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AD,BE,CF 为 ΔABC 的角平分线,过 D,E,F 的圆又与各边交于 X,Y,Z。求证 DX=EY+FZ

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发表于 2023-4-21 08:10 | 显示全部楼层 |阅读模式

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 楼主| 发表于 2023-4-21 09:55 | 显示全部楼层
本帖最后由 天山草 于 2023-4-21 10:02 编辑

这道题并不难,我已做出。先看看友友们是怎样做的。

如果问,两圆有没有相切的可能?什么条件下两圆相切?这可能就难了,不知道如何下手,我没有做出来。
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 楼主| 发表于 2023-4-23 08:29 | 显示全部楼层
本帖最后由 天山草 于 2023-4-23 08:36 编辑


程序代码:
  1. Clear["Global`*"];
  2. (*三角形ABC各顶点坐标 \[DoubleLongRightArrow]*)
  3. \!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c; a = ((t^2 - 1) (t - c) + 2 I t (t - c))/(1 - c t + t^2);
  4. \!\(\*OverscriptBox[\(a\), \(_\)]\) = ((t^2 - 1) (t - c) - 2 I t (t - c))/(1 - c t + t^2);
  5. (*\[EmptyUpTriangle]ABC各顶角平分线与对边的交点坐标 D、E、F \[DoubleLongRightArrow]*)
  6. d = ((t^2 + 1) (c - t))/(c t - t^2 + 1);  \!\(\*OverscriptBox[\(d\), \(_\)]\) = d; e = ( 2 c (t + I) (t - c))/(-c^2 + t^2 + 1);  
  7. \!\(\*OverscriptBox[\(e\), \(_\)]\) = ( 2 c (t - I) (t - c))/(-c^2 + t^2 + 1); f = (c (t + I)^2)/( 2 c t - t^2 - 1);
  8. \!\(\*OverscriptBox[\(f\), \(_\)]\) = (c (t - I)^2)/(2 c t - t^2 - 1);
  9. (*\[EmptyUpTriangle]ABC的内心坐标 \[DoubleLongRightArrow]*) i = t + I;  \!\(\*OverscriptBox[\(i\), \(_\)]\) = t - I;
  10. (*过ABC三点的圆的圆心坐标:*)
  11. WX[a_, b_, c_] := (a \!\(\*OverscriptBox[\(a\), \(_\)]\) (c - b) + b \!\(\*OverscriptBox[\(b\), \(_\)]\)(a - c) + c \!\(\*OverscriptBox[\(c\), \(_\)]\)(b - a) )/(\!\(\*OverscriptBox[\(a\), \(_\)]\)(c - b) + \!\(\*OverscriptBox[\(b\), \(_\)]\)(a - c) + \!\(\*OverscriptBox[\(c\), \(_\)]\)(b - a));
  12. \!\(\*OverscriptBox[\(WX\), \(_\)]\)[a_, b_, c_] := -((a \!\(\*OverscriptBox[\(a\), \(_\)]\)(\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)) + b \!\(\*OverscriptBox[\(b\), \(_\)]\)(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(c\), \(_\)]\)) + c \!\(\*OverscriptBox[\(c\), \(_\)]\)(\!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\)))/(\!\(\*OverscriptBox[\(a\), \(_\)]\)(c - b) + \!\(\*OverscriptBox[\(b\), \(_\)]\)(a - c) + \!\(\*OverscriptBox[\(c\), \(_\)]\)(b - a)));
  13. n = Simplify@ WX[d, e, f]; \!\(\*OverscriptBox[\(n\), \(_\)]\) = Simplify@ \!\(\*OverscriptBox[\(WX\), \(_\)]\)[d, e, f];
  14. (*复斜率定义:*) k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\));
  15. Simplify@Solve[{(n - d) (\!\(\*OverscriptBox[\(n\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)) == (n - x) (\!\(\*OverscriptBox[\(n\), \(_\)]\) - \!\(\*OverscriptBox[\(x\), \(_\)]\)), k[b, c] == k[b, x], x != d}, {x, \!\(\*OverscriptBox[\(x\), \(_\)]\)}];
  16. x = (c (c^2 (5 t^2 - 3) + c (2 t - 6 t^3) + (t^2 + 1)^2))/( 2 (2 c^3 t - c^2 (t^2 + 1) - 2 c (t^3 + t) + (t^2 + 1)^2));
