关于三角形内心的向量等式

realnumber

三角形ABC,A,B,C对应的三边为a,b,c,P是三角形ABC所在平面内一点,求证:$a\vv{PA}+b\vv{PB}+c\vv{PC}=\vv{0}$成立的充要条件是"P是三角形ABC的内心".
   

hbghlyj

$a\vv{PA}+b\vv{PB}+c\vv{PC}=\vv{0}$
$⇔a\vv{OA}+b\vv{OB}+c\vv{OC}=(a+b+c)\vv{OP}$
$⇔\vv{OP}=\frac{a\vv{OA}+b\vv{OB}+c\vv{OC}}{a+b+c}$


平面$ABC$上任意一点$P$可以表成
$\vv{OP}=p\vv{OA}+q\vv{OB}+r\vv{OC},\;p+q+r=1$  (barycentric coordinate system)
$p:q:r=\S{PBC}:\S{PCA}:\S{PAB}$   (有向面积)

当$P$为$A$, $p:q:r=1:0:0$
当$P$为$BC$中点, $p:q:r=0:1:1$
当$P$为重心, $p:q:r=1:1:1$	当$P$为$A$关于$BC$的对称点, $p:q:r=-1:1:1$
当$P$为内心, $p:q:r=a:b:c$	当$P$为$A$-旁心, $p:q:r=-a:b:c$
当$P$为共轭重心, $p:q:r=a^2:b^2:c^2$	当$P$为$A$-旁共轭重心, $p:q:r=-a^2:b^2:c^2$
⋯	⋯

realnumber

充分性证明
设N为此内切圆与AB的切点
$a\vv{PA}+b\vv{PB}+c\vv{PC}=-(a+b+c)\vv{AP}+b\vv{AB}+c\vv{AC}$ 以下证明这个向量和$\vv{AB}$作数量积为0.
$\vv{AP}\cdot \vv{AB}=(\vv{AN}+\vv{NP})\cdot \vv{AB}=\vv{AN}\cdot \vv{AB}=0.5(b+c-a)c$
$(b\vv{AB}+c\vv{AC})\cdot \vv{AB}=bc^2+bc^2\cos A$ ，
这样$(a\vv{PA}+b\vv{PB}+c\vv{PC})\cdot \vv{AB}$化简可得0,同样与$\vv{AC}$数量积也为0，
因此  $ a\vv{PA}+b\vv{PB}+c\vv{PC}=\vv{0}$成立.

必要性证明

可证$\frac{\vv{AP} \cdot \vv{AB}}{\abs{\vv{AP}}\abs{\vv{AB}}}=\frac{\vv{AP}\cdot\vv{AC}}{\abs{\vv{AP}}\abs{\vv{AC}}}$,即角A被AP平分.同样可证被B分别BP平分.

hbghlyj

$\vv{OP}=\frac{a\vv{OA}+b\vv{OB}+c\vv{OC}}{a+b+c}$可以唯一确定点$P$

hbghlyj

realnumber 发表于 2022-6-22 04:02
链接打不开

搬运一下.

To explain why these coordinates are signed ratios of areas, let us assume that we work in the Euclidean space $\mathbf{E}^{3}$. Here, consider the basis, namely $\{\mathbf{i},\mathbf{j},\mathbf{k}\}$. Consider also the positively oriented triangle $ABC$ lying in the $Oxy$ plane. It is known that for any basis $\{\mathbf{e},\mathbf{f},\mathbf{g}\}$ of $\mathbf{E}^{3}$ and any free vector $\mathbf{h}$ one has

$$\mathbf{h}=\frac{1}{(\mathbf{e},\mathbf{f},\mathbf{g})}\cdot\left[(\mathbf{h},\mathbf{f},\mathbf{g})\mathbf{e}+(\mathbf{e},\mathbf{h},\mathbf{g})\mathbf{f}+(\mathbf{e},\mathbf{f},\mathbf{h})\mathbf{g}\right],$$

where $(\mathbf{e},\mathbf{f},\mathbf{g})=(\mathbf{e}\times\mathbf{f})\cdot\mathbf{g}$ stands for the mixed product of these three vectors.

