两道关于三角形的题

葡萄糖

1.一个三角形的内心到顶点的距离分别是x,y,z ，求该三角形的面积？？

倪举鹏

内心o, 三点（-5,0）（4cos(x),4sin(x)）,(3cos(y),3sin(y))  用向量夹角  列两个二元方程组

zuijianqiugen

参见《有关三角形内心、旁心、外心的半径方程  》
http://zuijianqiugen.blog.163.co ... 622014219104231469/

zuijianqiugen

参见《有关三角形内心、旁心、外心的半径方程》
http://zuijianqiugen.blog.163.co ... 622014219104231469/

数学星空

1. 若内心\(I\)到三顶点的距离为\(x,y,z\),则\(x\sin(\frac{A}{2})=r,y\sin(\frac{B}{2})=r,z\sin(\frac{C}{2})=r\)
   
   楼上所得到的半径方程：\(\frac{1}{r^3}=(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2})\frac{1}{r}+\frac{2}{xyz}\)

   本质就是\(\sin(\frac{A}{2})^2+\sin(\frac{B}{2})^2+\sin(\frac{C}{2})^2+2\sin(\frac{B}{2})\sin(\frac{C}{2})\sin(\frac{A}{2})-1\)

   \(＝\sin(\frac{A}{2})^2+\sin(\frac{B}{2})^2+（\cos(\frac{A}{2})\cos(\frac{B}{2})-\sin(\frac{A}{2})\sin(\frac{B}{2})^2+2\sin(\frac{B}{2})\sin(\frac{C}{2})（\cos(\frac{A}{2})\cos(\frac{B}{2})-\sin(\frac{A}{2})\sin(\frac{B}{2})\)
  
   \(＝\sin(\frac{A}{2})^2+\sin(\frac{B}{2})^2+（1-\sin(\frac{A}{2})^2)(1-\sin(\frac{B}{2})^2)-\sin(\frac{A}{2})^2\sin(\frac{B}{2})^2-1=0\)


2.若旁心\(I_1\)到三顶点的距离为\(x,y,z\),则\(x\sin(\frac{A}{2})=r,y\cos(\frac{B}{2})=r,z\cos(\frac{C}{2})=r\)

  楼上所得到的半径方程：\(\frac{1}{r^3}=(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2})\frac{1}{r}-\frac{2}{xyz}\)

  本质就是\(\sin(\frac{A}{2})^2+\cos(\frac{B}{2})^2+\cos(\frac{C}{2})^2+2\sin(\frac{B}{2})\cos(\frac{C}{2})\cos(\frac{A}{2})-1\)

   \(＝\sin(\frac{A}{2})^2+\cos(\frac{B}{2})^2+（\sin(\frac{A}{2})\cos(\frac{B}{2})+\cos(\frac{A}{2})\sin(\frac{B}{2})^2+2\sin(\frac{B}{2})\cos(\frac{C}{2})（\sin(\frac{A}{2})\cos(\frac{B}{2})+\cos(\frac{A}{2})\sin(\frac{B}{2})\)
  
   \(＝\sin(\frac{A}{2})^2+\cos(\frac{B}{2})^2+（1-\sin(\frac{A}{2})^2)(1-\cos(\frac{B}{2})^2)-2\sin(\frac{A}{2})^2\cos(\frac{B}{2})^2-1=0\)


3.对于锐角三角形\(\triangle ABC\)若外心\(O\)到三边的距离为\(x,y,z\),则\(x=R\cos(A),y=R\cos(B),y=R\cos(B)\)

  楼上所得到的半径方程：\(R^3=(x^2+y^2+z^2)R-2xyz\)

  本质就是\(\cos(A)^2+\cos(B)^2+\cos(C)^2-2\cos(A)\cos(B)\cos(C)-1\)

   \(＝\cos(A)^2+\cos(B)^2+(\cos(A)\cos(B)-\sin(A)\sin(B))^2－2\cos(A)\cos(B)((\cos(A)\cos(B)-\sin(A)\sin(B))-1\)
  
