有三个相似形的几何题

天山草

下面这个几何题来自网上，原题做法比较复杂。下面用不用太耗费脑细胞的解析法做。



构图和设点： 将B点置于坐标系原点，BC 与实轴重合，C 点坐标为 1。

用 mathematica 写的计算程序如下：



程序运行结果：（大概需要运行 5 秒钟）

天山草

1. Clear["Global`*"];
   
2. (*将B点置于坐标系原点，C点坐标为1，AB的复斜率为kAB=u^4，AC的复斜率为kAC=1/v^4*)
   
3. \!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1; a = (u^4 (v^4 - 1))/( u^4 v^4 - 1);
   
4. \!\(\*OverscriptBox[\(a\), \(_\)]\) = (v^4 - 1)/( u^4 v^4 - 1); (*\[EmptyUpTriangle]ABC的顶点坐标*)
   
5. i = (u^2 (v^2 - 1))/(u^2 v^2 - 1); \!\(\*OverscriptBox[\(i\), \(_\)]\) = (v^2 - 1)/( u^2 v^2 - 1); (*\[EmptyUpTriangle]ABC的内心坐标*) t1 = ( u^2 (u^2 + 1) (v^2 - 1))/(2 (u^2 v^2 - 1));  
   
6. \!\(\*OverscriptBox[\(t1\), \(_\)]\) = ((u^2 + 1) (v^2 - 1))/( 2 u^2 (u^2 v^2 - 1));   (*在此构图下，有上面的已知公式可引用*)
   
7. e = \[Lambda] a; \!\(\*OverscriptBox[\(e\), \(_\)]\) = \[Lambda] \!\(\*OverscriptBox[\(a\), \(_\)]\); (*设E点坐标，\[Lambda]是0到1之间的正实数*)Print["E = ", e];
   
8. Jx[p_, a_, b_] := (\!\(\*OverscriptBox[\(a\), \(_\)]\) b - a \!\(\*OverscriptBox[\(b\), \(_\)]\) + \!\(\*OverscriptBox[\(p\), \(_\)]\) (a - b))/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\));(* P 点关于镜线的像，镜线过已知点 A、B *)
   
9. \!\(\*OverscriptBox[\(Jx\), \(_\)]\)[p_, a_, b_] := (a \!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\) b + p (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)))/(a - b);
   
10. t = Simplify@Jx[t1, e, i]; \!\(\*OverscriptBox[\(t\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jx\), \(_\)]\)[t1, e, i]; Print["T = ", t];
    
11. k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\));(*复斜率定义*)
    
12. Jd[k1_, a1_, k2_, a2_] := -((k2 (a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\)) - k1 (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
    
13. \!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k1_, a1_, k2_, a2_] := -((a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\) - (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
    
14. (*复斜率等于k1且经过A1点的直线，与复斜率等于k2且经过A2点的直线，两直线的交点*)
    
15. f = Simplify@Jd[k[a, c], a, k[e, t], e]; \!\(\*OverscriptBox[\(f\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k[a, c], a, k[e, t], e];
    
16. Print["F = ", f];Yuan[zz_, aa_, bb_, cc_] :=  Simplify[( (zz - cc) (aa - bb))/( (zz - bb) (aa - cc)) - ( (\!\(\*OverscriptBox[\(zz\), \(_\)]\) - \!\(\*OverscriptBox[\(cc\), \(_\)]\)) (\!\(\*OverscriptBox[\(aa\), \(_\)]\) - \!\(\*OverscriptBox[\(bb\), \(_\)]\)))/( (\!\(\*OverscriptBox[\(zz\), \(_\)]\) - \!\(\*OverscriptBox[\(bb\), \(_\)]\)) (\!\(\*OverscriptBox[\(aa\), \(_\)]\) -
    
17. \!\(\*OverscriptBox[\(cc\), \(_\)]\)))];  (* 此表达式 = 0 即是经过三点 AA、BB、CC \的圆方程，要求的未知点坐标是 ZZ *)
    
18. W1 = {j, \!\(\*OverscriptBox[\(j\), \(_\)]\)} /.    Simplify@    Solve[{Yuan[j, a, e, f] == 0, Yuan[j, a, b, c] == 0, j != a}, {j,
    
19. \!\(\*OverscriptBox[\(j\), \(_\)]\)}] // Flatten;
    
20. j = Part[W1, 1]; \!\(\*OverscriptBox[\(j\), \(_\)]\) = Part[W1, 2];  Print["J = ", j];
    
