借助 mathematica 在复平面上用解析几何方法作几何题示例（3）

TSC999 · 发表于 2023-6-19 10:37

例 31：

程序代码：

Clear["Global`*"]; (*设B在坐标原点，C在1点，\[Angle]B=60^\[Degree]，复斜率kAB=u^2，kAC=1/v^2，kCD=w *)
\!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1;
(*由\[Angle]B=60^\[Degree]求 u 值:*)Simplify@ Solve[{E^(2 I (60 \[Pi] )/180) == u^2}, {u}]; u = Power[-1, (3)^-1];
(*设kAC、kCD的复斜率分别为 1/v^2 和 w:*) kAC = 1/v^2; kCD = w;
(*以下坐标是此构图下的已知公式，直接引用不再推导:*)
a = Simplify[(u^2 (v^2 - 1))/(u^2 v^2 - 1)]; \!\(\*OverscriptBox[\(a\), \(_\)]\) = Simplify[(v^2 - 1)/(u^2 v^2 - 1)];
(*D点坐标:*)d = Simplify[-((u^2 v^2 (w - 1))/(u^2 v^2 - 1))]; \!\(\*OverscriptBox[\(d\), \(_\)]\) = Simplify[(1 - w)/(w (u^2 v^2 - 1))];
(*四个点确定的两条直线的交点:*)
Jd1[a_, b_, c_, d_] := ((\!\(\*OverscriptBox[\(c\), \(_\)]\) d - c \!\(\*OverscriptBox[\(d\), \(_\)]\)) (a - b) - (\!\(\*OverscriptBox[\(a\), \(_\)]\) b - a \!\(\*OverscriptBox[\(b\), \(_\)]\)) (c - d))/((a - b) (\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)) - (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)) (c - d));
\!\(\*OverscriptBox[\(Jd1\), \(_\)]\)[a_, b_, c_, d_] := ((c \!\(\*OverscriptBox[\(d\), \(_\)]\) - \!\(\*OverscriptBox[\(c\), \(_\)]\) d) (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)) - (a \!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\) b) (\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)))/((\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)) (c - d) - (a - b) (\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)));
(*P点坐标:*)p = Simplify@Jd1[a, b, c, d]; \!\(\*OverscriptBox[\(p\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd1\), \(_\)]\)[a, b, c, d];
(*Q点坐标:*)q = Simplify@Jd1[a, d, b, c]; \!\(\*OverscriptBox[\(q\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd1\), \(_\)]\)[a, d, b, c];
(*线段BA、BQ、BC、BP的长度:*)
BA = Simplify[Sqrt[(b - a) (\!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\))]]; BA = ( v^2 - 1)/((-1)^(2/3) v^2 - 1) Sqrt[((-1)^(2/3)) ];
BQ = q; BC = 1;
BP = Simplify[Sqrt[(b - p) (\!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(p\), \(_\)]\))]]; BP = -(( w - 1)/((-1)^(2/3) - w)) Sqrt[((-1)^(2/3)) ];
Print["\!\(\*FractionBox[\(BA\), \(BQ\)]\) = ", FullSimplify@ComplexExpand[BA/BQ]];
Print["\!\(\*FractionBox[\(BC\), \(BP\)]\) = ", FullSimplify@ComplexExpand[BC/BP]];
Print["因此 \!\(\*FractionBox[\(BA\), \(BQ\)]\) +\!\(\*FractionBox[\(\(\\\ \)\(BC\)\), \(BP\)]\) = ", Simplify[BA/BQ + BC/BP]];

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TSC999 · 发表于 2023-6-19 10:43

例 32：

程序代码：

Clear["Global`*"]; (*设B在坐标原点，C在1点，BA/BQ+BC/BP=1，复斜率 kAB=u^2，kAC=1/v^2，kCD=w *)
\!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1;
(*设kAC、kCD的复斜率分别为 1/v^2 和 w:*) kAC = 1/v^2; kCD = w;
(*以下坐标是此构图下的已知公式，直接引用不再推导:*)
a = (u^2 (v^2 - 1))/(u^2 v^2 - 1); \!\(\*OverscriptBox[\(a\), \(_\)]\) = (v^2 - 1)/( u^2 v^2 - 1); d = -((u^2 v^2 (w - 1))/(u^2 v^2 - 1)); \!\(\*OverscriptBox[\(d\), \(_\)]\) = (1 - w)/(w (u^2 v^2 - 1));
(*四个点确定的两条直线的交点:*)
Jd1[a_, b_, c_, d_] := ((\!\(\*OverscriptBox[\(c\), \(_\)]\) d - c \!\(\*OverscriptBox[\(d\), \(_\)]\)) (a - b) - (\!\(\*OverscriptBox[\(a\), \(_\)]\) b - a \!\(\*OverscriptBox[\(b\), \(_\)]\)) (c - d))/((a - b) (\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)) - (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)) (c - d));
\!\(\*OverscriptBox[\(Jd1\), \(_\)]\)[a_, b_, c_, d_] := ((c \!\(\*OverscriptBox[\(d\), \(_\)]\) - \!\(\*OverscriptBox[\(c\), \(_\)]\) d) (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)) - (a \!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\) b) (\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)))/((\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)) (c - d) - (a - b) (\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)));
(*P点坐标:*)p = Simplify@Jd1[a, b, c, d]; \!\(\*OverscriptBox[\(p\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd1\), \(_\)]\)[a, b, c, d];
(*Q点坐标:*)q = Simplify@Jd1[a, d, b, c]; \!\(\*OverscriptBox[\(q\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd1\), \(_\)]\)[a, d, b, c];
Print["P = ", p]; Print["Q = ", q];
(*线段BA、BQ、BC、BP的长度:*)
BA = Simplify[Sqrt[(b - a) (\!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\))]]; BA = (u (v^2 - 1))/(
u^2 v^2 - 1);
BQ = q; BC = 1;
BP = Simplify[Sqrt[(b - p) (\!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(p\), \(_\)]\))]]; BP = -((u (w - 1))/(u^2 - w));
\[Lambda]1 = Simplify[BA/BQ]; \[Lambda]2 = Simplify[BC/BP];
Print["\[Lambda]1 + \[Lambda]2 = ", Simplify[\[Lambda]1 + \[Lambda]2]];
Simplify@Solve[{\[Lambda]1 + \[Lambda]2 == 1}, {u}]; u = Power[-1, (3)^-1];
Simplify@Solve[{E^(2 I \[Angle]B) == u^2, 0 < \[Angle]B < \[Pi]/2}, {\[Angle]B}] // Flatten

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TSC999 · 发表于 2023-6-26 06:29

本帖最后由 TSC999 于 2023-6-27 23:50 编辑 例 33：
证明另一条公切线通过 D 点.png

解答：

程序代码：

Clear["Global`*"]; (*设B在坐标原点，C在1点， kAB=u^4，kAC=1/v^4 *)
\!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1; a = (u^4 (v^4 - 1))/( u^4 v^4 - 1);
\!\(\*OverscriptBox[\(a\), \(_\)]\) = (v^4 - 1)/(u^4 v^4 - 1); i = ( u^2 (v^2 - 1))/(u^2 v^2 - 1);
\!\(\*OverscriptBox[\(i\), \(_\)]\) = (v^2 - 1)/(u^2 v^2 - 1);
(*\[EmptyUpTriangle]ABC 的各边长度：*) BC = 1; AB = (u^2 (v^4 - 1))/( u^4 v^4 - 1); AC = v^2(u^4 - 1)/(u^4 v^4 - 1);
AI = Simplify[Sqrt[(a - i) (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(i\), \(_\)]\))]]; AI = -(( u v (u^2 - 1) (v^2 - 1) I)/(u^4 v^4 - 1));
BI = Simplify[Sqrt[(b - i) (\!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(i\), \(_\)]\))]]; BI = (u (v^2 - 1))/( u^2 v^2 - 1);
CI = Simplify[Sqrt[(c - i) (\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(i\), \(_\)]\))]]; CI = (v (u^2 - 1) )/( u^2 v^2 - 1);
o1 = (i BC + c BI + b CI)/(BC + BI + CI); \!\(\*OverscriptBox[\(o1\), \(_\)]\) = (\!\(\*OverscriptBox[\(i\), \(_\)]\) BC + \!\(\*OverscriptBox[\(c\), \(_\)]\) BI + \!\(\*OverscriptBox[\(b\), \(_\)]\) CI)/(BC + BI + CI); o2 = ( i AB + b AI + a BI)/(AB + AI + BI);
\!\(\*OverscriptBox[\(o2\), \(_\)]\) = (\!\(\*OverscriptBox[\(i\), \(_\)]\) AB + \!\(\*OverscriptBox[\(b\), \(_\)]\) AI + \!\(\*OverscriptBox[\(a\), \(_\)]\) BI)/(AB + AI + BI);
o3 = (i AC + c AI + a CI)/(AC + AI + CI); \!\(\*OverscriptBox[\(o3\), \(_\)]\) = (\!\(\*OverscriptBox[\(i\), \(_\)]\) AC + \!\(\*OverscriptBox[\(c\), \(_\)]\) AI + \!\(\*OverscriptBox[\(a\), \(_\)]\) CI)/(AC + AI + CI);
Print["a = ", Simplify[a], "， i = ", Simplify[i]];Print["o1 = ", Simplify[o1], "， o2 = ", Simplify[o2], "， o3 = ",
Simplify[o3]];
k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
W1 = {d, \!\(\*OverscriptBox[\(d\), \(_\)]\)} /. Simplify@Solve[{k[b, c] == k[b, d], k[o1, d] == -1}, {d, \!\(\*OverscriptBox[\(d\), \(_\)]\)}] // Flatten ;
d = Part[W1, 1]; \!\(\*OverscriptBox[\(d\), \(_\)]\) = d; Print["d = ", d];
W2 = {n, \!\(\*OverscriptBox[\(n\), \(_\)]\)} /. Simplify@Solve[{k[o2, o3] == k[o2, n], k[a, i] == k[a, n]}, {n, \!\(\*OverscriptBox[\(n\), \(_\)]\)}] // Flatten ;
n = Part[W2, 1]; \!\(\*OverscriptBox[\(n\), \(_\)]\) = Part[W2, 2]; Print["n = ", n];
Jx[p_, a_, b_] := (\!\(\*OverscriptBox[\(a\), \(_\)]\) b - a \!\(\*OverscriptBox[\(b\), \(_\)]\) + \!\(\*OverscriptBox[\(p\), \(_\)]\) (a - b))/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\));(*P点的镜线，镜线过已知点 A、B*)
\!\(\*OverscriptBox[\(Jx\), \(_\)]\)[p_, a_, b_] := (a \!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\) b + p (
\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)))/(a - b);
a1 = Simplify@Jx[a, o2, o3]; \!\(\*OverscriptBox[\(a1\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jx\), \(_\)]\)[a, o2, o3]; Print["a1 = ", a1];
Print["ND 的复斜率 kND = ", Simplify[k[n, d]]];
Print["NA1 的复斜率 kNA1 = ", Simplify[k[n, a1]]];
Print["由于 kND = kNA1，所以 N、D、A1 三点共线，即直线 NDA1 是两圆的另一条公切线。"];

