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 sixintegers?.?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.

?

Output

For each line of input you should produce one line of output. This line contains six floating point numbers??separated by a single space. These six floating-point actually means that the Cartesian coordinates of D, E and F are??respectively. Errors less than???will be accepted.

?

Sample Input???Output for Sample Input

2 1 1 2 2 1 2 0 0 100 0 50 50

1.316987 1.816987 1.183013 1.683013 1.366025 1.633975 56.698730 25.000000 43.301270 25.000000 50.000000 13.397460 ?

---

題意：作三角形ABC每個內角的三等分線，相交成三角形DEF，則DEF是等邊三角形。給出A、B、C 3個的位置確定D、E、F 3個點的位置。

分析：本題沒什么算法可言，只要根據題意計算。考慮到對稱性，只需要知道如何求D點即可。首先需要計算角ABC的值a，然后把射線BC逆時針旋轉a/3，得到直線BD，同理可以得到直線CD，求出交點即可。

#include<cstdio> #include<cmath> #include<algorithm> 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) {return a.x < b.x || (a.x == b.x && 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; } //點乘 double Length(Vector A) { return sqrt(Dot(A, A)); } //向量的模 double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); } //兩個向量的夾角 double Cross(Vector A, Vector B) { return A.x * B.y - A.y * B.x; } //叉乘 double Area(Point A, Point B, Point C) { return Cross(B - A, C - A); } //三個點組成的三角形的面積 Vector Rotate(Vector A, double rad) { //向量A逆時針旋轉rad弧度后的坐標return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad)); }//求直線交點 Point GetLineIntersection(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 getD(Point A, Point B, Point C) {Vector v1 = C - B;double a1 = Angle(A-B, v1);v1 = Rotate(v1, a1/3);Vector v2 = B - C;double a2 = Angle(A-C, v2);v2 = Rotate(v2, -a2/3);return GetLineIntersection(B, v1, C, v2); } int main() {int T;Point A, B, C, D, E, F;scanf("%d",&T);while(T--) {scanf("%lf%lf%lf%lf%lf%lf",&A.x, &A.y, &B.x, &B.y, &C.x, &C.y);D = getD(A, B, C);E = getD(B, C, A);F = getD(C, A, B);printf("%lf %lf %lf %lf %lf %lf\n", D.x, D.y, E.x, E.y, F.x, F.y);}return 0; }

總結

以上是生活随笔為你收集整理的UVA 11178 Morley’s Theorem（莫雷定理 计算几何）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。