费尔巴哈定理 向量证明

hbghlyj

arxiv.org/abs/1610.03962


$\newcommand\0{R} \newcommand\1{r} \newcommand\2{r_c} \newcommand\ip[2]{\langle#1,#2\rangle} \newcommand\4{\Delta} \newcommand\7{d_a} \newcommand\8{d_b} \newcommand\9{d_c} \renewcommand\v[1]{\mathbf{#1}} \newcommand\eh{\frac12} \newcommand\nv[1]{\|\v #1\|} \newcommand\nn[1]{\|#1\|} \newcommand\ipv[2]{\ip{\v #1}{\v #2}} \newcommand\qq[1]{\kern1pt\overline{\kern-1pt#1\kern-0.5pt}\kern0.5pt} \newcommand\vv[1]{\overrightarrow{#1}} \newcommand\bb[2]{|#1#2|} \renewcommand\gg[2]{\ell(#1,#2)} $Let $\triangle ABC$ be a triangle with circumradius $\0$ and sidelengths $a$, $b$ and $c$. We choose a coordinate system, which has its origin in the circumcenter $O$ of the triangle. We denote the vectors from $O$ to the vertices $A$, $B$ and $C$ by $\v u$, $\v v$ and $\v w$ respectively. Then we have $a=\|\v v-\v w\|$, $b=\|\v u-\v w\|$, $c=\|\v u-\v v\|$ and $\|\v u\|=\|\v v\|=\|\v w\|=R$. The semiperimeter $\eh(a+b+c)$ is denoted by $s$. Let $N$ be the nine-point center and $H$ the orthocenter of the triangle. Set $\v n=\eh\v u+\eh\v v+\eh\v w$. Then we have $\|\v n-(\eh\v u+\eh\v v)\|=\eh\|\v w\|=\eh\0$. Similarly we get $\|\v n-(\eh\v u+\eh\v w)\|=\eh\0$ and $\|\v n-(\eh\v v+\eh\v w)\|=\eh\0$. Therefore $\v n$ is the vector from $O$ to $N$ and $\eh \0$ is the nine-point radius. Let $I$ be the incenter and $I_c$ the excenter opposite $C$. Set $\v m=\frac a{2s}\v u+\frac b{2s}\v v+\frac c{2s}\v w$. The equations of the angle bisectors through $A$ and $B$ are $$ \v u+\lambda(\tfrac1c(\v v-\v u)+\tfrac1b(\v w-\v u)) \ \ \ \text{ and }\ \ \ \v v+\mu(\tfrac1c(\v u-\v v)+\tfrac1a(\v w-\v v)) $$ Choosing $\lambda=\frac{bc}{2s}$ and $\mu=\frac{ac}{2s}$ we get $\v m$ in both cases. This shows that $\v m$ is the vector from $O$ to the incenter $I$. Set $\v m_c=\frac a{2(s-c)}\v u+\frac b{2(s-c)}\v v-\frac c{2(s-c)}\v w$. The equations of the external angle bisectors through $A$ and $B$ are $$ \v u+\lambda(\tfrac1c(\v v-\v u)-\tfrac1b(\v w-\v u)) \ \ \ \text{ and }\ \ \ \v v+\mu(\tfrac1c(\v u-\v v)-\tfrac1a(\v w-\v v)) $$ Choosing $\lambda=\frac{bc}{2(s-c)}$ and $\mu=\frac{ac}{2(s-c)}$ we get $\v m_c$ in both cases. This shows that $\v m_c$ is the vector from $O$ to the excenter $I_c$. We use the following lemma for the computation of distances. Lemma For any $\7$, $\8$ and $\9$ in $\mathbb R$ we have $$\|\7\v u+\8\v v+\9\v w\|^2=\0^2(\7+\8+\9)^2-(a^2\8\9+b^2\7\9+c^2\7\8) $$ Proof. We have $c^2=\|\v u-\v v\|^2=\ip{\v u-\v v}{\v u-\v v}=\|\v u\|^2+\|\v v\|^2-2\ip{\v u}{\v v}$. Since we have also $\|\v u\|=\|\v v\|=\0$ we get $\ip{\v u}{\v v}=\0^2-\eh c^2$. Similarly we get $\ip{\v u}{\v w}=\0^2-\eh b^2$ and $\ip{\v v}{\v w}=\0^2-\eh a^2$. Using this and $\|\v u\|=\|\v v\|=\|\v w\|=\0$ we compute \begin{align*} \|\7\v u+&\8\v v+\9\v w\|^2 \\ &=\7^2\|\v u\|^2+\8^2\|\v v\|^2+\9^2\|\v w\|^2 +2\7\8\ip{\v u}{\v v}+2\7\9\ip{\v u}{\v w}+2\8\9\ip{\v v}{\v w} \\ &=\0^2(\7^2+\8^2+\9^2+2\7\8+2\7\9+2\8\9)-c^2\7\8-b^2\7\9-a^2\8\9 \end{align*} This is the desired result. We use Heron's formula $\4^2=s(s-a)(s-b)(s-c)$ for the area $\4$ of the