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TSC999 · 发表于 2023-6-26 06:57

例 34：

解答：

程序代码：

Clear["Global`*"];(*令C点为坐标原点，A点坐标为 1，B点坐标为 \[ImaginaryI] *)
\!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 0; \!\(\*OverscriptBox[\(a\), \(_\)]\) = a = 1; b = I;
\!\(\*OverscriptBox[\(b\), \(_\)]\) = - I; n = (Sqrt[2] - 1) I; \!\(\*OverscriptBox[\(n\), \(_\)]\) = - (Sqrt[2] - 1) I;
(*圆心坐标：*) o = 1/2 + 1/2 I;
\!\(\*OverscriptBox[\(o\), \(_\)]\) = 1/2 - 1/2 I;(*下面有证明O与G重合*)
Foot[p_, x_, y_] := ( \!\(\*OverscriptBox[\(x\), \(_\)]\) y - x \!\(\*OverscriptBox[\(y\), \(_\)]\) + (x - y) \!\(\*OverscriptBox[\(p\), \(_\)]\) + (\!\(\*OverscriptBox[\(x\), \(_\)]\) - \!\(\*OverscriptBox[\(y\), \(_\)]\)) p)/(2 (\!\(\*OverscriptBox[\(x\), \(_\)]\) -
\!\(\*OverscriptBox[\(y\), \(_\)]\))); (* 从P点向XY直线引垂线，垂足的复坐标 *)
\!\(\*OverscriptBox[\(Foot\), \(_\)]\)[p_, x_, y_] := (x \!\(\*OverscriptBox[\(y\), \(_\)]\) - \!\(\*OverscriptBox[\(x\), \(_\)]\) y + (
\!\(\*OverscriptBox[\(x\), \(_\)]\) - \!\(\*OverscriptBox[\(y\), \(_\)]\)) p + (x - y) \!\(\*OverscriptBox[\(p\), \(_\)]\))/(2 (x - y));
k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\));(*复斜率定义*)
d = Simplify@Foot[c, a, n]; \!\(\*OverscriptBox[\(d\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Foot\), \(_\)]\)[c, a, n];
f = Simplify@Foot[b, c, d]; \!\(\*OverscriptBox[\(f\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Foot\), \(_\)]\)[b, c, d];
Simplify@Solve[{k[a, d] == k[a, h], k[c, h] == -k[c, o]}, {h, \!\(\*OverscriptBox[\(h\), \(_\)]\)}];
h = ((1 - I) (Sqrt[2] - 1))/(Sqrt[2] - 2); \!\(\*OverscriptBox[\(h\), \(_\)]\) = ((1 + I) (Sqrt[2] - 1))/( Sqrt[2] - 2);
e = Simplify@Foot[f, h, b]; \!\(\*OverscriptBox[\(e\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Foot\), \(_\)]\)[f, h, b];
Simplify@Solve[{k[a, b] == k[a, g], k[e, f] == k[e, g]}, {g, \!\(\*OverscriptBox[\(g\), \(_\)]\)}];
g = 1/2 + I/2; \!\(\*OverscriptBox[\(g\), \(_\)]\) = 1/2 - I/2;
Simplify@Solve[{k[a, d] == k[a, i], k[e, f] == k[e, i]}, {i, \!\(\*OverscriptBox[\(i\), \(_\)]\)}];
i = ((1294 + 816 I) - (915 + 577 I) Sqrt[2])/(956 - 676 Sqrt[2]);
\!\(\*OverscriptBox[\(i\), \(_\)]\) = ((816 + 577 I) - (577 + 408 I) Sqrt[2])/((239 + 577 I) - (169 + 408 I) Sqrt[2]);
AI = FullSimplify[Sqrt[(a - i) (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(i\), \(_\)]\))]]; GI = FullSimplify[Sqrt[(g - i) (\!\(\*OverscriptBox[\(g\), \(_\)]\) - \!\(\*OverscriptBox[\(i\), \(_\)]\))]];
Print["AI = ", AI, " \[TildeTilde] ", N[AI]]; Print["GI = ", GI, " \[TildeTilde] ", N[GI]];

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TSC999 · 发表于 2023-6-26 17:53

例 35：

解答：

程序代码：

Clear["Global`*"]; (*设B在坐标原点，C在1点， kAB=u^2，kAC=1/v^2 *)
\!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1; a = (u^2 (v^2 - 1))/( u^2 v^2 - 1);
\!\(\*OverscriptBox[\(a\), \(_\)]\) = (v^2 - 1)/(u^2 v^2 - 1);d = ((u + 1) (v - 1))/(2 u v - 2);
\!\(\*OverscriptBox[\(d\), \(_\)]\) = d; e = ((v - 1) (2 u v + u - 1))/(2 v (u v - 1));
\!\(\*OverscriptBox[\(e\), \(_\)]\) = ((v - 1) ((u - 1) v - 2))/( 2 - 2 u v);
f = (u (u + 1) (v - 1))/(2 u v - 2); \!\(\*OverscriptBox[\(f\), \(_\)]\) = ((u + 1) (v - 1))/( 2 u (u v - 1)); i = (u (v - 1))/(u v - 1);
\!\(\*OverscriptBox[\(i\), \(_\)]\) = (v - 1)/(u v - 1); h = ( u^2 + 1)/(1 - u^2 v^2);
\!\(\*OverscriptBox[\(h\), \(_\)]\) = ((u^2 + 1) v^2)/(u^2 v^2 - 1);
k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
Simplify@Solve[{k[e, f] == k[e, t], k[d, t] == -k[e, f]}, {t, \!\(\*OverscriptBox[\(t\), \(_\)]\)}];
t = ((u + 1) (v - 1) (u v + u + v - 1))/(4 v (u v - 1)); \!\(\*OverscriptBox[\(t\), \(_\)]\) = -(((u + 1) (v - 1) (u (v - 1) - v - 1))/(4 u (u v - 1)));
Simplify@Solve[{k[d, f] == k[d, m], k[e, m] == -k[d, f]}, {m, \!\(\*OverscriptBox[\(m\), \(_\)]\)}];
m = ((v - 1) (u^2 v (v + 1) + u (-v^2 + 2 v + 1) + v - 1))/( 4 v (u v - 1));
\!\(\*OverscriptBox[\(m\), \(_\)]\) = -(((v - 1) (u^2 (v - 1) v - u (v^2 + 2 v - 1) - v - 1))/(4 u v (u v - 1)));
Simplify@Solve[{k[d, e] == k[d, n], k[f, n] == -k[d, e]}, {n, \!\(\*OverscriptBox[\(n\), \(_\)]\)}];
n = ((u + 1) (v - 1) (u^2 v + u v + u - 1))/(4 u v (u v - 1)); \!\(\*OverscriptBox[\(n\), \(_\)]\) = -(((u + 1) (v - 1) (u^2 v -
u (v + 1) - 1))/(4 u (u v - 1)));
Simplify@Solve[{k[e, m] == k[e, p], k[d, t] == k[d, p]}, {p, \!\(\*OverscriptBox[\(p\), \(_\)]\)}];
p = ((v - 1) (u^2 v + u + v - 1))/(2 v (u v - 1)); \!\(\*OverscriptBox[\(p\), \(_\)]\) = -(((v - 1) (u^2 (v - 1) - u v - 1))/(2 u (u v - 1)));
Simplify[k[t, d]/k[i, t] == k[t, h]/ k[d, t]] (*\[Angle]ITD=\[Angle]DTH*)
Simplify[k[m, p]/k[h, m] == k[m, i]/k[p, m]] (*\[Angle]HMP=\[Angle]PMI*)
Simplify[k[n, p]/k[h, n] == k[n, i]/k[p, n]] (*\[Angle]HNP=\[Angle]PNI*)
Simplify[k[t, p]/k[m, t] == k[t, n]/k[p, t]] (*\[Angle]MTP=\[Angle]PTN*)
Simplify[k[n, p]/k[t, n] == k[n, m]/k[p, n]] (*\[Angle]TNP=\[Angle]PNM*)

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TSC999 · 发表于 2023-6-28 15:19

