines contain six integers X A ;Y A ;X B ;Y
B ;X C ;Y C . This six integers actually indicates that the Cartesian
coordinates of point A, B and C are ( X A ;Y A ),( X B ;Y B ) and ( X
C ;Y C ) respectively. You can assume that the area of triangle ABC is
not equal to zero, 0  X A ;Y A ;X B ;Y B ;X C ;Y C  1000 and the
points A, B and C are in counter clockwise order. Output For each line
of input you should produce one line of output. This line contains
six oating point numbers X D ;Y D ;X E ;Y E ;X F ;Y F separated by a
single space. These six oating-point actually means that the
Cartesian coordinates of D, E and F are ( X D ;Y D ),( X E ;Y E ) ,( X
F ;Y F ) respectively. Errors less than 10

算出来两边夹角，把其中一边旋转夹角的三分之一，再求两条直线交点。

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;struct v{    double x,y;    void rd()    {        scanf("%lf%lf",&x,&y);    }    void out()    {        printf("%.6f %.6f ",x,y);    }    v operator + (const v &v1)    {        return (v){x+v1.x,y+v1.y};    }    v operator - (const v &v1)    {        return (v){x-v1.x,y-v1.y};    }    v operator * (const double &k)    {        return (v){k*x,k*y};    }    v operator / (const double &k)    {        return (v){x/k,y/k};    }}a,b,c;double dot(v v1,v v2){    return v1.x*v2.x+v1.y*v2.y;}double cross(v v1,v v2){    return v1.x*v2.y-v1.y*v2.x;}double len(v v1){    return sqrt(dot(v1,v1));}double angle(v v1,v v2){    return acos(dot(v1,v2)/len(v1)/len(v2));}v rot(v v1,double x){    return (v){v1.x*cos(x)-v1.y*sin(x),v1.x*sin(x)+v1.y*cos(x)};}struct line{    v p,a;};v intersec(line l1,line l2){    v v1=l1.p-l2.p;    double t=cross(l2.a,v1)/cross(l1.a,l2.a);    return l1.p+l1.a*t;}v solve(v a,v b,v c){    double a1=angle(a-c,b-c),a2=angle(c-b,a-b);    line l1=(line){c,rot(b-c,-a1/3)},l2=(line){b,rot(c-b,a2/3)};    return intersec(l1,l2);}int main(){    int T;    scanf("%d",&T);    while (T--)    {        a.rd();        b.rd();        c.rd();        solve(a,b,c).out();        solve(b,c,a).out();        solve(c,a,b).out();        putchar('\n');    }}