21. W2 = {p, \!\(\*OverscriptBox[\(p\), \(_\)]\)} /.     Simplify@     Solve[{Yuan[p, a, b, c] == 0, k[i, p] == k[i, j], p != j}, {p,
    
22. \!\(\*OverscriptBox[\(p\), \(_\)]\)}] // Flatten;
    
23. p = Part[W2, 1]; \!\(\*OverscriptBox[\(p\), \(_\)]\) = Part[W2, 2];  Print["P = ", p];
    
24. W3 = {q, \!\(\*OverscriptBox[\(q\), \(_\)]\)} /.     Simplify@     Solve[{Yuan[q, a, e, f] == 0, k[i, q] == k[i, j], q != j}, {q,
    
25. \!\(\*OverscriptBox[\(q\), \(_\)]\)}] // Flatten;q = Part[W3, 1];
    
26. \!\(\*OverscriptBox[\(q\), \(_\)]\) = Part[W3, 2];  Print["Q = ", q];
    
27. W4 = {m, \!\(\*OverscriptBox[\(m\), \(_\)]\)} /.     Simplify@     Solve[{Yuan[m, a, b, c] == 0, k[i, m] == k[i, a], m != a}, {m,
    
28. \!\(\*OverscriptBox[\(m\), \(_\)]\)}] // Flatten;
    
29. m = Part[W4, 1]; \!\(\*OverscriptBox[\(m\), \(_\)]\) = Part[W4, 2];  Print["M = ", m];
    
30. W5 = {n, \!\(\*OverscriptBox[\(n\), \(_\)]\)} /.     Simplify@     Solve[{Yuan[n, a, e, f] == 0, k[i, n] == k[i, a], n != a}, {n,
    
31. \!\(\*OverscriptBox[\(n\), \(_\)]\)}] // Flatten; n = Part[W5, 1]; \!\(\*OverscriptBox[\(n\), \(_\)]\) = Part[W5, 2];  Print["N = ", n];
    
32. S[a_, b_] := (a - b) (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\));(*两点间的距离平方*)
    
33. Print["测试\[EmptyUpTriangle]BJE∽\[EmptyUpTriangle]CJF是否成立:"]
    
34. Simplify[S[j, e]/S[j, f] == S[e, b]/S[f, c] == S[j, b]/  S[j, c]](*测试\[EmptyUpTriangle]BJE∽\[EmptyUpTriangle]CJF是否成立*)
    
35. Print["测试\[EmptyUpTriangle]QEF∽\[EmptyUpTriangle]PCB是否成立:"]
    
36. Simplify[S[q, e]/S[p, c] == S[q, f]/S[p, b] == S[e, f]/  S[b, c]](*测试\[EmptyUpTriangle]QEF∽\[EmptyUpTriangle]PCB是否成立*)
    
37. Print["测试\[EmptyUpTriangle]NEF∽\[EmptyUpTriangle]MBC是否成立:"]
    
38. Simplify[S[e, n]/S[b, m] == S[n, f]/S[m, c] == S[e, f]/  S[b, c]](*测试\[EmptyUpTriangle]NEF∽\[EmptyUpTriangle]MBC是否成立*)   
    
39. Print["测试 NE = NF 是否成立:"]
    
40. Simplify[S[n, e] == S[n, f]](*测试 NE = NF 是否成立*)
    
41. Print["测试 \[ExponentialE]^2\[ImaginaryI]\[Angle]BJI = \[ExponentialE]^2\[ImaginaryI]\[Angle]FJI 是否成立，如果成立则 \[Angle]BJI = \[Angle]FJI:"];
    
42. Simplify[k[j, i]/k[j, b] == k[j, f]/k[j, i]]

复制代码

denglongshan

用线段平方比例固然可以，但用向量商更直接

denglongshan

假设\(\triangle ABC\)与\(\triangle A_1B_1C_1\)顺相似，∠B=∠B1则\(\frac{\overrightarrow{BA}}{\overrightarrow{BC}}=\frac{\overrightarrow{B_1A_1}}{\overrightarrow{B_1C_1}}\)
假设\(\triangle ABC\)与\(\triangle A_1B_1C_1\)逆相似，∠B=∠B1，则\(\overline{\left( \frac{\overrightarrow{BA}}{\overrightarrow{BC}}\right)}=\frac{\overrightarrow{B_1A_1}}{\overrightarrow{B_1C_1}}\)