例 36：

程序代码：

Clear["Global`*"]; (*设B在坐标原点，C在1点， kAB=u^2，kAC=1/v^2 *)
\!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1; a = (u^2 (v^2 - 1))/( u^2 v^2 - 1);
\!\(\*OverscriptBox[\(a\), \(_\)]\) = (v^2 - 1)/( u^2 v^2 - 1);(* \[EmptyUpTriangle]ABC各顶点坐标 *)
g = (a + b + c)/3; \!\(\*OverscriptBox[\(g\), \(_\)]\) = (\!\(\*OverscriptBox[\(a\), \(_\)]\) + \!\(\*OverscriptBox[\(b\), \(_\)]\) +
\!\(\*OverscriptBox[\(c\), \(_\)]\))/3; (*重心坐标*)
r = \[Lambda] g; \!\(\*OverscriptBox[\(r\), \(_\)]\) = \[Lambda] \!\(\*OverscriptBox[\(g\), \(_\)]\);(* 再设 \[Mu]=GC/GS，则 c=\[Mu]s+(1-\[Mu])g *)
Simplify@Solve[{c == \[Mu] s + (1 - \[Mu]) g, \!\(\*OverscriptBox[\(c\), \(_\)]\) == \[Mu] \!\(\*OverscriptBox[\(s\), \(_\)]\) + (1 - \[Mu])
\!\(\*OverscriptBox[\(g\), \(_\)]\)}, {s, \!\(\*OverscriptBox[\(s\), \(_\)]\)}];(* 求 S 点坐标与 \[Mu] 的关系 *)
s = -((\[Mu] + u^2 (-(-\[Mu] + (2 \[Mu] + 1) v^2 + 1)) + 2)/( 3 \[Mu] (u^2 v^2 - 1)));
\!\(\*OverscriptBox[\(s\), \(_\)]\) = (-2 \[Mu] + v^2 (\[Mu] + (\[Mu] + 2) u^2 - 1) - 1)/(3 \[Mu] (u^2 v^2 - 1));
k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\));(*复斜率定义*)
Simplify@Solve[{k[b, g] == k[b, d], k[a, d] == k[a, s]}, {d, \!\(\*OverscriptBox[\(d\), \(_\)]\)}];(* 求 D 点坐标 *)
d = ((\[Mu] + 2) (u^2 (2 v^2 - 1) - 1))/(3 (\[Mu] + 1) (u^2 v^2 - 1)); \!\(\*OverscriptBox[\(d\), \(_\)]\) = ((\[Mu] + 2) ((u^2 + 1) v^2 - 2))/(3 (\[Mu] + 1) (u^2 v^2 - 1));
Simplify@Solve[{k[b, a]/k[b, s] == k[c, r]/k[c, a] == k[g, d]/ k[g, s]}, {\[Lambda], \[Mu]}];(* 由已知三个角相等的条件求 \[Lambda]、\[Mu] 与 u、v 的关系 *)
\[Lambda] = (3 (u^4 v^2 (v^2 - 2) + u^2 (v^4 + 1) - 2 v^2 + 1))/( u^4 v^2 (2 v^2 - 1) + 2 u^2 (v^4 - 3 v^2 + 1) - v^2 +
2); \[Mu] = (-2 u^4 (v^4 + v^2) + u^2 (v^4 + 6 v^2 + 1) - 2 (v^2 + 1))/(u^4 (v^4 + v^2) + u^2 (-5 v^4 + 6 v^2 - 5) + v^2 + 1);
(* 将 \[Lambda]、\[Mu] 代回 S、R 的原先坐标，消去其中的 \[Lambda]、\[Mu]： *)
s = Simplify[-((\[Mu] + u^2 (-(-\[Mu] + (2 \[Mu] + 1) v^2 + 1)) + 2)/( 3 \[Mu] (u^2 v^2 - 1)))];
\!\(\*OverscriptBox[\(s\), \(_\)]\) = Simplify[(-2 \[Mu] + v^2 (\[Mu] + (\[Mu] + 2) u^2 - 1) - 1)/( 3 \[Mu] (u^2 v^2 - 1))];
r = Simplify[\[Lambda] g]; \!\(\*OverscriptBox[\(r\), \(_\)]\) = Simplify[\[Lambda] \!\(\*OverscriptBox[\(g\), \(_\)]\)];
Print["S = ", s]; Print["R = ", r];
Simplify[k[a, s]/k[a, r] k[a, c]/k[a, b] == k[g, c]/k[g, b]](* 如果此式相等则表明\[Angle]RAS+\[Angle]BAC=\[Angle]BGC *)

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TSC999 · 发表于 2023-7-2 10:37

例 37：

解答：

程序代码：

Clear["Global`*"](*ABCD的外接圆为O，圆心O为坐标原点。BC边平行于实轴，AB、AC、BD的复斜率分别为 \
u^2、v^2、w^2 *)
\!\(\*OverscriptBox[\(o\), \(_\)]\) = o = 0; a = I u v; \!\(\*OverscriptBox[\(a\), \(_\)]\) = 1/a; b = I u/v;
\!\(\*OverscriptBox[\(b\), \(_\)]\) = 1/b; c = I v/u; \!\(\*OverscriptBox[\(c\), \(_\)]\) = 1/c; kBC = 1; kAB = u^2; kAC = v^2; kBD = w^2;
(* D点坐标：*)Simplify@Solve[{(b - d)/(1/b - 1/d) == kBD}, {d}]; d = I v w^2/u;
\!\(\*OverscriptBox[\(d\), \(_\)]\) = 1/d;(*以下求各线段的复斜率：*)
kAD = Simplify[(a - d)/(1/a - 1/d)]; kAD = v^2 w^2; kAP = Simplify[Sqrt[kAB kAD]]; kAP = -u v w;
kBQ = u; kDC = Simplify[(d - c)/(1/d - 1/c)]; kDC = ( v^2 w^2)/u^2; kCP = Simplify[Sqrt[kBC kDC]]; kCP = (v w)/u;
kDQ = Simplify[Sqrt[kAD kDC]]; kDQ = -((v^2 w^2)/u); kBY = Simplify[Sqrt[kBC kAB]]; kBY = -u;
kDY = Simplify[Sqrt[kAD kDC]]; kDY = ( v^2 w^2)/u; kAX = -kAP; kCX = -kCP;
(*过A1点、复斜率等于k1的直线，与过A2点、复斜率等于k2的直线的交点：*)
Jd[k1_, a1_, k2_, a2_] := -((k2 (a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\)) - k1 (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k1_, a1_, k2_, a2_] := -((a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\) - (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
p = Simplify@Jd[kAP, a, kCP, c]; \!\(\*OverscriptBox[\(p\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[kAP, a, kCP, c];
q=Simplify@Jd[kDQ, d, kBQ, b]; \!\(\*OverscriptBox[\(q\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[kDQ, d, kBQ, b];
x = Simplify@Jd[kAX, a, kCX, c]; \!\(\*OverscriptBox[\(x\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[kAX, a, kCX, c];
y = Simplify@Jd[kBY, b, kDY, d]; \!\(\*OverscriptBox[\(y\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[kBY, b, kDY, d];
Print["A = ", a, "， B = ", b, "， C = ", c, "， D = ", d];
Print["P = ", p, "， Q = ", q, "， X = ", x, "， Y = ", y];
k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
Print["P、Q、X、Y 四点是否共圆：", Simplify[k[y, q]/k[y, p] == k[x, q]/k[x, p]]];

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TSC999 · 发表于 2023-7-3 18:37

例 38：

程序代码：

\!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1; (*令B为坐标原点，C为1点，则A、P点坐标如下：*)
a = (1 - E^(-2 I (24 + 54)/180 \[Pi] ))/( 1 - E^(-2 I (24 + 54 + 24 + 30)/180 \[Pi] )); \!\(\*OverscriptBox[\(a\), \(_\)]\) = (
1 - E^(2 I (24 + 54)/180 \[Pi] ))/( 1 - E^(2 I (24 + 54 + 24 + 30)/180 \[Pi] )); (* A-B-C 逆时针环绕*)
p = (1 - E^(-2 I (24)/180 \[Pi] ))/( 1 - E^(-2 I (24 + 24)/180 \[Pi] )); \!\(\*OverscriptBox[\(p\), \(_\)]\) = (1 - E^(2 I (24)/180 \[Pi] ))/(1 - E^(2 I (24 + 24)/180 \[Pi] )); (* P-B-C 逆时针环绕*)
\[Angle][a_, b_, c_] := ArcTan[(I (\!\(\*OverscriptBox[\(a\), \(_\)]\) (c - b) + \!\(\*OverscriptBox[\(b\), \(_\)]\) (a - c) + \!\(\*OverscriptBox[\(c\), \(_\)]\) (b - a)))/(\!\(\*OverscriptBox[\(a\), \(_\)]\) (b - c) + \!\(\*OverscriptBox[\(b\), \(_\)]\) (a - 2 b + c) +
\!\(\*OverscriptBox[\(c\), \(_\)]\) (b - a))];(*从AB边逆时针转到BC边的有向角\[Angle]ABC的角度值*)
Print["\[Angle]BAP = ", FullSimplify[\[Angle][b, a, p]]];

