证明题

chenjiahao

如图，A点位于(0,0),B点位于(0,2),C点位于(7,-4)，
A,B,C都在\(\bigcirc ABC\)上，
分别作\(AB{,}BC\)
过C点作\(\bigcirc ABC\)的切线\(CI\)，
过A点作\(CI\)的垂线\(AI\),垂足I，
D为\(CI\)与\(\bigcirc ABC\)的交点,
过D点作\(AC\)的垂线\(DE\),垂足E，
过E点作\(BC\)的垂线\(EF\),垂足F，
H为\(EF\)与\(\bigcirc ABC\)的交点,
G为\(DE\)与\(BC\)的交点,
连结\(DH{,}FG\)


①证明：\(DH=FG\)
②C还位于哪个点时，\(DH=FG\)仍成立

天山草

当 C 点坐标为 (7, -4) 时并不能保证 HD = GF。因为此时 HD=0.21486149533989896， 而  GF=0.2466819194792003。二者并不相等。

只有当 C 点坐标为 (7, -3.822726165554712) 时才有 HD = GF = 0.247079464822713，

或者当 C 点坐标为 (7, -5.780787773863573) 时也有 HD = GF = 0.22026169856688738。

当然，如果 C 点的横坐标由 7 改为其它值时，符合题目要求的 C 点的纵坐标也将变化。

天山草

用 MMA 写的程序代码：

1. Clear["Global`*"];
   
2. \!\(\*OverscriptBox[\(a\), \(_\)]\) = a = 0; b = 0 + 2 I; \!\(\*OverscriptBox[\(b\), \(_\)]\) = 0 - 2 I; c = 7 + t I;
   
3. \!\(\*OverscriptBox[\(c\), \(_\)]\) = 7 - t I;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)); (* \[EmptyUpTriangle]ABC的外心*)
   
4. \!\(\*OverscriptBox[\(WX\), \(_\)]\)[a_, b_, c_] := (\!\(\*OverscriptBox[\(a\), \(_\)]\) (\!\(\*OverscriptBox[\(b\), \(_\)]\) (a - b) +
   
5. \!\(\*OverscriptBox[\(c\), \(_\)]\) (c - a)) + \!\(\*OverscriptBox[\(b\), \(_\)]\) \!\(\*OverscriptBox[\(c\), \(_\)]\) (b - c))/(
   
6. \!\(\*OverscriptBox[\(a\), \(_\)]\) (c - b) + \!\(\*OverscriptBox[\(b\), \(_\)]\) (a - c) + \!\(\*OverscriptBox[\(c\), \(_\)]\) (b - a));
   
7. o = Simplify@WX[a, b, c]; \!\(\*OverscriptBox[\(o\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(WX\), \(_\)]\)[a, b, c];
   
8. Jd[k1_, a1_, k2_, a2_] := -((k2 (a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\)) - k1 (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
   
9. \!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k1_, a1_, k2_, a2_] := -((a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\) - (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
   
10. k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\));
    
11. \!\(\*OverscriptBox[\(k\), \(_\)]\)[a_, b_] := (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\))/(a - b);(*复斜率定义*)
    
12. i = Simplify@Jd[-k[o, c], c, k[o, c], a]; \!\(\*OverscriptBox[\(i\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[-k[o, c], c, k[o, c], a];
    
13. W1 = Simplify@Solve[{(o - a) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\)) == (o - d) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)), k[a, d] == k[d, i], d != a}, {d, \!\(\*OverscriptBox[\(d\), \(_\)]\)}];
    
14. {d, \!\(\*OverscriptBox[\(d\), \(_\)]\)} = {d, \!\(\*OverscriptBox[\(d\), \(_\)]\)} /. Last[W1];
    
15. e = Simplify@Jd[-k[a, c], d, k[a, c], a]; \!\(\*OverscriptBox[\(e\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[-k[a, c], d, k[a, c], a];
    
16. f = Simplify@Jd[-k[b, c], e, k[b, c], b]; \!\(\*OverscriptBox[\(f\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[-k[b, c], e, k[b, c], b];
    
17. g = Simplify@Jd[k[b, c], b, k[d, e], d]; \!\(\*OverscriptBox[\(g\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k[b, c], b, k[d, e], d];
    
18. W = {h, \!\(\*OverscriptBox[\(h\), \(_\)]\)} /. Simplify@Solve[{(o - a) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\)) == (o - h) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(h\), \(_\)]\)), k[f, e] == k[e, h]}, {h, \!\(\*OverscriptBox[\(h\), \(_\)]\)}] // Factor // Flatten ;h = Part[W, 3]; \!\(\*OverscriptBox[\(h\), \(_\)]\) = Part[W, 4];
    
19. NSolve[{(h - d) (\!\(\*OverscriptBox[\(h\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)) ==
    
20. (g - f) (\!\(\*OverscriptBox[\(g\), \(_\)]\) - \!\(\*OverscriptBox[\(f\), \(_\)]\)),  t \[Element] \[DoubleStruckCapitalR]}, {t}]
    
21. Print["当 s = 7、t = -3.822726165554712 时："]
    
22. {D -> d} /. {s -> 7, t -> -3.822726165554712`}
    
23. {H -> h} /. {s -> 7, t -> -3.822726165554712`}
    
24. {G -> g} /. {s -> 7, t -> -3.822726165554712`}
    
25. {F -> f} /. {s -> 7, t -> -3.822726165554712`}
    
26. {HD -> Abs[(h - d)], GF -> Abs[(g - f)]} /. {s -> 7,  t -> -3.822726165554712`}
    
27. Print["测试 HD = GF 是否成立："]
    
28. Simplify[Abs[(h - d)] == Abs[(g - f)]] /. {s -> 7,   t -> -3.822726165554712`}
    
29. Print["当 s = 7、t = -5.780787773863573 时："]
    
30. {D -> d} /. {s -> 7, t -> -5.780787773863573`}
    
31. {H -> h} /. {s -> 7, t -> -5.780787773863573`}
    
32. {G -> g} /. {s -> 7, t -> -5.780787773863573`}
    
33. {F -> f} /. {s -> 7, t -> -5.780787773863573`}
    
34. {HD -> Abs[(h - d)], GF -> Abs[(g - f)]} /. {s -> 7,  t -> -5.780787773863573`}
    
35. Print["测试 HD = GF 是否成立："]
    
36. Simplify[Abs[(h - d)] == Abs[(g - f)]] /. {s -> 7,  t -> -5.780787773863573`}

复制代码

程序运行结果：



说明： t 应为实数，在 s = 7 的条件下，实数解只有两个。其它许多非实数解都是无效解。

天山草

原题给的 C 点坐标 ( 7, -4 ) 为什么不能保证  HD = GF 成立？

见以下计算：



可见从小数第二位起 HD 就与 GF 不一样了，因此它们并不相等。