泰博定理三的证明

denglongshan · 发表于 2017-8-16 20:25

图上是相切的，注意标记4的那行，差了一个很长的等式。

denglongshan · 发表于 2017-8-16 20:37

denglongshan 发表于 2017-8-16 12:25
图上是相切的，注意标记4的那行，差了一个很长的等式。

，注意标记5的那行，也差了一个很长的等式。

ataorj · 发表于 2017-8-16 21:50

我看到你的图片文字模糊不清。
你的研究我也不懂，祝你顺利！
但愿你没搞错到底是哪两圆

ataorj · 发表于 2017-8-17 11:18

泰博定理三点共线几何证明
三点共线其实并非限于本题目实例,而是更一般情形:如图角度和字母对象除x,y外都已知,证明E,F,H共线
证明:
tgA=b/c
x=d*tgA-c
y=c-e/tgA
证x/b=y/a即可[说明:则△FDE∽△FGH,∠DFE=∠GFH;而D,F,G又共线]
x/b=(db/c-c)/b=d/c-c/b
y/a=(c-ce/b)/a=c/a-ce/(ab)
即证d/c-c/b=c/a-ce/(ab)
dab=cc(a+b-e)
dab=ccd
即证ab=cc
而a=c/tgA=cc/b
则确实ab=cc,证毕.

ataorj · 发表于 2017-8-17 11:32

以此为基础估计也容易求出圆半径和圆心距离,则可证明圆相切.(暂时不考虑)
==================
另外为前文附三角形内心坐标公式推导:
三角形内心坐标公式推导
设:在三角形ABC中,三顶点的坐标为：A(Ax,Ay),B(Bx,By),C(Cx,Cy); BC=a,CA=b,AB=c;AD,BE,CF为三角平分线;D,E,F在三边上,内心为P
==========
CE=h,AE=k
k/h=c/a,k/h+1=c/a+1,b/h=(a+c)/a
h=ab/(a+c)
Ex=Cx-(Cx-Ax)h/b=Cx-(Cx-Ax)a/(a+c)
Ey=Cy-(Cy-Ay)h/b=Cy-(Cy-Ay)a/(a+c)
PE/PB=h/a,(PE+PB)/PB=(h+a)/a
Px=Bx+(Ex-Bx)a/(h+a)
=(hBx+aEx)/(h+a)
=(bBx+Cx(a+c)-a(Cx-Ax))/(b+a+c)
=(bBx+cCx+aAx)/(b+a+c)

同理:
Py=(bBy+cCy+aAy)/(b+a+c)