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TSC999 · 发表于 2023-7-4 07:25

例 39：

解答：

程序代码：

Clear["Global`*"]; (*设B在坐标原点，C在1点， kAB=u^2，kAC=1/v^2 *)
\!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1; a = (u^2 (v^2 - 1))/( u^2 v^2 - 1);
\!\(\*OverscriptBox[\(a\), \(_\)]\) = (v^2 - 1)/(u^2 v^2 - 1); AB = ( u (v^2 - 1))/(u^2 v^2 - 1); AC = (v (u^2 - 1))/( u^2 v^2 - 1); BC = 1; BM = 1/2;
\!\(\*OverscriptBox[\(m\), \(_\)]\) = m = 1/2; e = \[Mu] a; \!\(\*OverscriptBox[\(e\), \(_\)]\) = \[Mu] \!\(\*OverscriptBox[\(a\), \(_\)]\); (* \[Mu]=BE/BA *)
BE = Simplify[Sqrt[(b - e) (\!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(e\), \(_\)]\))]]; BE = (\[Mu] u (v^2 - 1))/(
u^2 v^2 - 1); AE = Simplify[AB - BE];
WX[a1_, b1_, c1_] := (a1 \!\(\*OverscriptBox[\(a1\), \(_\)]\) (c1 - b1) + b1 \!\(\*OverscriptBox[\(b1\), \(_\)]\) (a1 - c1) + c1
\!\(\*OverscriptBox[\(c1\), \(_\)]\) (b1 - a1) )/(\!\(\*OverscriptBox[\(a1\), \(_\)]\) (c1 - b1) + \!\(\*OverscriptBox[\(b1\), \(_\)]\) (a1 - c1) + \!\(\*OverscriptBox[\(c1\), \(_\)]\) (b1 - a1));(*三角形 A1B1C1 的外心坐标：*)
\!\(\*OverscriptBox[\(WX\), \(_\)]\)[a1_, b1_, c1_] := -((a1 \!\(\*OverscriptBox[\(a1\), \(_\)]\) (\!\(\*OverscriptBox[\(c1\), \(_\)]\) - \!\(\*OverscriptBox[\(b1\), \(_\)]\)) + b1 \!\(\*OverscriptBox[\(b1\), \(_\)]\) (\!\(\*OverscriptBox[\(a1\), \(_\)]\) - \!\(\*OverscriptBox[\(c1\), \(_\)]\)) + c1 \!\(\*OverscriptBox[\(c1\), \(_\)]\) (\!\(\*OverscriptBox[\(b1\), \(_\)]\) - \!\(\*OverscriptBox[\(a1\), \(_\)]\)))/(\!\(\*OverscriptBox[\(a1\), \(_\)]\) (c1 - b1) + \!\(\*OverscriptBox[\(b1\), \(_\)]\) (a1 - c1) +
\!\(\*OverscriptBox[\(c1\), \(_\)]\) (b1 - a1)));
o = Simplify@ WX[a, e, m]; \!\(\*OverscriptBox[\(o\), \(_\)]\) = Simplify@ \!\(\*OverscriptBox[\(WX\), \(_\)]\)[a, e, m];
k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\));(*复斜率定义*)
W = {g, \!\(\*OverscriptBox[\(g\), \(_\)]\)} /. Simplify@Solve[{(o - a) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\)) == (o - g) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(g\), \(_\)]\)), k[a, c] == k[a, g]}, {g, \!\(\*OverscriptBox[\(g\), \(_\)]\) }] // Flatten ;
g = Part[W, 1]; \!\(\*OverscriptBox[\(g\), \(_\)]\) = Part[W, 2]; Print["G = ", g];
CG = Simplify[Sqrt[(c - g) (\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(g\), \(_\)]\))]]; CG = (u^4 v^4 - 2 u^2 (\[Mu] + \[Mu] v^4 + (1 - 2 \[Mu]) v^2) + 1)/( 2 (u^2 - 1) v (u^2 v^2 - 1)); AG = Simplify[AC - CG];
W1 = {q, \!\(\*OverscriptBox[\(q\), \(_\)]\)} /. Simplify@Solve[{(o - a) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\)) == (o - q) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(q\), \(_\)]\)), k[b, c] == k[b, q], q != m}, {q,
\!\(\*OverscriptBox[\(q\), \(_\)]\) }] // Flatten ;
q = Part[W1, 1]; \!\(\*OverscriptBox[\(q\), \(_\)]\) = Part[W1, 2]; Print["Q = ", q];
q = (2 \[Mu] u^2 (v^2 - 1)^2)/(u^2 v^2 - 1)^2; \!\(\*OverscriptBox[\(q\), \(_\)]\) = ( 2 \[Mu] u^2 (v^2 - 1)^2)/(u^2 v^2 - 1)^2; BQ = Simplify[Sqrt[(b - q) (\!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(q\), \(_\)]\))]];
BQ = 2 (\[Mu] u^2 (v^2 - 1)^2)/(u^2 v^2 - 1)^2;
p = Simplify[\[Eta] a + (1 - \[Eta]) m]; \!\(\*OverscriptBox[\(p\), \(_\)]\) = Simplify[\[Eta] \!\(\*OverscriptBox[\(a\), \(_\)]\) + (1 - \[Eta])
\!\(\*OverscriptBox[\(m\), \(_\)]\)];(*\[Eta]=MP/MA*) Print["P = ", p];
W2 = {d, \!\(\*OverscriptBox[\(d\), \(_\)]\)} /. Simplify@Solve[{k[g, p] == k[g, d], k[a, b] == k[a, d]}, {d, \!\(\*OverscriptBox[\(d\), \(_\)]\) }] // Flatten ;
d = Part[W2, 1]; \!\(\*OverscriptBox[\(d\), \(_\)]\) = Part[W2, 2]; Print["D = ", d];
AD = Simplify[Sqrt[(a - d) (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\))]]; AD = (-(\[Eta] - 1) u (v^2 -
1) (-2 \[Mu] u^2 + v^4 (u^4 - 2 \[Mu] u^2) - 2 v^2 (u^4 - (2 \[Mu] + 1) u^2 + 1) + 1))/( 2 (u^2 v^2 - 1) (-2 \[Mu] u^2 + v^4 (u^4 - 2 \[Mu] u^2) - v^2 (\[Eta] + (\[Eta] + 1) u^4 - 2 u^2 (\[Eta] + 2 \[Mu]) + 1) + 1));
W3 = {f, \!\(\*OverscriptBox[\(f\), \(_\)]\)} /. Simplify@Solve[{k[e, p] == k[e, f], k[a, c] == k[a, f]}, {f, \!\(\*OverscriptBox[\(f\), \(_\)]\) }] // Flatten ;
f = Part[W3, 1]; \!\(\*OverscriptBox[\(f\), \(_\)]\) = Part[W3, 2]; Print["F = ", f];
AF = Simplify[Sqrt[(a - f) (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(f\), \(_\)]\))]]; AF = ((\[Eta] - 1) (\[Mu] -
1) (u^2 - 1) v)/((\[Eta] - 2 \[Mu] + 1) (u^2 v^2 - 1));
MP = Simplify[Sqrt[(m - p) (\!\(\*OverscriptBox[\(m\), \(_\)]\) - \!\(\*OverscriptBox[\(p\), \(_\)]\))]]; MP = (-\[Eta] Sqrt[-((u^2 -
2) v^2 + 1) (u^2 (v^2 - 2) + 1)])/(2 (u^2 v^2 - 1)) ;
MA = Simplify[Sqrt[(m - a) (\!\(\*OverscriptBox[\(m\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\))]]; MA = -Sqrt[-((u^2 - 2) v^2 + 1) (u^2 (v^2 - 2) + 1)]/(2 (u^2 v^2 - 1));
Print["\!\(\*FractionBox[\(AB\\\ BE + AC\\\ CG\), SuperscriptBox[\(BC\), \(2\)]]\)=\!\(\*FractionBox[\(1\), \(2\)]\) 是否成立： ",
Simplify[(AB BE + AC CG)/BC^2 == 1/2]];
Print["\!\(\*FractionBox[\(AB\), \(AD\)]\)+\!\(\*FractionBox[\(AC\), \(AG\)]\)==\!\(\*FractionBox[\(AB\), \(AE\)]\)+\!\(\*FractionBox[\(AC\), \(AF\)]\)==\!\(\*FractionBox[\(2 MA\), \(MA - MP\)]\) 是否成立： ",
Simplify[AB/AD + AC/AG == AB/AE + AC/AF == (2 MA)/(MA - MP)]];

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TSC999 · 发表于 2023-7-4 12:56

本帖最后由 TSC999 于 2023-7-4 14:35 编辑 例 40：
证明圆E与圆O相切题.png

解答：

程序代码：

Clear["Global`*"];(*B为原点，内切圆半径为1，BC边与横轴重合。T、C 点坐标为两个实变量*)
\!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c; a = ((t^2 - 1) (t - c) + 2 I t (t - c))/(1 - c t + t^2);
\!\(\*OverscriptBox[\(a\), \(_\)]\) = ((t^2 - 1) (t - c) - 2 I t (t - c))/(1 - c t + t^2);
(*\[EmptyUpTriangle]ABC的外心坐标：*) o = (I (t + I)^2 (c - t + I)^2)/( 4 (t^2 + 1) - 4 c t);
\!\(\*OverscriptBox[\(o\), \(_\)]\) = (-I (t - I)^2 (c - t - I)^2)/(4 (t^2 + 1) - 4 c t);
\!\(\*OverscriptBox[\(oo\), \(_\)]\) = \!\(\*OverscriptBox[\(o\), \(_\)]\); (*转存一下，为的是以后从 Overscript[o, _] \
中消掉 c。后面的 Overscript[ee, _]=Overscript[e, _] 同理*)
(*\[EmptyUpTriangle]ABC的外接圆半径：*)R = ((1 + t^2) (1 + (t - c)^2) )/( 4 c t - 4 t^2 - 4);
(*三角形\[EmptyUpTriangle]ABC垂心坐标:*) h = (I (t^2 - 1) (-c + t + I)^2)/( 2 c t - 2 (t^2 + 1));
\!\(\*OverscriptBox[\(h\), \(_\)]\) = -((I (t^2 - 1) (c - t + I)^2)/(2 c t - 2 (t^2 + 1)));
d = Simplify[(a + h)/2]; \!\(\*OverscriptBox[\(d\), \(_\)]\) = Simplify[(\!\(\*OverscriptBox[\(a\), \(_\)]\) + \!\(\*OverscriptBox[\(h\), \(_\)]\))/2];
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的外心坐标*)
\!\(\*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)));
e = Simplify@WX[d, o, h]; \!\(\*OverscriptBox[\(e\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(WX\), \(_\)]\)[d, o, h];
\!\(\*OverscriptBox[\(ee\), \(_\)]\) = \!\(\*OverscriptBox[\(e\), \(_\)]\);(*\[EmptyUpTriangle]HOD的外心坐标*)
k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
(* 求AB与OE的交点E1：*)
W = {e1, \!\(\*OverscriptBox[\(e1\), \(_\)]\)} /. Simplify@Solve[{k[a, b] == k[a, e1], k[o, e1] == k[o, e]}, {e1, \!\(\*OverscriptBox[\(e1\), \(_\)]\)}] // Flatten;
e1 = Part[W, 1]; \!\(\*OverscriptBox[\(e1\), \(_\)]\) = Part[W, 2];
W1 = {c} /. Simplify@Solve[{e == e1, \!\(\*OverscriptBox[\(e\), \(_\)]\) == \!\(\*OverscriptBox[\(e1\), \(_\)]\)}, {c}] // Flatten; (*当E与E1重合时C点坐标与T点坐标的关系*)
c = Part[W1, 3];(*因为实数c>t>0，故选取第三个解*)
Print["C = ", c]; o = Simplify[o]; \!\(\*OverscriptBox[\(o\), \(_\)]\) = Simplify[\!\(\*OverscriptBox[\(oo\), \(_\)]\)]; e = Simplify[e];
\!\(\*OverscriptBox[\(e\), \(_\)]\) = Simplify[\!\(\*OverscriptBox[\(ee\), \(_\)]\)]; R = FullSimplify[R];
Print["O = ", o]; Print["E = ", e]; Print["R = ", R];
(*求圆E上O点的对径点F的坐标：*)
Jx[p_, a_, k_] := -k \!\(\*OverscriptBox[\(a\), \(_\)]\) + k \!\(\*OverscriptBox[\(p\), \(_\)]\) + a; (*P点的镜线，镜线过已知点A且复斜率为k*)
\!\(\*OverscriptBox[\(Jx\), \(_\)]\)[p_, a_, k_] := (k \!\(\*OverscriptBox[\(a\), \(_\)]\) - a + p)/k;
f = Simplify@Jx[o, e, -k[o, e]]; \!\(\*OverscriptBox[\(f\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jx\), \(_\)]\)[o, e, -k[o, e]];
Print["F = ", Simplify[f]];
Print["如果圆E的直径OF等于圆O的半径，则二圆相切。测试是否相切： ", Simplify[(f - o) (\!\(\*OverscriptBox[\(f\), \(_\)]\) - \!\(\*OverscriptBox[\(o\), \(_\)]\)) == R^2]];

