UVA 11178 Morley's Theorem

大意：做三角形ABC每个内角额三等分线，相交成三角形DEF，则三角形DEF是等边三角形。你的任务是根据A、B、C三个点的位置确定D、E、F的位置。

思路：求出<ABC的大小a1，然后将向量BC旋转a1/3，求出<ACB的大小a1，然后将向量CB旋转a2/3，然后求两线段的交点D即可，E、F同理。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
using namespace std;

struct Point
{
	double x, y;
	Point(double x = 0, double y = 0) : x(x), y(y) { }
};

typedef Point Vector;

Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); }

Vector operator - (Point A, Point B) { return Vector(A.x-B.x, A.y-B.y); }

Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }

Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }

bool operator < (const Point &a, const Point &b)
{
	if(a.x != b.x) return a.x < b.x;
	return a.y < b.y;
}

const double eps = 1e-10;

int dcmp(double x)
{
	if(fabs(x) < eps) return 0; return x < 0 ? -1 : 1;
}

bool operator == (const Point &a, const Point &b)
{
	return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
}

double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }

double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
double Area2(Point A, Point B, Point C) { return Cross(B-A, C-A); }

Vector Rotate(Vector A, double rad)
{
	return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
}

Point GetIntersection(Point P, Vector v, Point Q, Vector w)
{
	Vector u = P-Q;
	double t = Cross(w, u) / Cross(v, w);
	return P+v*t;
}

Point A, B, C, D, E, F;

Point read_point()
{
	Point A;
	scanf("%lf%lf", &A.x, &A.y);
	return A;
}

Point getD(Point A, Point B, Point C)
{
	Vector v1 = C-B, v2 = A-B;
	double a1 = Angle(v1, v2);
	v1 = Rotate(v1, a1/3);
	
	Vector v3 = B-C, v4 = A-C;
	double a2 = Angle(v3, v4);
	v3 = Rotate(v3, -a2/3);
	return GetIntersection(B, v1, C, v3);
}

void solve()
{
	A = read_point();
	B = read_point();
	C = read_point();
	D = getD(A, B, C);
	E = getD(B, C, A);
	F = getD(C, A, B);
	printf("%.6lf %.6lf %.6lf %.6lf %.6lf %.6lf\n", D.x, D.y, E.x, E.y, F.x, F.y);
}

int main()
{
	int T;
	scanf("%d", &T);
	while(T--)
	{
		solve();
	}
	return 0;
}