triangle. This can also be written as $16\4^2=2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4$. Furthermore, we use the formulas $\1=\frac\4s$, $\2=\frac\4{s-c}$ and $\0=\frac{abc}{4\4}$ for the inradius $\1$, the exradius $\2$ and the circumradius $\0$. In particular we have $\1\0=\frac{abc}{4s}$ and $\2\0=\frac{abc}{4(s-c)}$. Theorem 0 (Euler) $|OI|^2=\0^2-2\0\1$ Proof. We have $\overrightarrow{OI}=\v m=\7\v u+\8\v v+\9\v w$ with $\7=\frac a{2s}$, $\8=\frac b{2s}$ and $\9=\frac c{2s}$. We get $\7+\8+\9=1$ and $a^2\8\9+b^2\7\9+c^2\7\8=\frac{abc}{4s^2}(a+b+c)=\frac{abc}{2s}=2\1\0$. The Lemma gives the result. Theorem 1 (Feuerbach) $|IN|=\eh\0-\1$ Proof. We have $\overrightarrow{IN}=\v n-\v m=\7\v u+\8\v v+\9\v w$ with $\7=\frac{s-a}{2s}$, $\8=\frac{s-b}{2s}$ and $\9=\frac{s-c}{2s}$. We get $\7+\8+\9=\eh$ and \begin{align*} a^2\8\9&+b^2\7\9+c^2\7\8 \\ &=\tfrac{1}{16s^2}(a^2(a^2-(b-c)^2)+b^2(b^2-(a-c)^2)+c^2(c^2-(a-b)^2)) \\ &=\tfrac{1}{16s^2}(a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 +2abc(a+b+c)) \\ &=-\tfrac{\4^2}{s^2}+\tfrac{abc}{4s}=-\1^2+\1\0 \end{align*} The Lemma implies $|IN|^2=\frac14\0^2+\1^2-\1\0=(\eh\0-\1)^2$. By Theorem 0 we have $\eh\0\ge\1$. Hence $|IN|=\eh\0-\1$ follows. Theorem 2 (Feuerbach) $|I_cN|=\eh\0+\2$ Proof. We have $\overrightarrow{I_cN}=\v n-\v m_c=\7\v u+\8\v v+\9\v w$ with $\7=\frac12-\frac a{2(s-c)}=-\frac{s-b}{2(s-c)}$, $\8=\frac12-\frac b{2(s-c)}=-\frac{s-a}{2(s-c)}$ and $\9=\frac12+\frac c{2(s-c)}=\frac{s}{2(s-c)}$. We get $\7+\8+\9=\eh$ and \begin{align*} a^2\8\9&+b^2\7\9+c^2\7\8 \\ &=\tfrac{1}{16(s-c)^2}(a^2(a^2-(b+c)^2)+b^2(b^2-(a+c)^2)+c^2(c^2-(a-b)^2)) \\ &=\tfrac{1}{16(s-c)^2}(a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 -2abc(a+b-c)) \\ &=-\tfrac{\4^2}{(s-c)^2}-\tfrac{abc}{4(s-c)}=-\2^2-\2\0 \end{align*} The Lemma implies $|I_cN|^2=\frac14\0^2+\2^2+\2\0=(\eh\0+\2)^2$. We get $|I_cN|=\eh\0+\2$. Theorem 1 implies that the nine-point circle is tangent to the incircle and Theorem 2 implies that the nine-point circle is tangent to the excircle opposite $C$. We get also Theorem $|OH|^2=9\0^2-(a^2+b^2+c^2)$ Proof. Since $N$ is the midpoint between $O$ and $H$, we get that $\v u+\v v+\v w$ is the vector from $O$ to $H$. Hence the desired result follows from the Lemma with $\7=\8=\9 =1$. Classical proofs of of Feuerbach's Theorem can be found in [F] and [J]. In [S] a proof is given, which uses vector computations. The proof in this paper is still simpler than those in [S]. Bibliography [F] K. W. Feuerbach, Eigenschaften einiger merkwürdigen Punkte des geradlinigen Dreiecks und mehrerer durch sie bestimmten Linien und Figuren: Eine analytisch-trigonometrische Abhandlung, Riegel & Wiesner Nürnberg, 1822. [J] R. Johnson, Advanced Euclidean Geometry. Dover Publications, 2013. [S] Michael J. G. Scheer, A Simple Vector Proof of Feuerbach's Theorem. Forum Geometricorum $\textbf{11}~(2011), 205 – 210$.

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《关于三角形的几何量 — kuing》

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hbghlyj 发表于 2022-8-13 23:14
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