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楼主 |
发表于 2023-12-21 14:56
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原题没有错。解法如下:
- Clear["Global`*"];
- \!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; c = -2 Sqrt[2] I; \!\(\*OverscriptBox[\(c\), \(_\)]\) = 2 Sqrt[2] I; d = -2 + 2 I;
- \!\(\*OverscriptBox[\(d\), \(_\)]\) = -2 - 2 I; e = -Sqrt[2] I; \!\(\*OverscriptBox[\(e\), \(_\)]\) = Sqrt[2] I; R = 2 Sqrt[2];
- \!\(\*OverscriptBox[\(f\), \(_\)]\) = f;
- k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
- W1 = {g, \!\(\*OverscriptBox[\(g\), \(_\)]\)} /. Simplify@Solve[{(b - g) (\!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(g\), \(_\)]\)) == R^2, k[e, f] == k[e, g]}, {g, \!\(\*OverscriptBox[\(g\), \(_\)]\)}] // Flatten;
- g = Part[W1, 3]; \!\(\*OverscriptBox[\(g\), \(_\)]\) = Part[W1, 4];
- W2 = {g1, \!\(\*OverscriptBox[\(g1\), \(_\)]\)} /. Simplify@Solve[{(b - g1) (\!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(g1\), \(_\)]\)) == R^2, k[d, f] == k[d, g1]}, {g1, \!\(\*OverscriptBox[\(g1\), \(_\)]\)}] // Flatten;
- g1 = Part[W2, 1]; \!\(\*OverscriptBox[\(g1\), \(_\)]\) = Part[W2, 2];
- W3 = {a, \!\(\*OverscriptBox[\(a\), \(_\)]\)} /. Simplify@Solve[{(b - a) (\!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\)) == R^2, k[d, e] == k[d, a]}, {a, \!\(\*OverscriptBox[\(a\), \(_\)]\)}] // Flatten;
- a = Part[W3, 1]; \!\(\*OverscriptBox[\(a\), \(_\)]\) = Part[W3, 2];
- W = {f} /. Simplify@Solve[{(c - a) (\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\)) == (g - g1) (\!\(\*OverscriptBox[\(g\), \(_\)]\) - \!\(\*OverscriptBox[\(g1\), \(_\)]\)), k[b, f] == 1}, {f}] // Flatten;
- f = Part[W, 2];f = f // ComplexExpand // FullSimplify; \!\(\*OverscriptBox[\(f\), \(_\)]\) = f;
- Print["F = ", f, " \[TildeTilde] ", N[f]];
- S[a_, b_, c_] := (\!\(\*OverscriptBox[\(a\), \(_\)]\) (b - c) + \!\(\*OverscriptBox[\(b\), \(_\)]\) (c - a) + \!\(\*OverscriptBox[\(c\), \(_\)]\) (a - b))/(4 I);(*已知三角形各顶点坐标,求其有向面积,ABC逆时针向环绕*)
- SS = FullSimplify@S[d, e, f];
- Print["\[EmptyUpTriangle]DEF 的面积 = ", SS, " \[TildeTilde] ", N[SS, 50]];
复制代码
运行结果:
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