三角型的性质

三角形顶点坐标:A(x1; y1);B(x2; y2);C(x3; y3);

④重心G(x4;y4);x4=(x1+x2+x3)/3;y4=(y1+y2+y3)/3;

⑤外心W(x5;y5);根据外心到各顶点的距离相等:AG=BG;AG=CG;即:

Sqrt[(x1 - x5)^2 + (y1 - y5)^2] == Sqrt[(x2 - x5)^2 + (y2 - y5)^2],

Sqrt[(x1 - x5)^2 + (y1 - y5)^2] == Sqrt[(x3 - x5)^2 + (y3 - y5)^2]

解得:x5 = (x2^2 y1 - x3^2 y1 - x1^2 y2 + x3^2 y2 - y1^2 y2 + y1 y2^2 + x1^2 y3 - x2^2 y3 + y1^2 y3 - y2^2 y3 - y1 y3^2 + y2 y3^2)/(2 (x2 y1 - x3 y1 - x1 y2 + x3 y2 + x1 y3 - x2 y3));

y5 = -(-x1^2 x2 + x1 x2^2 + x1^2 x3 - x2^2 x3 - x1 x3^2 + x2 x3^2 - x2 y1^2 + x3 y1^2 + x1 y2^2 - x3 y2^2 - x1 y3^2 + x2 y3^2)/(2 (x2 y1 - x3 y1 - x1 y2 + x3 y2 + x1 y3 - x2 y3));

⑥内心N(x6;y6);根据内心到各边的距离相等:先求内心到各边垂线垂足与顶点的距离;

1/2 (Sqrt[(x1 - x2)^2 + (y1 - y2)^2] + Sqrt[(x1 - x3)^2 + (y1 - y3)^2] - Sqrt[(x2 - x3)^2 + (y2 - y3)^2]);

1/2 (Sqrt[(x1 - x2)^2 + (y1 - y2)^2] - Sqrt[(x1 - x3)^2 + (y1 - y3)^2] + Sqrt[(x2 - x3)^2 + (y2 - y3)^2]);

1/2 (-Sqrt[(x1 - x2)^2 + (y1 - y2)^2] + Sqrt[(x1 - x3)^2 + (y1 - y3)^2] + Sqrt[(x2 - x3)^2 + (y2 - y3)^2]);

计算内心到个顶点的距离;根据勾股定理计算内心到各边的距离,根据距离相等列方程:

(x1 - x6)^2 - 1/4 (Sqrt[(x1 - x2)^2 + (y1 - y2)^2] + Sqrt[(x1 - x3)^2 + (y1 - y3)^2] - Sqrt[(x2 - x3)^2 + (y2 - y3)^2])^2 + (y1 - y6)^2 == (x2 - x6)^2 - 1/4 (Sqrt[(x1 - x2)^2 + (y1 - y2)^2] - Sqrt[(x1 - x3)^2 + (y1 - y3)^2] + Sqrt[(x2 - x3)^2 + (y2 - y3)^2])^2
+ (y2 - y6)^2,(x1 - x6)^2 - 1/4 (Sqrt[(x1 - x2)^2 + (y1 - y2)^2] + Sqrt[(x1 - x3)^2 + (y1 - y3)^2] - Sqrt[(x2 - x3)^2 + (y2 - y3)^2])^2 + (y1 - y6)^2 == (x3 - x6)^2 - 1/4 (-Sqrt[(x1 - x2)^2 + (y1 - y2)^2] + Sqrt[(x1 - x3)^2 + (y1 - y3)^2] + Sqrt[(x2 - x3)^2
+ (y2 - y3)^2])^2 + (y3 - y6)^2