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TSC999 · 发表于 2023-7-7 21:04

例 41：

程序代码：

Clear["Global`*"];(*设 BD=x，CE=y，令 B 点在坐标原点，C 点在 1 点*)
\!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1; a = 1/2 + I Sqrt[3]/2;
\!\(\*OverscriptBox[\(a\), \(_\)]\) = 1/2 - I Sqrt[3]/2;d = x Cos[\[Pi]/3] + I x Sin[\[Pi]/3];
\!\(\*OverscriptBox[\(d\), \(_\)]\) = x Cos[\[Pi]/3] - I x Sin[\[Pi]/3];e = (1 - y Cos[\[Pi]/3]) + I y Sin[\[Pi]/3];
\!\(\*OverscriptBox[\(e\), \(_\)]\) = (1 - y Cos[\[Pi]/3]) - I y Sin[\[Pi]/3];
(*由三角形三顶点求内切圆半径：*)
R[a_, b_, c_] := (I (\!\(\*OverscriptBox[\(a\), \(_\)]\) (c - b) + \!\(\*OverscriptBox[\(b\), \(_\)]\) (a - c) + \!\(\*OverscriptBox[\(c\), \(_\)]\) (b - a)))/(2 (Sqrt[(a - b) (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\))] + Sqrt[(a - c) (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(c\), \(_\)]\))] + Sqrt[(b - c) (\!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(c\), \(_\)]\))]));(*ABC逆时针环绕*)
RA = Simplify[R[a, d, e]]; RA = (Sqrt[3] (x - 1) (y - 1))/( 2 (Sqrt[x^2 - x (y + 1) + y^2 - y + 1] + (1 - x) + (1 - y)));
RB = Simplify@R[b, e, d]; RB = (Sqrt[3] x (1 - y))/( 2 (Sqrt[x^2 - x (y + 1) + y^2 - y + 1] + x + Sqrt[y^2 - y + 1]));
RC = Simplify@R[c, e, b]; RC = (Sqrt[3] y)/( 2 (y + Sqrt[y^2 - y + 1] + 1));
Simplify@Solve[{RA == RB == RC}, {x, y}];
x = (2 - Power[2, (3)^-1] + 3 Power[4, (3)^-1])/10; y = (1 + 2 Power[2, (3)^-1] - Power[4, (3)^-1])/5;
d = x Cos[\[Pi]/3] + I x Sin[\[Pi]/3]; \!\(\*OverscriptBox[\(d\), \(_\)]\) = x Cos[\[Pi]/3] - I x Sin[\[Pi]/3];
e = (1 - y Cos[\[Pi]/3]) + I y Sin[\[Pi]/3]; \!\(\*OverscriptBox[\(e\), \(_\)]\) = (1 - y Cos[\[Pi]/3]) - I y Sin[\[Pi]/3];
Print["BD = (2-\!\(\*RadicalBox[\(2\), \(3\)]\)+3\!\(\*RadicalBox[\(4\), \(3\)]\))/10 \[TildeTilde] ", N[x]];
Print["CE = (1+2 \!\(\*RadicalBox[\(2\), \(3\)]\)-\!\(\*RadicalBox[\(4\), \(3\)]\))/5 \[TildeTilde] ", N[y]];
Print["DB 是否等于 DE：", Simplify[(b - d) (\!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)) == (e - d) (\!\(\*OverscriptBox[\(e\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\))]];

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TSC999 · 发表于 2023-7-7 21:30

例 42：

程序代码：

Clear["Global`*"];
\!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 6; \!\(\*OverscriptBox[\(m\), \(_\)]\) = m = 2; a = u + v I; \!\(\*OverscriptBox[\(a\), \(_\)]\) = u - v I;
(*从AB边逆时针转到BC边的有向角\[Angle]ABC的角度值：*)
\[Angle][a_, b_, c_] := ArcTan[(I (\!\(\*OverscriptBox[\(a\), \(_\)]\) (c - b) + \!\(\*OverscriptBox[\(b\), \(_\)]\) (a - c) + \!\(\*OverscriptBox[\(c\), \(_\)]\) (b - a)))/(\!\(\*OverscriptBox[\(a\), \(_\)]\) (b - c) + \!\(\*OverscriptBox[\(b\), \(_\)]\) (a - 2 b + c) + \!\(\*OverscriptBox[\(c\), \(_\)]\) (b - a))];
\[Angle]CBA = Simplify[\[Angle][c, b, a]]; \[Angle]ACB = Simplify[\[Angle][a, c, b]];
Simplify@Solve[{ \[Angle]ACB == 2 \[Angle]CBA}, v]; v = Sqrt[3] Sqrt[(u - 4) u];a = u + v I; \!\(\*OverscriptBox[\(a\), \(_\)]\) = u - v I;
d = (a + c)/2; \!\(\*OverscriptBox[\(d\), \(_\)]\) = (\!\(\*OverscriptBox[\(a\), \(_\)]\) + \!\(\*OverscriptBox[\(c\), \(_\)]\))/2;
BM = 2; MD = Simplify[Sqrt[(m - d) (\!\(\*OverscriptBox[\(m\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\))]]; MD = u - 1;
AD = Simplify[Sqrt[(a - d) (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\))]] ; AD = u - 3;
Print["MD = BM+AD 是否成立： ", Simplify[MD == BM + AD]];

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TSC999 · 发表于 2023-7-10 21:25