  17. \!\(\*OverscriptBox[\(x\), \(_\)]\) = ( c (c^2 (5 t^2 - 3) + c (2 t - 6 t^3) + (t^2 + 1)^2))/( 2 (2 c^3 t - c^2 (t^2 + 1) - 2 c (t^3 + t) + (t^2 + 1)^2));
  18. Simplify@Solve[{(n - d) (\!\(\*OverscriptBox[\(n\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)) == (n - y) (\!\(\*OverscriptBox[\(n\), \(_\)]\) - \!\(\*OverscriptBox[\(y\), \(_\)]\)), k[a, c] == k[c, y], y != e}, {y, \!\(\*OverscriptBox[\(y\), \(_\)]\)}];
  19. y = (c^3 t (t^2 + 8 I t + 1) - c^2 (t^4 + 13 I t^3 - 5 t^2 - 3 I t + 2) - c (t^5 - 4 I t^4 + 8 t^3 + 6 I t^2 - t + 2 I) + t (t - I)^2 (t + I)^3)/(2 (2 c^3 t^2 + c^2 (-5 t^3 + 2 I t^2 + t) + c (4 t^4 - 3 I t^3 - t^2 + I t - 1) - t^5 + I t^4 + t - I));
  20. \!\(\*OverscriptBox[\(y\), \(_\)]\) = (  c^3 t (t^2 - 8 I t + 1) - c^2 (t^4 - 13 I t^3 - 5 t^2 + 3 I t + 2) + c (-t^5 - 4 I t^4 - 8 t^3 + 6 I t^2 + t + 2 I) +  t (t - I)^3 (t + I)^2)/(2 (2 c^3 t^2 + c^2 (-5 t^3 - 2 I t^2 + t) +  c (4 t^4 + 3 I t^3 - t^2 - I t - 1) - t^5 - I t^4 + t + I));
  21. Simplify@Solve[{(n - d) (\!\(\*OverscriptBox[\(n\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)) == (n - z) (\!\(\*OverscriptBox[\(n\), \(_\)]\) - \!\(\*OverscriptBox[\(z\), \(_\)]\)), k[b, a] == k[b, z], z != f}, {z, \!\(\*OverscriptBox[\(z\), \(_\)]\)}];
  22. z = -(((t + I) (c^3 (3 - 5 t^2) + c^2 t (11 t^2 - 5) - c (7 t^4 + 1) + t (t^2 + 1)^2))/( 2 (t - I) (c^3 t - c^2 (t^2 - 1) - c (t^3 + t) + t^4 - 1)));
  23. \!\(\*OverscriptBox[\(z\), \(_\)]\) = -(((t - I) (c^3 (3 - 5 t^2) + c^2 t (11 t^2 - 5) - c (7 t^4 + 1) + t (t^2 + 1)^2))/(2 (t + I) (c^3 t - c^2 (t^2 - 1) - c (t^3 + t) + t^4 - 1)));
  24. DX = Simplify[x - d] // Factor;
  25. EY = Simplify[Sqrt[(e - y) (\!\(\*OverscriptBox[\(e\), \(_\)]\) - \!\(\*OverscriptBox[\(y\), \(_\)]\))]] // Factor;
  26. EY = -(1/2) (7 c^4 t^3 - c^4 t - 20 c^3 t^4 - 10 c^3 t^2 + 2 c^3 + 18 c^2 t^5 + 12 c^2 t^3 - 6 c^2 t - 4 c t^6 + 2 c t^4 +
  27.      8 c t^2 + 2 c - t^7 - 3 t^5 - 3 t^3 - t)/(2 c^4 t^2 - 3 c^3 t^3 + c^3 t - c^2 t^4 - 2 c^2 t^2 - c^2 + 3 c t^5 + 2 c t^3 - c t -  t^6 - t^4 + t^2 + 1);
  28. FZ = Simplify[Sqrt[(f - z) (\!\(\*OverscriptBox[\(f\), \(_\)]\) - \!\(\*OverscriptBox[\(z\), \(_\)]\))]] // Factor;
  29. FZ = 1/2 (8 c^4 t^3 - 8 c^4 t - 25 c^3 t^4 + 8 c^3 t^2 + c^3 + 27 c^2 t^5 + 10 c^2 t^3 - c^2 t - 11 c t^6 - 13 c t^4 - c t^2 + c + t^7 + 3 t^5 + 3 t^3 + t)/(2 c^4 t^2 - 3 c^3 t^3 + c^3 t - c^2 t^4 - 2 c^2 t^2 - c^2 + 3 c t^5 + 2 c t^3 - c t - t^6 - t^4 + t^2 + 1);
  30. FullSimplify[DX] // Factor
  31. FullSimplify[EY + FZ] // Factor
  32. Simplify[DX == EY + FZ]
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点评

同感,同问,X、Y和Z地位相同  发表于 2023-4-23 21:05
这题感觉有点奇怪,按理说 DX、EY、FZ 是对等的,怎么知道该谁等于另两个的和?  发表于 2023-4-23 09:31
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发表于 2023-4-24 20:27 | 显示全部楼层
结论不准确