解得:x6 = (x2^2 y1 - x3^2 y1 - x1^2 y2 + x3^2 y2 - y1^2 y2 + y1 y2^2 + x1^2 y3 - x2^2 y3 + y1^2 y3 - y2^2 y3 - y1 y3^2 + y2 y3^2 + y2 Sqrt[x1^2 - 2 x1 x2 + x2^2 + y1^2 - 2 y1 y2 + y2^2] Sqrt[x1^2 - 2 x1 x3 + x3^2 + y1^2 - 2 y1 y3 + y3^2] - Sqrt[x1^2 - 2 x1 x2
+ x2^2 + y1^2 - 2 y1 y2 + y2^2] y3 Sqrt[x1^2 - 2 x1 x3 + x3^2 + y1^2 - 2 y1 y3 + y3^2] - y1 Sqrt[x1^2 - 2 x1 x2 + x2^2 + y1^2 - 2 y1 y2 + y2^2] Sqrt[x2^2 - 2 x2 x3 + x3^2 + y2^2 - 2 y2 y3 + y3^2] + Sqrt[x1^2 - 2 x1 x2 + x2^2 + y1^2 - 2 y1 y2 + y2^2] y3 Sqrt[x2^2
- 2 x2 x3 + x3^2 + y2^2 - 2 y2 y3 + y3^2] + y1 Sqrt[x1^2 - 2 x1 x3 + x3^2 + y1^2 - 2 y1 y3 + y3^2] Sqrt[x2^2 - 2 x2 x3 + x3^2 + y2^2 - 2 y2 y3 + y3^2] - y2 Sqrt[x1^2 - 2 x1 x3 + x3^2 + y1^2 - 2 y1 y3 + y3^2] Sqrt[x2^2 - 2 x2 x3 + x3^2 + y2^2 - 2 y2 y3 + y3^2])/(2
(x2 y1 - x3 y1 - x1 y2 + x3 y2 + x1 y3 - x2 y3));

y6 = -(-x1^2 x2 + x1 x2^2 + x1^2 x3 - x2^2 x3 - x1 x3^2 + x2 x3^2 - x2 y1^2 + x3 y1^2 + x1 y2^2 - x3 y2^2 - x1 y3^2 + x2 y3^2 + x2 Sqrt[x1^2 - 2 x1 x2 + x2^2 + y1^2 - 2 y1 y2 + y2^2] Sqrt[x1^2 - 2 x1 x3 + x3^2 + y1^2 - 2 y1 y3 + y3^2] - x3 Sqrt[x1^2 - 2 x1
x2 + x2^2 + y1^2 - 2 y1 y2 + y2^2] Sqrt[x1^2 - 2 x1 x3 + x3^2 + y1^2 - 2 y1 y3 + y3^2] - x1 Sqrt[x1^2 - 2 x1 x2 + x2^2 + y1^2 - 2 y1 y2 + y2^2] Sqrt[x2^2 - 2 x2 x3 + x3^2 + y2^2 - 2 y2 y3 + y3^2] + x3 Sqrt[x1^2 - 2 x1 x2 + x2^2 + y1^2 - 2 y1 y2 + y2^2] Sqrt[x2^2
- 2 x2 x3 + x3^2 + y2^2 - 2 y2 y3 + y3^2] + x1 Sqrt[x1^2 - 2 x1 x3 + x3^2 + y1^2 - 2 y1 y3 + y3^2] Sqrt[x2^2 - 2 x2 x3 + x3^2 + y2^2 - 2 y2 y3 + y3^2] - x2 Sqrt[x1^2 - 2 x1 x3 + x3^2 + y1^2 - 2 y1 y3 + y3^2] Sqrt[x2^2 - 2 x2 x3 + x3^2 + y2^2 - 2 y2 y3 + y3^2])/(2
(x2 y1 - x3 y1 - x1 y2 + x3 y2 + x1 y3 - x2 y3));

⑦垂心H(x7;y7);分别做高线:

AH⊥BC;BH⊥AC;(y1 - y7)/(x1 - x7) (y2 - y3)/(x2 - x3) == -1,(y2 - y7)/(x2 - x7) (y1 - y3)/(x1 - x3) == -1

解得:x7 = -(x1 x2 y1 - x1 x3 y1 - x1 x2 y2 + x2 x3 y2 + y1^2 y2 - y1 y2^2 + x1 x3 y3 - x2 x3 y3 - y1^2 y3 + y2^2 y3 + y1 y3^2 - y2 y3^2)/(-x2 y1 + x3 y1 + x1 y2 - x3 y2 - x1 y3 + x2 y3);

y7 = -(x1^2 x2 - x1 x2^2 - x1^2 x3 + x2^2 x3 + x1 x3^2 - x2 x3^2 + x1 y1 y2 - x2 y1 y2 - x1 y1 y3 + x3 y1 y3 + x2 y2 y3 - x3 y2 y3)/(x2 y1 - x3 y1 - x1 y2 + x3 y2 + x1 y3 - x2 y3);

外接圆圆心

a=((x1*x1+y1*y1-x2*x2-y2*y2)*(y1-y3)-(x1*x1+y1*y1-x3*x3-y3*y3)*(y1-y2))/(2.0*((y1-y3)*(x1-x2)-(y1-y2)*(x1-x3)));

b=((x1*x1+y1*y1-x2*x2-y2*y2)*(x1-x3)-(x1*x1+y1*y1-x3*x3-y3*y3)*(x1-x2))/(2.0*((x1-x3)*(y1-y2)-(x1-x2)*(y1-y3)));