例 43：

注：D 点位置确定以后，由于△ABC 和△DEF 有共同的垂心 H，所以 E 和 F 点的位置也就随之而定了。
解答：
证明HKO共线题解答.png

程序代码：

Clear["Global`*"];(*令ABCDEF的外接圆为单位圆，圆心O为坐标原点。BC边平行于实轴，AB、AC、DB的复斜率分别为 \
u^2、v^2、w^2 *)
\!\(\*OverscriptBox[\(o\), \(_\)]\) = o = 0; a = I u v; \!\(\*OverscriptBox[\(a\), \(_\)]\) = 1/a; b = I u/v;
\!\(\*OverscriptBox[\(b\), \(_\)]\) = 1/b; c = I v/u; \!\(\*OverscriptBox[\(c\), \(_\)]\) = 1/c;
Simplify@Solve[{(b - d)/(1/b - 1/d) == w^2}, {d}];(* DB的复斜率为 w^2 *)
d = I v w^2/u; \!\(\*OverscriptBox[\(d\), \(_\)]\) = 1/d;
h = (I ((v^2 + 1) u^2 + v^2))/(u v); \!\(\*OverscriptBox[\(h\), \(_\)]\) = -((I (u^2 + v^2 + 1))/(u v));
W1 = {e, f} /. Simplify@Solve[{(e - h)/(1/e - \!\(\*OverscriptBox[\(h\), \(_\)]\)) == -((f - d)/(1/f - 1/d)), ( f - h)/(1/f - \!\(\*OverscriptBox[\(h\), \(_\)]\)) == -((e - d)/(1/e - 1/d)), e != 0, f != 0}, {e, f}] // Flatten ;
t1 = \[Sqrt](((v^2 + 1) (w^2 - 1) u^4 + (w^2 v^4 - (w^4 - 4 w^2 + 1) v^2 + w^2) u^2 - v^2 (v^2 + 1) w^2 (w^2 - 1))^2 -
4 u^2 v^2 w^2 ((w^2 - 1) u^2 + (v^2 + 1) w^2) (u^2 (v^2 + 1) - v^2 (w^2 - 1)));
e = -((I (t - u^4 (v^2 + 1) (w^2 - 1) + u^2 (-v^4 w^2 + v^2 (w^4 - 4 w^2 + 1) - w^2) + v^2 (v^2 + 1) w^2 (w^2 - 1)))/(
2 u v ((w^2 - 1) u^2 + (v^2 + 1) w^2))) ; \!\(\*OverscriptBox[\(e\), \(_\)]\) = 1/e;
f = (I (t + u^4 (v^2 + 1) (w^2 - 1) + u^2 (v^4 w^2 - v^2 (w^4 - 4 w^2 + 1) + w^2) - v^2 (v^2 + 1) w^2 (w^2 - 1)))/(
2 u v ((w^2 - 1) u^2 + (v^2 + 1) w^2)) ; \!\(\*OverscriptBox[\(f\), \(_\)]\) = 1/f;
(*四个点确定的两条直线的交点:*)
Jd1[a_, b_, c_, d_] := ((\!\(\*OverscriptBox[\(c\), \(_\)]\) d - c \!\(\*OverscriptBox[\(d\), \(_\)]\)) (a - b) - (\!\(\*OverscriptBox[\(a\), \(_\)]\) b - a \!\(\*OverscriptBox[\(b\), \(_\)]\)) (c - d))/((a - b) (\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)) - (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)) (c - d));
\!\(\*OverscriptBox[\(Jd1\), \(_\)]\)[a_, b_, c_, d_] := ((c \!\(\*OverscriptBox[\(d\), \(_\)]\) - \!\(\*OverscriptBox[\(c\), \(_\)]\) d) (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)) - (a \!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\) b) (\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)))/((\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)) (c - d) - (a - b) (\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)));
p = Simplify@Jd1[a, d, c, f]; \!\(\*OverscriptBox[\(p\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd1\), \(_\)]\)[a, d, c, f];
q = Simplify@Jd1[b, e, c, f]; \!\(\*OverscriptBox[\(q\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd1\), \(_\)]\)[b, e, c, f];
r = Simplify@Jd1[a, d, b, e]; \!\(\*OverscriptBox[\(r\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd1\), \(_\)]\)[a, d, b, e];
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));(*三角形 ABC 的外心坐标：*)
\!\(\*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)));
kk = Simplify@WX[p, q, r]; \!\(\*OverscriptBox[\(kk\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(WX\), \(_\)]\)[p, q, r];
K[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
k = Simplify[kk /. t -> t1]; \!\(\*OverscriptBox[\(k\), \(_\)]\) = Simplify[\!\(\*OverscriptBox[\(kk\), \(_\)]\) /. t -> t1];
Print["OH 的复斜率 kOH = ", Simplify[K[o, h]]];Print["OK 的复斜率 kOK = ", Simplify[K[o, k]]];
Print["由于 kOH = kOK，所以 O、K、H 三点共线。 "];

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TSC999 · 发表于 2023-7-12 15:19

本帖最后由 TSC999 于 2023-7-12 15:28 编辑 例 44：
证明 I 也是三角形 DEF 的内心.png

Clear["Global`*"];(*B为原点，内切圆半径为1，BC边与实轴重合。T、C 点坐标为两个变量*)
\!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c; a = ((t^2 - 1) (t - c) + 2 I t (t - c))/(1 - c t + t^2); \!\(\*OverscriptBox[\(a\), \(_\)]\) = ((t^2 - 1) (t - c) - 2 I t (t - c))/(1 - c t + t^2);
(*\[EmptyUpTriangle]ABC的内心坐标 \[DoubleLongRightArrow]*) i = t + I; \!\(\*OverscriptBox[\(i\), \(_\)]\) = t - I;
h = (I (t^2 - 1) (-c + t + I)^2)/(2 c t - 2 (t^2 + 1)); \!\(\*OverscriptBox[\(h\), \(_\)]\) = -((I (t^2 - 1) (c - t + I)^2)/(2 c t - 2 (t^2 + 1)));
d = \[Lambda] a + (1 - \[Lambda]) h;(*HD/HA=\[Lambda]*)\!\(\*OverscriptBox[\(d\), \(_\)]\) = \[Lambda] \!\(\*OverscriptBox[\(a\), \(_\)]\) + (1 - \[Lambda]) \!\(\*OverscriptBox[\(h\), \(_\)]\);
e = h/\[Mu];(*BH/BE=\[Mu]*)\!\(\*OverscriptBox[\(e\), \(_\)]\) = \!\(\*OverscriptBox[\(h\), \(_\)]\)/\[Mu];
f = \[Nu] h + (1 - \[Nu]) c;(*CF/CH=\[Nu]*)\!\(\*OverscriptBox[\(f\), \(_\)]\) = \[Nu] \!\(\*OverscriptBox[\(h\), \(_\)]\) + (1 - \[Nu]) \!\(\*OverscriptBox[\(c\), \(_\)]\);AD2 = Simplify[(a - d) (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\))]; BE2 = Simplify[(b - e) (\!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(e\), \(_\)]\))]; CF2 = Simplify[(c - f) (\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(f\), \(_\)]\))];
Simplify@Solve[{AD2 == 4}, {\[Lambda]}]; \[Lambda] = ( c^2 (t^2 - 1) - 2 c (t^2 - 1) t + t^4 - 2 t^2 - 3)/( c^2 (t^2 - 1) - 2 c (t^3 + t) + (t^2 + 1)^2);
Simplify@Solve[{BE2 == 4}, {\[Mu]}]; \[Mu] = ( (1 - t^2) (c^2 - 2 c t + t^2 + 1))/(4 (1 + t^2 - c t));
Simplify@Solve[{CF2 == 4}, {\[Nu]}]; \[Nu] = (4 (c t - t^2 - 1))/((t^2 + 1) (c^2 - 2 c t + t^2 - 1));
d = Simplify[\[Lambda] a + (1 - \[Lambda]) h]; \!\(\*OverscriptBox[\(d\), \(_\)]\) = Simplify[\[Lambda] \!\(\*OverscriptBox[\(a\), \(_\)]\) + (1 - \[Lambda]) \!\(\*OverscriptBox[\(h\), \(_\)]\)]; e = Simplify[h/\[Mu]];
\!\(\*OverscriptBox[\(e\), \(_\)]\) = Simplify[\!\(\*OverscriptBox[\(h\), \(_\)]\)/\[Mu]]; f = Simplify[\[Nu] h + (1 - \[Nu]) c];
\!\(\*OverscriptBox[\(f\), \(_\)]\) = Simplify[\[Nu] \!\(\*OverscriptBox[\(h\), \(_\)]\) + (1 - \[Nu]) \!\(\*OverscriptBox[\(c\), \(_\)]\)];
Print["D = ", d]; Print["E = ", e]; Print["F = ", f];
k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
Print["I 点是否在\[Angle]EDF 的平分线上：", Simplify[k[d, i]^2 == k[d, e] k[d, f]]]
Print["I 点是否在\[Angle]DFE 的平分线上：", Simplify[k[f, i]^2 == k[f, d] k[f, e]]]
Print["因此 I 也是 \[EmptyUpTriangle]DEF 的内心"]

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TSC999 · 发表于 2023-7-13 17:33

本帖最后由 TSC999 于 2023-7-13 17:41 编辑 例 45：
证明 DZ 是圆的切线.png

程序代码：

Clear["Global`*"](*令△ABC的外接圆为单位圆，圆心O为坐标原点，且BC边平行于实轴，AB、AC、AD的复斜率分别为 u^2、v^2、t^2 *)
\!\(\*OverscriptBox[\(o\), \(_\)]\) = o = 0; a = I u v; \!\(\*OverscriptBox[\(a\), \(_\)]\) = 1/a; b = I u/v;
\!\(\*OverscriptBox[\(b\), \(_\)]\) = 1/b; c = I v/u; \!\(\*OverscriptBox[\(c\), \(_\)]\) = 1/c; kAB = u^2; kAC = v^2; kAD = t^2;
(*过A1点、复斜率等于k1的直线，与过A2点、复斜率等于k2的直线的交点：*)
Jd[k1_, a1_, k2_, a2_] := -((k2 (a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\)) - k1 (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k1_, a1_, k2_, a2_] := -((a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\) - (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
d = Simplify@Jd[1, b, t^2, a]; \!\(\*OverscriptBox[\(d\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[1, b, t^2, a];
w = Simplify@Jd[-1, d, k[a, o], a]; \!\(\*OverscriptBox[\(w\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[-1, d, k[a, o], a];
x = Simplify@Jd[-1, d, v^2, c]; \!\(\*OverscriptBox[\(x\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[-1, d, v^2, c];
y = Simplify@Jd[-1, d, u^2, b] ; \!\(\*OverscriptBox[\(y\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[-1, d, u^2, b] ;
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));(*三角形 ABC 的外心坐标：*)
\!\(\*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)));
o1 = Simplify@WX[a, x, y]; \!\(\*OverscriptBox[\(o1\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(WX\), \(_\)]\)[a, x, y];
\!\(\*OverscriptBox[\(to1\), \(_\)]\) = \!\(\*OverscriptBox[\(o1\), \(_\)]\);
Jx[p_, a_, b_] := (\!\(\*OverscriptBox[\(a\), \(_\)]\) b - a \!\(\*OverscriptBox[\(b\), \(_\)]\) + \!\(\*OverscriptBox[\(p\), \(_\)]\) (a - b))/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\));(*P点的镜线P1，镜线是AB*)
\!\(\*OverscriptBox[\(Jx\), \(_\)]\)[p_, a_, b_] := (a \!\(\*OverscriptBox[\(b\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\) b + p (
\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)))/(a - b);
z = Simplify@Jx[a, o, o1]; \!\(\*OverscriptBox[\(z\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jx\), \(_\)]\)[a, o, o1];
\!\(\*OverscriptBox[\(tz\), \(_\)]\) = \!\(\*OverscriptBox[\(z\), \(_\)]\);
mdw = (d + w)/2; \!\(\*OverscriptBox[\(mdw\), \(_\)]\) = (\!\(\*OverscriptBox[\(d\), \(_\)]\) + \!\(\*OverscriptBox[\(w\), \(_\)]\))/2;
Simplify@Solve[{k[o, mdw] == 1}, {t}];
t = Sqrt[u^4 v^4 + u^2 (v^2 - 1) - v^2]/Sqrt[u^4 v^2 + u^2 v^2 (v^2 - 1) - 1];
o1 = Simplify[o1]; \!\(\*OverscriptBox[\(o1\), \(_\)]\) = Simplify[\!\(\*OverscriptBox[\(to1\), \(_\)]\)]; Print["o1 = ", o1];
z = Simplify[z]; \!\(\*OverscriptBox[\(z\), \(_\)]\) = Simplify[\!\(\*OverscriptBox[\(tz\), \(_\)]\)]; Print["z = ", z];
Print["OZ 的复斜率 kOZ = ", Simplify[k[o, z]]];
Print["O1Z 的复斜率 kO1Z = ", Simplify[k[o1, z]]];
Print["由于 kOZ = -kO1Z，所以 OZ 垂直于 O1Z，即 OZ 或 DZ 是圆O1 的切线。"];