Take $\mathbf{e}=\vec{AB},\,\mathbf{f}=\vec{AC},\,\mathbf{g}=\mathbf{k},\,\mathbf{h}=\vec{AP}$, where $P$ is an arbitrary point in the plane $Oxy$, and remark that

$$(\mathbf{e},\mathbf{f},\mathbf{h})=(\vec{AB}\times\vec{AC})\cdot\vec{AP}=(\vert\vec{AB}\times\vec{AC}\vert\mathbf{k})\cdot\vec{AP} = 0.$$

A subtle point regarding our choice of free vectors: $\mathbf{e}$ is, in fact, the equipollence class of the bound vector $\vec{AB}$.

We have obtained that

$$\vec{AP}=m_B\cdot\vec{AB}+m_C\cdot\vec{AC},\,\mbox{ where }\,m_B=\frac{(\vec{AP},\vec{AC},\mathbf{k})}{(\vec{AB},\vec{AC},\mathbf{k})},\,m_C=\frac{(\vec{AB},\vec{AP},\mathbf{k})}{(\vec{AB},\vec{AC},\mathbf{k})}.$$

Given the positive (counterclockwise) orientation of triangle $ABC$, the denominator of both $m_B$ and $m_C$ is precisely the double of the area of the triangle $ABC$. Also,

$$(\vec{AP},\vec{AC},\mathbf{k})=(\vec{PC},\vec{PA},\mathbf{k})\,\mbox{ and }\,(\vec{AB},\vec{AP},\mathbf{k})=(\vec{PA},\vec{PB},\mathbf{k})$$

and so the numerators of $m_B$ and $m_C$ are the doubles of the signed areas of triangles $APC$ and respectively $ABP$.

Further, we deduce that

$$\vec{OP}=(1-m_B-m_C)\cdot\vec{OA}+m_B\cdot\vec{OB}+m_C\cdot\vec{OC}$$

which means that the numbers $1-m_B-m_C$, $m_B$ and $m_C$ are the barycentric coordinates of $P$. Similarly, the third barycentric coordinate reads as

$$m_A = 1 - m_B - m_C = \frac{(\vec{PB},\vec{PC},\mathbf{k})}{(\vec{AB},\vec{AC},\mathbf{k})}.$$

This $m$-letter notation of the barycentric coordinates comes from the fact that the point $P$ may be interpreted as the center of mass for the masses $m_A$, $m_B$, $m_C$ which are located in $A$, $B$ and $C$.

Switching back and forth between the barycentric coordinates and other coordinate systems makes some problems much easier to solve.

hbghlyj

对于$\Bbb R^3$的任意一组基${\bf e},{\bf f},{\bf g}$,对于任意一点${\bf h}∈\Bbb R^3$存在$p,q,r∈\Bbb R$使得$${\bf h}=p{\bf e}+q{\bf f}+r{\bf g}$$
两边做混合积 $(·,{\bf f},{\bf g})$ ,由混合积的性质,$$p({\bf e},{\bf f},{\bf g})=({\bf h},{\bf f},{\bf g})$$
所以$$p=\frac{(\mathbf{h},\mathbf{f},\mathbf{g})}{(\mathbf{e},\mathbf{f},\mathbf{g})}$$
于是
$$\mathbf{h}=\frac{1}{(\mathbf{e},\mathbf{f},\mathbf{g})}\cdot\left[(\mathbf{h},\mathbf{f},\mathbf{g})\mathbf{e}+(\mathbf{e},\mathbf{h},\mathbf{g})\mathbf{f}+(\mathbf{e},\mathbf{f},\mathbf{h})\mathbf{g}\right]$$

其实这就是3×3的Cramer法则.
那个混合积的性质就是行列式的性质:
     当两列相同时,行列式为0: $({\bf f},{\bf f},{\bf g})=({\bf g},{\bf f},{\bf g})=0$
     行列式的线性: $(p{\bf e}+q{\bf f}+r{\bf g},{\bf f},{\bf g})=p({\bf e},{\bf f},{\bf g})+q({\bf f},{\bf f},{\bf g})+r({\bf g},{\bf f},{\bf g})$
或者可以展开为外积,内积的性质.