   \(＝\cos(A)^2+\cos(B)^2+\cos(A)^2\cos(B)^2+(1-cos(A)^2)(1-cos(B)^2)-2cos(A)^2cos(B)^2-1=0\)


4.对于钝角三角形\(\triangle ABC\)(设\( \angle A>90^\circ \)),若外心\(O\)到三边的距离为\(x,y,z\),则\(-x=R\cos(A),y=R\cos(B),y=R\cos(B)\)

  楼上所得到的半径方程：\(R^3=(x^2+y^2+z^2)R+2xyz\)

  本质就是\(\cos(A)^2+\cos(B)^2+\cos(C)^2-2\cos(A)\cos(B)\cos(C)-1\)

   \(＝\cos(A)^2+\cos(B)^2+(\cos(A)\cos(B)-\sin(A)\sin(B))^2-2\cos(A)\cos(B)((\cos(A)\cos(B)-\sin(A)\sin(B))-1\)
  
   \(＝\cos(A)^2+\cos(B)^2+\cos(A)^2\cos(B)^2+(1-cos(A)^2)(1-cos(B)^2)-2cos(A)^2cos(B)^2-1=0\)

数学星空

若内心\(I\)到三顶点的距离为\(x,y,z\),面积为\(s\),周长\(L\),则有结论：

  1.  \(16x^4y^4z^4s^6+(xy+xz+yz)(xy+xz-yz)(xy-xz-yz)(xy-xz+yz)(x^2y^2+x^2z^2+y^2z^2)^2s^4-2x^2y^2z^2(x^8y^6-x^8y^4z^2-x^8y^2z^4+x^8z^6+x^6y^8-4x^6y^6z^2+4x^6y^4z^4-4x^6y^2z^6+x^6z^8-x^4y^8z^2+4x^4y^6z^4+4x^4y^4z^6-x^4y^2z^8-x^2y^8z^4-4x^2y^6z^6-x^2y^4z^8+y^8z^6+y^6z^8)s^2+x^4z^4y^4(y-z)^2(y+z)^2(x-z)^2(x+z)^2(x-y)^2(x+y)^2＝0\)

  2.  \(y^2x^2z^2L^6+(x^4y^4-10x^4y^2z^2+x^4z^4-10x^2y^4z^2-10x^2y^2z^4+y^4z^4)L^4+(-8x^6y^4+32x^6y^2z^2-8x^6z^4-8x^4y^6-16x^4y^4z^2-16x^4y^2z^4-8x^4z^6+32x^2y^6z^2-16x^2y^4z^4+32x^2y^2z^6-8y^6z^4-8y^4z^6)L^2+16(y-z)^2(y+z)^2(x-z)^2(x+z)^2(x-y)^2(x+y)^2＝0\)

若外心\(O\)到三角形\(\triangle ABC\)三边的距离为\(x,y,z\)，面积为\(s\),周长\(L\),则有结论：
  
  1.  \(L^6-12(z+x+y)^2L^4+(-64x^4+64x^3y+64x^3z+64x^2y^2-128x^2yz+64x^2z^2+64xy^3-128xy^2z-128xyz^2+64xz^3-64y^4+64y^3z+64y^2z^2+64yz^3-64z^4)L^2+256(y-z)^2(x-z)^2(x-y)^2＝0\)

  2.  \(s^6+(x^4-10x^2y^2-10x^2z^2+y^4-10y^2z^2+z^4)s^4+(-8x^6y^2-8x^6z^2+32x^4y^4-16x^4y^2z^2+32x^4z^4-8x^2y^6-16x^2y^4z^2-16x^2y^2z^4-8x^2z^6-8y^6z^2+32y^4z^4-8y^2z^6)s^2+16(y-z)^2(y+z)^2(x-z)^2(x+z)^2(x-y)^2(x+y)^2＝0\)