denglongshan · 发表于 2017-8-17 19:32

边长用根号表示软件计算有时算不出来。

denglongshan · 发表于 2017-9-10 22:46

若∠ADC是定值，则PM/PN是定值。

denglongshan · 发表于 2021-12-10 22:05

\!\(\*OverscriptBox["b", "_"]\) = b = -1;
\!\(\*OverscriptBox["c", "_"]\) = c = 1;
\!\(\*OverscriptBox["v", "_"]\) = 1/v;
\!\(\*OverscriptBox["u", "_"]\) = 1/u;
k[a_, b_] := (a - b)/(
\!\(\*OverscriptBox["a", "_"]\) -
\!\(\*OverscriptBox["b", "_"]\));
\!\(\*OverscriptBox["k", "_"]\)[a_, b_] := 1/k[a, b];(*复斜率定义*)
SquareDistance[a_, b_] := (a - b) (
\!\(\*OverscriptBox["a", "_"]\) -
\!\(\*OverscriptBox["b", "_"]\));
\!\(\*OverscriptBox["Jd", "_"]\)[k1_, a1_, k2_, a2_] := -((a1 - k1
\!\(\*OverscriptBox["a1", "_"]\) - (a2 - k2
\!\(\*OverscriptBox["a2", "_"]\)))/(
k1 - k2));(*复斜率等于k1,过点A1与复斜率等于k2,过点A2的直线交点*)
Jd[k1_, a1_, k2_, a2_] := -((k2 (a1 - k1
\!\(\*OverscriptBox["a1", "_"]\)) - k1 (a2 - k2
\!\(\*OverscriptBox["a2", "_"]\)))/(k1 - k2));
Waixin[a_, b_, c_] := (a
\!\(\*OverscriptBox["a", "_"]\) (b - c) + b
\!\(\*OverscriptBox["b", "_"]\) (c - a) + c
\!\(\*OverscriptBox["c", "_"]\) (a - b) )/(
\!\(\*OverscriptBox["a", "_"]\) (b - c) +
\!\(\*OverscriptBox["b", "_"]\) (c - a) +
\!\(\*OverscriptBox["c", "_"]\) (a - b));
\!\(\*OverscriptBox["Waixin", "_"]\)[a_, b_, c_] := -((a
\!\(\*OverscriptBox["a", "_"]\) (
\!\(\*OverscriptBox["b", "_"]\) -
\!\(\*OverscriptBox["c", "_"]\)) + b
\!\(\*OverscriptBox["b", "_"]\) (
\!\(\*OverscriptBox["c", "_"]\) -
\!\(\*OverscriptBox["a", "_"]\)) + c
\!\(\*OverscriptBox["c", "_"]\) (
\!\(\*OverscriptBox["a", "_"]\) -
\!\(\*OverscriptBox["b", "_"]\)) )/(
\!\(\*OverscriptBox["a", "_"]\) (b - c) +
\!\(\*OverscriptBox["b", "_"]\) (c - a) +
\!\(\*OverscriptBox["c", "_"]\) (a - b)));
XiangjiaoxuanLianxin[o1_, a_, o2_, b_] := (a
\!\(\*OverscriptBox["a", "_"]\) - b
\!\(\*OverscriptBox["b", "_"]\) +
\!\(\*OverscriptBox["b", "_"]\) o2 + b
\!\(\*OverscriptBox["o2", "_"]\) -
\!\(\*OverscriptBox["a", "_"]\) o1 - a
\!\(\*OverscriptBox["o1", "_"]\) +
\!\(\*OverscriptBox["o2", "_"]\) o1 - o2
\!\(\*OverscriptBox["o1", "_"]\))/(2 (
\!\(\*OverscriptBox["o2", "_"]\) -
\!\(\*OverscriptBox["o1", "_"]\)));(*圆（O1，A）与圆（O2，B)连心线与公共弦的交点*)
\!\(\*OverscriptBox["XiangjiaoxuanLianxin", "_"]\)[o1_, a_, o2_,
b_] := (a
\!\(\*OverscriptBox["a", "_"]\) - b
\!\(\*OverscriptBox["b", "_"]\) +
\!\(\*OverscriptBox["b", "_"]\) o2 + b
\!\(\*OverscriptBox["o2", "_"]\) -
\!\(\*OverscriptBox["a", "_"]\) o1 - a
\!\(\*OverscriptBox["o1", "_"]\) + o2
\!\(\*OverscriptBox["o1", "_"]\) -
\!\(\*OverscriptBox["o2", "_"]\) o1)/(2 (o2 - o1));
a = Jd[v^2, b, u^2, c];
\!\(\*OverscriptBox["a", "_"]\) =
\!\(\*OverscriptBox["Jd", "_"]\)[v^2, b, u^2, c]; p = Jd[v, b, u, c];
\!\(\*OverscriptBox["p", "_"]\) =
\!\(\*OverscriptBox["Jd", "_"]\)[v, b, u, c];(*BA直线的复斜率等于v^2*)
d = Jd[w^2, a, 1, c];
\!\(\*OverscriptBox["d", "_"]\) =
\!\(\*OverscriptBox["Jd", "_"]\)[w^2, a, 1, c];(*DA直线的复斜率等于w^2*)
f = Jd[w, p, 1, c];
\!\(\*OverscriptBox["f", "_"]\) =
\!\(\*OverscriptBox["Jd", "_"]\)[w, p, 1, c]; g = Jd[-w, p, 1, c];
\!\(\*OverscriptBox["g", "_"]\) =
\!\(\*OverscriptBox["Jd", "_"]\)[-w, p, 1, c];
m = Jd[-1, f, -w, d];
\!\(\*OverscriptBox["m", "_"]\) =
\!\(\*OverscriptBox["Jd", "_"]\)[-1, f, -w, d]; n = Jd[-1, g, w, d];
\!\(\*OverscriptBox["n", "_"]\) =
\!\(\*OverscriptBox["Jd", "_"]\)[-1, g, w, d];
q = Waixin[a, b, c];
\!\(\*OverscriptBox["q", "_"]\) =
\!\(\*OverscriptBox["Waixin", "_"]\)[a, b, c];
Simplify[{1, a,
\!\(\*OverscriptBox["a", "_"]\), p,
\!\(\*OverscriptBox["p", "_"]\), , f,
\!\(\*OverscriptBox["f", "_"]\), g,
\!\(\*OverscriptBox["g", "_"]\)}]
Simplify[{2, m,
\!\(\*OverscriptBox["m", "_"]\), n,
\!\(\*OverscriptBox["n", "_"]\), , k[m, n], k[m, p]}]
Simplify[{3, q,
\!\(\*OverscriptBox["q", "_"]\)}]
Simplify[Solve[{SquareDistance[q, z] == SquareDistance[q, a],
SquareDistance[m, z] == SquareDistance[m, f]}, {z,
\!\(\*OverscriptBox["z", "_"]\)}]](*求两圆切点*)
Simplify[Solve[{SquareDistance[q, z] == SquareDistance[q, a],
SquareDistance[n, z] == SquareDistance[n, g]}, {z,
\!\(\*OverscriptBox["z", "_"]\)}]]
z1 = XiangjiaoxuanLianxin[q, a, m, f]; z2 =
XiangjiaoxuanLianxin[q, a, n, g];(*验证两圆切点*)
Simplify[{z1, z2}]
Simplify[{4, q - b, SquareDistance[q, b], , m - f, , n - g, ,
m - q, (2 v^3)/(-u^2 + v^2) I - (m - f) I}]
Simplify[{k[q, b], Sqrt[k[q, b]], k[m, q], k[n, q]}]
(*Simplify[{m-f,,q,(m-f-q)^2,SquareDistance[q,m],,n-g,,q,(n-g-q)^2,\
SquareDistance[q,n]}]*)

复制代码

denglongshan · 发表于 2021-12-12 22:17

因为涉及到内心，即角度关系

denglongshan · 发表于 2021-12-12 22:42

https://tieba.baidu.com/p/1626437383

沢山定理可以用来证明泰博定理，不过如果不知道如何构造，似乎不适宜用代数方法。

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泰博定理三的证明

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