hbghlyj

另一种理解:$$\{{\bf e}^1,{\bf e}^2,{\bf e}^3\}=\left\{{\mathbf{f}\times\mathbf{g}\over (\mathbf{e},\mathbf{f},\mathbf{g})},{\mathbf{g}\times\mathbf{e}\over (\mathbf{e},\mathbf{f},\mathbf{g})},{\mathbf{e}\times\mathbf{f}\over (\mathbf{e},\mathbf{f},\mathbf{g})}\right\}$$是$\{{\bf e},{\bf f},{\bf g}\}$的对偶基[序号相同的向量的内积为1,序号不同的向量互相正交,即${\bf e}^i·{\bf e}_j=δ_{ij}$]

于是,把${\bf h}=p{\bf e}+q{\bf f}+r{\bf g}$两边和${\bf e}^1$作内积,得$${\bf e}^1·{\bf h}=p$$ 也可以理解为:
把$\{{\bf e}^1,{\bf e}^2,{\bf e}^3\}$作为列向量的矩阵是把$\{{\bf e},{\bf f},{\bf g}\}$作为列向量的矩阵的逆,即$[{\bf e}^1,{\bf e}^2,{\bf e}^3]=[{\bf e}_1,{\bf e}_2,{\bf e}_3]^{-1}$.

In 3-dimensional Euclidean space, for a given basis {e₁, e₂, e₃}, you can find the biorthogonal (dual) basis {e¹, e², e³} by formulas below:

$$\mathbf {e} ^{1}=\left({\frac {\mathbf {e} _{2}\times \mathbf {e} _{3}}{V}}\right)^{\mathsf {T}},\ \mathbf {e} ^{2}=\left({\frac {\mathbf {e} _{3}\times \mathbf {e} _{1}}{V}}\right)^{\mathsf {T}},\ \mathbf {e} ^{3}=\left({\frac {\mathbf {e} _{1}\times \mathbf {e} _{2}}{V}}\right)^{\mathsf {T}}.$$

en.wikipedia.org/wiki/Dual_basis
en.wikipedia.org/wiki/Biorthogonal_system

kuing

发个N年前写的短文好了：

hbghlyj

2维的对偶基.交互式演示:

对应于矩阵$\Phi$的两个列向量和矩阵$\tilde\Phi$的两个列向量.

带准星的点可以拖动.

下方的Description:

In mathematics, a pair of biorthogonal bases (a basis and its dual basis) can provide a representation for vectors in the plane; this is an alternative to what can be done with a single orthonormal basis. While in an orthonormal basis the basis vectors are mutually orthogonal, in a pair of biorthogonal bases the first vector in the basis (solid black in figure) is orthogonal to the second vector in the dual basis (dashed red in figure). Similarly, the second vector in the basis (dashed black in figure) is orthogonal to the first vector in the dual basis (solid red in figure). Furthermore, the corresponding vectors in the two bases have an inner product equal to 1. As each basis can be represented by a matrix whose columns are the basis vectors, the two bases are given when they exist (which is when the matrices are invertible).

For an orthonormal basis represented by a matrix $\Phi$, the following is true:
$$\Phi\Phi^{\sf T}=I$$
while for the two matrices $\Phi$ and $\tilde\Phi$, representing bases in a biorthogonal pair, the counterpart is$$\Phi\tilde\Phi^{\sf T}=I$$

3维的对偶基.
对应于矩阵$\Phi$的两个列向量和矩阵$\tilde\Phi$的两个列向量.
只看方向,是一个三面角的三条棱和三个面的法向量.

正交基的对偶基是自己. 反之,与自己对偶的基是正交基.
encyclopediaofmath.org/wiki/Biorthogonal_system

isee

亦可参考  向量求直角三角形内切圆半径
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