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TSC999 · 发表于 2023-7-13 22:12

例 46：

程序代码：

Clear["Global`*"];(*令B为坐标原点，C为 1 点*)
\!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1; a = 1/2 + Sqrt[3]/2 I;
\!\(\*OverscriptBox[\(a\), \(_\)]\) = 1/2 - Sqrt[3]/2 I; \!\(\*OverscriptBox[\(d\), \(_\)]\) = d;
e = (1 - d/2) + d Sqrt[3]/2 I; \!\(\*OverscriptBox[\(e\), \(_\)]\) = (1 - d/2) - d Sqrt[3]/2 I;
k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
(*过A1点、复斜率等于k1的直线，与过A2点、复斜率等于k2的直线的交点：*)
Jd[k1_, a1_, k2_, a2_] := -((k2 (a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\)) - k1 (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k1_, a1_, k2_, a2_] := -((a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\) - (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
f = Simplify@Jd[k[a, d], d, k[b, e], e]; \!\(\*OverscriptBox[\(f\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k[a, d], d, k[b, e], e];
g = Simplify@Jd[k[c, f], f, k[d, e], e]; \!\(\*OverscriptBox[\(g\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k[c, f], f, k[d, e], e];
h = Simplify@Jd[k[b, g], d, k[a, g], e]; \!\(\*OverscriptBox[\(h\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k[b, g], d, k[a, g], e];
(*从AB边逆时针转到BC边的有向角\[Angle]ABC的角度值:*)
\[Angle][a_, b_, c_] := ArcTan[(I (\!\(\*OverscriptBox[\(a\), \(_\)]\) (c - b) + \!\(\*OverscriptBox[\(b\), \(_\)]\) (a - c) + \!\(\*OverscriptBox[\(c\), \(_\)]\) (b - a)))/(\!\(\*OverscriptBox[\(a\), \(_\)]\) (b - c) + \!\(\*OverscriptBox[\(b\), \(_\)]\) (a - 2 b + c) + \!\(\*OverscriptBox[\(c\), \(_\)]\) (b - a))];
Print["\[Angle]ACH = ", Simplify[\[Angle][a, c, h]]];
Print["\[Angle]HCB = ", Simplify[\[Angle][h, c, b]]];
Print["因此 CH 平分 \[Angle]ACB。"];

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TSC999 · 发表于 2023-7-15 08:15

例 47：

程序代码：

Clear["Global`*"];(*令△ABC的外接圆为单位圆，圆心O为坐标原点，且BC边平行于实轴，AB、AC的复斜率分别为 u^2、v^2 *)
\!\(\*OverscriptBox[\(o\), \(_\)]\) = o = 0; a = I u v; \!\(\*OverscriptBox[\(a\), \(_\)]\) = 1/a; b = I u/v;
\!\(\*OverscriptBox[\(b\), \(_\)]\) = 1/b; c = I v/u; \!\(\*OverscriptBox[\(c\), \(_\)]\) = 1/c;
K[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
W1 = {d, \!\(\*OverscriptBox[\(d\), \(_\)]\)} /. Simplify@Solve[{(o - b) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)) == (o - d) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)), K[a, d] == K[b, c]}, {d, \!\(\*OverscriptBox[\(d\), \(_\)]\)}] // Flatten ;
d = Part[W1, 1]; \!\(\*OverscriptBox[\(d\), \(_\)]\) = Part[W1, 2];
W2 = {p, \!\(\*OverscriptBox[\(p\), \(_\)]\)} /. Simplify@Solve[{K[a, p] == K[a, c], K[b, p] == K[b, d]}, {p, \!\(\*OverscriptBox[\(p\), \(_\)]\)}] // Flatten ; p = Part[W2, 1];
\!\(\*OverscriptBox[\(p\), \(_\)]\) = Part[W2, 2];
W3 = {e, \!\(\*OverscriptBox[\(e\), \(_\)]\)} /. Simplify@Solve[{K[a, e] == K[a, b], K[e, p] == -K[b, d]}, {e, \!\(\*OverscriptBox[\(e\), \(_\)]\)}] // Flatten; e = Part[W3, 1]; \!\(\*OverscriptBox[\(e\), \(_\)]\) = Part[W3, 2];
W4 = {k, \!\(\*OverscriptBox[\(k\), \(_\)]\)} /. Simplify@Solve[{(o - b) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)) == (o - k) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(k\), \(_\)]\)), K[d, e] == K[d, k]}, {k, \!\(\*OverscriptBox[\(k\), \(_\)]\)}] // Flatten;
k = Part[W4, 1]; \!\(\*OverscriptBox[\(k\), \(_\)]\) = Part[W4, 2];
W5 = {f, \!\(\*OverscriptBox[\(f\), \(_\)]\)} /. Simplify@Solve[{(o - b) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)) == (o - f) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(f\), \(_\)]\)), K[b, d] == -K[a, f]}, {f, \!\(\*OverscriptBox[\(f\), \(_\)]\)}] // Flatten;
f = Part[W5, 1]; \!\(\*OverscriptBox[\(f\), \(_\)]\) = Part[W5, 2];
W6 = {h, \!\(\*OverscriptBox[\(h\), \(_\)]\)} /. Simplify@Solve[{K[b, h] == K[b, d], K[k, h] == K[k, f]}, {h, \!\(\*OverscriptBox[\(h\), \(_\)]\)}] // Flatten;
h = Part[W6, 1]; \!\(\*OverscriptBox[\(h\), \(_\)]\) = Part[W6, 2];
Simplify[K[h, k] K[h, c] == K[h, b]^2]

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TSC999 · 发表于 2023-7-17 15:17

本帖最后由 TSC999 于 2023-7-17 15:30 编辑 例 48：
证明 FE、GH、BC 共点.png

程序代码：

Clear["Global`*"]; (*设B在坐标原点，C在1点， kAB=u^2，kAC=1/v^2 *)
\!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1; a = (u^2 (v^2 - 1))/( u^2 v^2 - 1);
\!\(\*OverscriptBox[\(a\), \(_\)]\) = (v^2-1)/(u^2 v^2-1); \!\(\*OverscriptBox[\(d\), \(_\)]\) = d; h = (u^2 + 1)/(1 - u^2 v^2);
\!\(\*OverscriptBox[\(h\), \(_\)]\) = ((u^2 + 1) v^2)/(u^2 v^2 - 1);
k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
W1 = {e, \!\(\*OverscriptBox[\(e\), \(_\)]\)} /. Simplify@Solve[{(d - e) (\!\(\*OverscriptBox[\(d\), \(_\)]\) - \!\(\*OverscriptBox[\(e\), \(_\)]\)) == (1 - d)^2, k[a, e] == k[a, c], e != c}, {e, \!\(\*OverscriptBox[\(e\), \(_\)]\)}] // Flatten;
e = Part[W1, 1]; \!\(\*OverscriptBox[\(e\), \(_\)]\) = Part[W1, 2];
W2 = {f, \!\(\*OverscriptBox[\(f\), \(_\)]\)} /. Simplify@Solve[{(d - f) (\!\(\*OverscriptBox[\(d\), \(_\)]\) - \!\(\*OverscriptBox[\(f\), \(_\)]\)) == d^2, k[a, f] == k[a, b], f != b}, {f, \!\(\*OverscriptBox[\(f\), \(_\)]\)}] // Flatten;
f = Part[W2, 1]; \!\(\*OverscriptBox[\(f\), \(_\)]\) = Part[W2, 2];
(*AF的中点：*)mAF = (a + f)/2; \!\(\*OverscriptBox[\(mAF\), \(_\)]\) = (\!\(\*OverscriptBox[\(a\), \(_\)]\) + \!\(\*OverscriptBox[\(f\), \(_\)]\))/2; (*AE的中点：*)mAE = (a + e)/2; \!\(\*OverscriptBox[\(mAE\), \(_\)]\) = (\!\(\*OverscriptBox[\(a\), \(_\)]\) + \!\(\*OverscriptBox[\(e\), \(_\)]\))/2;
W3 = {o, \!\(\*OverscriptBox[\(o\), \(_\)]\)} /. Simplify@Solve[{k[mAF, o] == -k[a, f], k[mAE, o] == -k[a, e]}, {o, \!\(\*OverscriptBox[\(o\), \(_\)]\)}] // Flatten;
o = Part[W3, 1]; \!\(\*OverscriptBox[\(o\), \(_\)]\) = Part[W3, 2]; (*设经过 AEF 的圆的圆心为 O *)
W4 = {g, \!\(\*OverscriptBox[\(g\), \(_\)]\)} /. Simplify@Solve[{(o - a) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\)) == (o - g) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(g\), \(_\)]\)), k[a, g] == k[a, d]}, {g, \!\(\*OverscriptBox[\(g\), \(_\)]\)}] // Flatten;
g = Part[W4, 1]; \!\(\*OverscriptBox[\(g\), \(_\)]\) = Part[W4, 2];
W5 = {x, \!\(\*OverscriptBox[\(x\), \(_\)]\)} /. Simplify@Solve[{k[b, c] == k[c, x], k[f, e] == k[e, x]}, {x, \!\(\*OverscriptBox[\(x\), \(_\)]\)}] // Flatten;
x = Part[W5, 1]; \!\(\*OverscriptBox[\(x\), \(_\)]\) = Part[W5, 2];(*设 BC 与 FE 的交点为 X*)
W6 = {x1, \!\(\*OverscriptBox[\(x1\), \(_\)]\)} /. Simplify@Solve[{k[b, c] == k[c, x1], k[g, h] == k[h, x1]}, {x1, \!\(\*OverscriptBox[\(x1\), \(_\)]\)}] // Flatten;
x1 = Part[W6, 1]; \!\(\*OverscriptBox[\(x1\), \(_\)]\) = Part[W6, 2];(*设 BC 与 GH 的交点为 X1*)
Print["x 是否等于 x1： ", Simplify[x == x1]];
Print["由于 x = x1，所以 FE、GH、BC 交于一点 X。"];

