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【公式编辑测试】解析几何-三角形外心重心垂心内心公式

2021-07-19 19:31:00 用户名、


坐标原点 o o o,三角形外心 O O O,重心 G G G,垂心 H H H,内心 I I I
来自老早以前的纸笔记本上,不保证正确性

三角形重心 G G G

向量

o G ⃗ = 1 3 ( o A ⃗ + o B ⃗ + o C ⃗ ) \vec{oG}=\frac{1}{3}(\vec{oA}+\vec{oB}+\vec{oC}) oG =31(oA +oB +oC )

直角坐标

x G = 1 3 ( x 1 + x 2 + x 3 ) x_G=\frac{1}{3}({x_1+x_2+x_3}) xG=31(x1+x2+x3)
y G = 1 3 ( y 1 + y 2 + y 3 ) y_G=\frac{1}{3}({y_1+y_2+y_3}) yG=31(y1+y2+y3)

三角形外心 O O O

向量

s i n ( 2 A ) ⋅ O A ⃗ + s i n ( 2 B ) ⋅ O B ⃗ + + s i n ( 2 C ) ⋅ O C ⃗ = 0 ⃗ sin(2A)\cdot\vec{OA}+sin(2B)\cdot\vec{OB}++sin(2C)\cdot\vec{OC}=\vec{0} sin(2A)OA +sin(2B)OB ++sin(2C)OC =0

o O ⃗ = s i n ( 2 A ) ⋅ o A ⃗ + s i n ( 2 B ) ⋅ o B ⃗ + + s i n ( 2 C ) ⋅ o C ⃗ s i n ( 2 A ) + s i n ( 2 B ) + s i n ( 2 C ) \vec{oO}=\frac{sin(2A)\cdot\vec{oA}+sin(2B)\cdot\vec{oB}++sin(2C)\cdot\vec{oC}}{sin(2A)+sin(2B)+sin(2C)} oO =sin(2A)+sin(2B)+sin(2C)sin(2A)oA +sin(2B)oB ++sin(2C)oC

直角坐标

x O = ∣ x 2 2 − x 3 2 2 + y 2 2 − y 3 2 2 y 2 − y 3 x 3 2 − x 1 2 2 + y 3 2 − y 1 2 2 y 3 − y 1 ∣ ∣ x 2 − x 3 y 2 − y 3 x 3 − x 1 y 3 − y 1 ∣ = [ x 1 2 ( y 2 − y 3 ) + x 2 2 ( y 3 − y 1 ) + x 3 2 ( y 1 − y 2 ) ] − ( y 1 − y 2 ) ( y 2 − y 3 ) ( y 3 − y 1 ) 2 [ x 1 ( y 2 − y 3 ) + x 2 ( y 3 − y 1 ) + x 3 ( y 1 − y 2 ) ] \begin{aligned} x_O& =\frac{ \left|\begin{array}{cccc} \frac{x_2^2-x_3^2}{2}+\frac{y_2^2-y_3^2}{2} & y_2-y_3 \\ \frac{x_3^2-x_1^2}{2}+\frac{y_3^2-y_1^2}{2} & y_3-y_1 \\ \end{array}\right| }{ \left|\begin{array}{cccc} x_2-x_3 & y_2-y_3 \\ x_3-x_1 & y_3-y_1 \\ \end{array}\right| } \\ &=\frac{[x_1^2(y_2-y_3)+x_2^2(y_3-y_1)+x_3^2(y_1-y_2)]-(y_1-y_2)(y_2-y_3)(y_3-y_1)} {2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]} \end{aligned} xO=x2x3x3x1y2y3y3y12x22x32+2y22y322x32x12+2y32y12y2y3y3y1=2[x1(y2y3)+x2(y3y1)+x3(y1y2)][x12(y2y3)+x22(y3y1)+x32(y1y2)](y1y2)(y2y3)(y3y1)

y O = ∣ x 2 − x 3 x 2 2 − x 3 2 2 + y 2 2 − y 3 2 2 x 3 − x 1 x 3 2 − x 1 2 2 + y 3 2 − y 1 2 2 ∣ ∣ x 2 − x 3 y 2 − y 3 x 3 − x 1 y 3 − y 1 ∣ = − [ y 1 2 ( x 2 − x 3 ) + y 2 2 ( x 3 − x 1 ) + y 3 2 ( x 1 − x 2 ) ] + ( x 1 − x 2 ) ( x 2 − x 3 ) ( x 3 − x 1 ) 2 [ x 1 ( y 2 − y 3 ) + x 2 ( y 3 − y 1 ) + x 3 ( y 1 − y 2 ) ] \begin{aligned} y_O&=\frac{ \left|\begin{array}{cccc} x_2-x_3 & \frac{x_2^2-x_3^2}{2}+\frac{y_2^2-y_3^2}{2} \\ x_3-x_1 & \frac{x_3^2-x_1^2}{2}+\frac{y_3^2-y_1^2}{2} \\ \end{array}\right| }{ \left|\begin{array}{cccc} x_2-x_3 & y_2-y_3 \\ x_3-x_1 & y_3-y_1 \\ \end{array}\right| } \\ &=\frac{-[y_1^2(x_2-x_3)+y_2^2(x_3-x_1)+y_3^2(x_1-x_2)]+(x_1-x_2)(x_2-x_3)(x_3-x_1)} {2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]} \end{aligned} yO=x2x3x3x1y2y3y3y1x2x3x3x12x22x32+2y22y322x32x12+2y32y12=2[x1(y2y3)+x2(y3y1)+x3(y1y2)][y12(x2x3)+y22(x3x1)+y32(x1x2)]+(x1x2)(x2x3)(x3x1)

