就是给出一个等边三角形的三个顶点坐标

然后每一个角的三等分线会交错成一个三角形，求出这个三角形的顶点坐标

一開始。我题意理解错了……还以为是随意三角形，所以代码可以处理随意三角形的情况

我的做法：

通过旋转点的位置得到这些三等分线的直线方程，然后用高斯消元求交点

我的代码：

#include<iostream>

#include<map>

#include<string>

#include<cstring>

#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<queue>

#include<vector>

#include<algorithm>

using namespace std;

struct dot

{

	double x,y;

	dot(){}

	dot(double a,double b){x=a;y=b;}

	dot operator -(const dot &a){return dot(x-a.x,y-a.y);}

	dot operator +(const dot &a){return dot(x+a.x,y+a.y);}

	double mod(){return sqrt(pow(x,2)+pow(y,2));}

	double mul(const dot &a){return x*a.x+y*a.y;}

};

void gauss(double a[10][10])

{

	int i,j,k,t,n=2;

	for(i=0;i<n;i++)

	{

		t=i;

		for(j=i+1;j<n;j++)

			if(fabs(a[j][i])>fabs(a[t][i]))

				t=i;

		if(i!=t)

			for(j=i;j<=n;j++)

				swap(a[i][j],a[t][j]);

		if(a[i][i]!=0)

			for(j=i+1;j<n;j++)

				for(k=n;k>=i;k--)

					a[j][k]-=a[j][i]/a[i][i]*a[i][k];

	}

	for(i=n-1;i>-1;i--)

	{

		for(j=i+1;j<n;j++)

			a[i][n]-=a[i][j]*a[j][n];

		a[i][n]/=a[i][i];

	}

}

dot ro(dot a,dot b,double c)

{

	a=a-b;

	a=dot(a.x*cos(c)-a.y*sin(c),a.x*sin(c)+a.y*cos(c));

	return a+b;

}

int main()

{

	pair<dot,dot>t;

	dot a[3];

	double b,c[10][10];

	int n,i;

	cin>>n;

	while(n--)

	{

		for(i=0;i<3;i++)

			scanf("%lf%lf",&a[i].x,&a[i].y);

		t.first=a[0]-a[1];t.second=a[2]-a[1];

		b=acos(t.first.mul(t.second)/t.first.mod()/t.second.mod())/3;

		t.first=a[1];t.second=ro(a[2],a[1],b);

		c[0][0]=t.first.y-t.second.y;c[0][1]=t.second.x-t.first.x;c[0][2]=t.second.x*t.first.y-t.second.y*t.first.x;

		t.first=a[1]-a[2];t.second=a[0]-a[2];

		b=2*acos(t.first.mul(t.second)/t.first.mod()/t.second.mod())/3;

		t.first=a[2];t.second=ro(a[0],a[2],b);

		c[1][0]=t.first.y-t.second.y;c[1][1]=t.second.x-t.first.x;c[1][2]=t.second.x*t.first.y-t.second.y*t.first.x;

		gauss(c);

		printf("%.6lf %.6lf ",c[0][2],c[1][2]);

		t.first=a[1]-a[2];t.second=a[0]-a[2];

		b=acos(t.first.mul(t.second)/t.first.mod()/t.second.mod())/3;

		t.first=a[2];t.second=ro(a[0],a[2],b);

		c[0][0]=t.first.y-t.second.y;c[0][1]=t.second.x-t.first.x;c[0][2]=t.second.x*t.first.y-t.second.y*t.first.x;

		t.first=a[1]-a[0];t.second=a[2]-a[0];

		b=2*acos(t.first.mul(t.second)/t.first.mod()/t.second.mod())/3;

		t.first=a[0];t.second=ro(a[1],a[0],b);

		c[1][0]=t.first.y-t.second.y;c[1][1]=t.second.x-t.first.x;c[1][2]=t.second.x*t.first.y-t.second.y*t.first.x;

		gauss(c);

		printf("%.6lf %.6lf ",c[0][2],c[1][2]);

		t.first=a[1]-a[0];t.second=a[2]-a[0];

		b=acos(t.first.mul(t.second)/t.first.mod()/t.second.mod())/3;

		t.first=a[0];t.second=ro(a[1],a[0],b);

		c[0][0]=t.first.y-t.second.y;c[0][1]=t.second.x-t.first.x;c[0][2]=t.second.x*t.first.y-t.second.y*t.first.x;

		t.first=a[0]-a[1];t.second=a[2]-a[1];

		b=2*acos(t.first.mul(t.second)/t.first.mod()/t.second.mod())/3;

		t.first=a[1];t.second=ro(a[2],a[1],b);

		c[1][0]=t.first.y-t.second.y;c[1][1]=t.second.x-t.first.x;c[1][2]=t.second.x*t.first.y-t.second.y*t.first.x;

		gauss(c);

		printf("%.6lf %.6lf\n",c[0][2],c[1][2]);

	}

}

原题：

Problem D
Morley’s Theorem
Input: Standard Input

Output: Standard Output

Morley’s theorem states that that the lines trisecting the angles of an arbitrary plane triangle meet at the vertices of an equilateral triangle. For example in the figure below the tri-sectors of angles A, B and C has intersected and created an equilateral
triangle DEF.

Of course the theorem has various generalizations, 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 intersected 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 find the Cartesian
coordinates of D, E and F given the coordinates of A, B, and C.

Input

First line of the input file contains an integer N (0<N<5001) which denotes the number of test cases to follow. Each of the next lines contain six integers
. This six integers actually indicates that the Cartesian coordinates of point A, B and C are
respectively. You can assume that the area of triangle ABC is not equal to zero,
and the points A, B and C are in counter clockwise order.