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

天山草

天山草

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。

天山草

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]

复制代码

denglongshan

结论不准确