三角形垂心 H H H

向量

0 ⃗ = t a n ( A ) ⋅ H A ⃗ + t a n ( B ) ⋅ H B ⃗ + t a n ( C ) ⋅ H C ⃗ \vec{0}=tan(A)\cdot \vec{HA}+tan(B)\cdot \vec{HB}+tan(C)\cdot \vec{HC} 0 =tan(A)HA +tan(B)HB +tan(C)HC

o H ⃗ = t a n ( A ) ⋅ o A ⃗ + t a n ( B ) ⋅ o B ⃗ + t a n ( C ) ⋅ o C ⃗ t a n ( A ) + t a n ( B ) + t a n ( C ) \vec{oH}=\frac{tan(A)\cdot \vec{oA}+tan(B)\cdot \vec{oB}+tan(C)\cdot \vec{oC}}{tan(A)+tan(B)+tan(C)} oH =tan(A)+tan(B)+tan(C)tan(A)oA +tan(B)oB +tan(C)oC

直角坐标

x H = ( y 1 − y 2 ) ( y 2 − y 3 ) ( y 3 − y 1 ) − [ x 1 x 2 ( y 1 − y 2 ) + x 2 x 3 ( y 2 − y 3 ) + x 3 x 1 ( y 3 − y 1 ) ] x 3 ( y 1 − y 2 ) + x 1 ( y 2 − y 3 ) + x 2 ( y 3 − y 1 ) x_H=\frac{(y_1-y_2)(y_2-y_3)(y_3-y_1)-[x_1x_2(y_1-y_2)+x_2x_3(y_2-y_3)+x_3x_1(y_3-y_1)]} {x_3(y_1-y_2)+x_1(y_2-y_3)+x_2(y_3-y_1)} xH=x3(y1y2)+x1(y2y3)+x2(y3y1)(y1y2)(y2y3)(y3y1)[x1x2(y1y2)+x2x3(y2y3)+x3x1(y3y1)]

y H = − ( x 1 − x 2 ) ( x 2 − x 3 ) ( x 3 − x 1 ) + [ y 1 y 2 ( x 1 − x 2 ) + y 2 y 3 ( x 2 − x 3 ) + y 3 y 1 ( x 3 − x 1 ) ] x 3 ( y 1 − y 2 ) + x 1 ( y 2 − y 3 ) + x 2 ( y 3 − y 1 ) y_H=\frac{-(x_1-x_2)(x_2-x_3)(x_3-x_1)+[y_1y_2(x_1-x_2)+y_2y_3(x_2-x_3)+y_3y_1(x_3-x_1)]} {x_3(y_1-y_2)+x_1(y_2-y_3)+x_2(y_3-y_1)} yH=x3(y1y2)+x1(y2y3)+x2(y3y1)(x1x2)(x2x3)(x3x1)+[y1y2(x1x2)+y2y3(x2x3)+y3y1(x3x1)]

三角形内心 I I I

向量

0 ⃗ = a ⋅ I A ⃗ + b ⋅ I B ⃗ + c ⋅ I C ⃗ \vec{0}=a\cdot \vec{IA}+b\cdot \vec{IB}+c\cdot \vec{IC} 0 =aIA +bIB +cIC

o I ⃗ = a ⋅ o A ⃗ + b ⋅ o B ⃗ + c ⋅ o C ⃗ a + b + c \vec{oI}=\frac{a\cdot \vec{oA}+b\cdot \vec{oB}+c\cdot \vec{oC}}{a+b+c} oI =a+b+caoA +boB +coC

直角坐标

x I = a x A + b x B + c x C a + b + c x_I=\frac{ax_A+bx_B+cx_C}{a+b+c} xI=a+b+caxA+bxB+cxC

y I = a y A + b y B + c y C a + b + c y_I=\frac{ay_A+by_B+cy_C}{a+b+c} yI=a+b+cayA+byB+cyC

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本文为[用户名、]所创,转载请带上原文链接,感谢
https://blog.51cto.com/u_15247503/2870087

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