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\frac{\square}{\square}\sqrt{\square}\square_{\baguet}^{\baguet}\overarc{\square}\ \dot{\baguet}\left(\square\right)\binom{\square}{\square}\begin{cases}\square\\\square\end{cases}\ \begin{bmatrix}\square&\square\\\square&\square\end{bmatrix}\to\Rightarrow\mapsto\alpha\ \theta\ \pi\times\div\pm\because\angle\ \infty
\frac{\square}{\square}\sqrt{\square}\sqrt[\baguet]{\square}\square_{\baguet}\square^{\baguet}\square_{\baguet}^{\baguet}\sum_{\baguet}^{\baguet}\prod_{\baguet}^{\baguet}\coprod_{\baguet}^{\baguet}\int_{\baguet}^{\baguet}\lim_{\baguet}\lim_{\baguet}^{\baguet}\bigcup_{\baguet}^{\baguet}\bigcap_{\baguet}^{\baguet}\bigwedge_{\baguet}^{\baguet}\bigvee_{\baguet}^{\baguet}
\underline{\square}\overline{\square}\overrightarrow{\square}\overleftarrow{\square}\overleftrightarrow{\square}\underrightarrow{\square}\underleftarrow{\square}\underleftrightarrow{\square}\dot{\baguet}\hat{\baguet}\vec{\baguet}\tilde{\baguet}
\left(\square\right)\left[\square\right]\left\{\square\right\}\left|\square\right|\left\langle\square\right\rangle\left\lVert\square\right\rVert\left\lfloor\square\right\rfloor\left\lceil\square\right\rceil\binom{\square}{\square}\boxed{\square}
\begin{cases}\square\\\square\end{cases}\begin{matrix}\square&\square\\\square&\square\end{matrix}\begin{pmatrix}\square&\square\\\square&\square\end{pmatrix}\begin{bmatrix}\square&\square\\\square&\square\end{bmatrix}\begin{Bmatrix}\square&\square\\\square&\square\end{Bmatrix}\begin{vmatrix}\square&\square\\\square&\square\end{vmatrix}\begin{Vmatrix}\square&\square\\\square&\square\end{Vmatrix}\begin{array}{l|l}\square&\square\\\hline\square&\square\end{array}
\to\gets\leftrightarrow\nearrow\searrow\downarrow\uparrow\updownarrow\swarrow\nwarrow\Leftarrow\Rightarrow\Leftrightarrow\rightharpoonup\rightharpoondown\impliedby\implies\Longleftrightarrow\leftharpoonup\leftharpoondown\longleftarrow\longrightarrow\longleftrightarrow\Uparrow\Downarrow\Updownarrow\hookleftarrow\hookrightarrow\mapsto
\alpha\beta\gamma\Gamma\delta\Delta\epsilon\varepsilon\zeta\eta\theta\Theta\iota\kappa\varkappa\lambda\Lambda\mu\nu\xi\Xi\pi\Pi\varpi\rho\varrho\sigma\Sigma\tau\upsilon\Upsilon\phi\Phi\varphi\chi\psi\Psi\omega\Omega\digamma\vartheta\varsigma\mathbb{C}\mathbb{H}\mathbb{N}\mathbb{P}\mathbb{Q}\mathbb{R}\mathbb{Z}\Re\Im\aleph\partial\nabla
\times\cdot\ast\div\pm\mp\circ\backslash\oplus\ominus\otimes\odot\bullet\varnothing\neq\equiv\not\equiv\sim\approx\simeq\cong\geq\leq\ll\gg\succ\prec\in\ni\cup\cap\subset\supset\not\subset\not\supset\notin\not\ni\subseteq\supseteq\nsubseteq\nsupseteq\sqsubset\sqsupset\sqsubseteq\sqsupseteq\sqcap\sqcup\wedge\vee\neg\forall\exists\nexists\uplus\bigsqcup\bigodot\bigotimes\bigoplus\biguplus\bigcap\bigcup\bigvee\bigwedge
\because\therefore\angle\parallel\perp\top\nparallel\measuredangle\sphericalangle\diamond\diamondsuit\doteq\propto\infty\bowtie\square\smile\frown\bigtriangledown\triangle\triangleleft\triangleright\bigcirc \wr\amalg\models\preceq\mid\nmid\vdash\dashv\nless\ngtr\ldots\cdots\vdots\ddots\surd\ell\flat\sharp\natural\wp\clubsuit\heartsuit\spadesuit\oint\lfloor\rfloor\lceil\rceil\lbrace\rbrace\lbrack\rbrack\vert\hbar\aleph\dagger\ddagger

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