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TSC999 · 发表于 2023-7-19 17:57

本帖最后由 TSC999 于 2023-7-20 10:36 编辑 例 49：
今年奥数题.png

据说这是 2023 年的一道奥数题。
程序代码：

Clear["Global`*"];(*u=\[ExponentialE]^(2\[ImaginaryI]\[Theta]);v=\[ExponentialE]^(2\[ImaginaryI]\[Beta]);*)
\!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1; a = 1/2 + I Sqrt[3]/2;
\!\(\*OverscriptBox[\(a\), \(_\)]\) = 1/a;
(*u=\[ExponentialE]^(2\[ImaginaryI]\[Theta]);v=\[ExponentialE]^(2\[ImaginaryI]\[Beta])*)
\[Angle]A1BC = \[Theta]; \[Angle]C1BA = \[Beta]; \[Angle]B1CA = \[Gamma]; \[Angle]BA1C = \[Pi] - 2 \[Theta];
\[Angle]AB1C = \[Pi] - 2 \[Gamma]; \[Angle]AC1B = \[Pi] - 2 \[Beta];
(*解 Solve[{\[Angle]BA1C+ \[Angle]AB1C+\[Angle]AC1B\[Equal](480\[Pi])/180},{\[Gamma]}]得到：*)
\[Gamma] = \[Pi]/6 - \[Beta] - \[Theta];(*现在只剩二个变量 \[Beta] 和 \[Theta]*)
kBA1 = u; kBC1 = E^(2 I (\[Pi]/3))/v; kCB1 = E^(2 I ((2 \[Pi])/3 + \[Gamma])); kCB1 = E^(2 I ((5 \[Pi])/6))/(u v);
mBC = (b + c)/2; \!\(\*OverscriptBox[\(mBC\), \(_\)]\) = (\!\(\*OverscriptBox[\(b\), \(_\)]\) + \!\(\*OverscriptBox[\(c\), \(_\)]\))/2; mAB = (a + b)/2; \!\(\*OverscriptBox[\(mAB\), \(_\)]\) = (\!\(\*OverscriptBox[\(a\), \(_\)]\) + \!\(\*OverscriptBox[\(b\), \(_\)]\))/2; mAC = (a + c)/2;
\!\(\*OverscriptBox[\(mAC\), \(_\)]\) = (\!\(\*OverscriptBox[\(a\), \(_\)]\) + \!\(\*OverscriptBox[\(c\), \(_\)]\))/2;
K[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
(*过A1点、复斜率等于k1的直线，与过A2点、复斜率等于k2的直线的交点：*)
Jd[k1_, a1_, k2_, a2_] := -((k2 (a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\)) - k1 (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k1_, a1_, k2_, a2_] := -((a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\) - (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
a1 = Simplify@Jd[kBA1, b, -K[b, c], mBC]; \!\(\*OverscriptBox[\(a1\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[kBA1, b, -K[b, c], mBC]; Print["A1 = ", a1];
b1 = Simplify@Jd[kCB1, c, -K[a, c], mAC]; \!\(\*OverscriptBox[\(b1\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[kCB1, c, -K[a, c], mAC]; Print["B1 = ", b1];
c1 = Simplify@Jd[kBC1, b, -K[a, b], mAB]; \!\(\*OverscriptBox[\(c1\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[kBC1, b, -K[a, b], mAB]; Print["C1 = ", c1];
a2 = Simplify@Jd[K[b, c1], c1, K[c, b1], b1]; \!\(\*OverscriptBox[\(a2\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[K[b, c1], c1, K[c, b1], b1];Print["A2 = ", a2];
c2 = Simplify@Jd[K[b, a1], a1, K[a, b1], b1]; \!\(\*OverscriptBox[\(c2\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[K[b, a1], a1, K[a, b1], b1];Print["C2 = ", c2];
b2 = Simplify@Jd[K[a, c1], c1, K[c, a1], a1]; \!\(\*OverscriptBox[\(b2\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[K[a, c1], c1, K[c, a1], a1];Print["B2 = ", b2];
Waixin[a_, b_, c_] := (a \!\(\*OverscriptBox[\(a\), \(_\)]\) (b - c) + b \!\(\*OverscriptBox[\(b\), \(_\)]\) (c - a) + c \!\(\*OverscriptBox[\(c\), \(_\)]\) (a - b) )/( \!\(\*OverscriptBox[\(a\), \(_\)]\) (b - c) + \!\(\*OverscriptBox[\(b\), \(_\)]\) (c - a) + \!\(\*OverscriptBox[\(c\), \(_\)]\) (a - b));
\!\(\*OverscriptBox[\(Waixin\), \(_\)]\)[a_, b_, c_] := -((a \!\(\*OverscriptBox[\(a\), \(_\)]\) (\!\(\*OverscriptBox[\(b\), \(_\)]\)-\!\(\*OverscriptBox[\(c\), \(_\)]\)) + b \!\(\*OverscriptBox[\(b\), \(_\)]\) (\!\(\*OverscriptBox[\(c\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\)) + c \!\(\*OverscriptBox[\(c\), \(_\)]\) (\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)) )/( \!\(\*OverscriptBox[\(a\), \(_\)]\) (b - c) + \!\(\*OverscriptBox[\(b\), \(_\)]\) (c - a) + \!\(\*OverscriptBox[\(c\), \(_\)]\) (a - b)));
o1 = Simplify@Waixin[a, a1, a2]; \!\(\*OverscriptBox[\(o1\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Waixin\), \(_\)]\)[a, a1, a2];
o2 = Simplify@Waixin[b, b1, b2]; \!\(\*OverscriptBox[\(o2\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Waixin\), \(_\)]\)[b, b1, b2];
o3 = Simplify@Waixin[c, c1, c2]; \!\(\*OverscriptBox[\(o3\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Waixin\), \(_\)]\)[c, c1, c2];
Print["O1 = ", o1]; Print["O2 = ", o2]; Print["O3 = ", o3];
XiangjiaoxuanLianxin[o1_, a_, o2_, b_] := (a \!\(\*OverscriptBox[\(a\), \(_\)]\) - b \!\(\*OverscriptBox[\(b\), \(_\)]\) + \!\(\*OverscriptBox[\(b\), \(_\)]\) o2 + b \!\(\*OverscriptBox[\(o2\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\) o1 - a \!\(\*OverscriptBox[\(o1\), \(_\)]\) + \!\(\*OverscriptBox[\(o2\), \(_\)]\) o1 - o2 \!\(\*OverscriptBox[\(o1\), \(_\)]\))/(2 (\!\(\*OverscriptBox[\(o2\), \(_\)]\) - \!\(\*OverscriptBox[\(o1\), \(_\)]\)));
\!\(\*OverscriptBox[\(XiangjiaoxuanLianxin\), \(_\)]\)[o1_, a_, o2_, b_] := (a \!\(\*OverscriptBox[\(a\), \(_\)]\) - b \!\(\*OverscriptBox[\(b\), \(_\)]\) + \!\(\*OverscriptBox[\(b\), \(_\)]\) o2 + b \!\(\*OverscriptBox[\(o2\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\) o1 - a \!\(\*OverscriptBox[\(o1\), \(_\)]\) + o2 \!\(\*OverscriptBox[\(o1\), \(_\)]\) - \!\(\*OverscriptBox[\(o2\), \(_\)]\) o1)/(2 (o2 - o1));
(*两圆连心线与公共弦的交点P，O1、O2是两圆圆心，A是O1上任一点，B是O2上任一点。如果两圆相切的，所求交点即是切点*)
m = Simplify@XiangjiaoxuanLianxin[o1, a, o2, b];
m1 = Simplify@XiangjiaoxuanLianxin[o1, a, o3, c];
SimpleFuc :=
Factor[#, Extension -> Sqrt[3]] &@(FactorTerms[Numerator[#]]/
FactorTerms[Denominator[#]]) &@Factor[#, Extension -> Sqrt[3]] &;
Print["M = ", SimpleFuc[m]];
Print["M1 = ", SimpleFuc[m1]];
Print["由于 M 和 M1 点的坐标相同，即圆O1、圆O2的根轴中心与圆O1、圆O3的根轴中心重合，因此三个圆只有两个公共交点。"];

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TSC999 · 发表于 2023-7-22 06:29

例 50：

程序代码：

Clear["Global`*"]; (*设A在坐标原点，C在1点， kAB=u^2，kBC=1/v^2 *)
\!\(\*OverscriptBox[\(a\), \(_\)]\) = a = 0; \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1; b = (u^2 (v^2 - 1))/( u^2 v^2 - 1);
\!\(\*OverscriptBox[\(b\), \(_\)]\) = (v^2 - 1)/( u^2 v^2 - 1);(*设 \[Lambda]=AF/AB, \[Mu]=AE/AC, \[Nu]=CD/CB *) o = (u^2 v^2)/(u^2 v^2 - 1); \!\(\*OverscriptBox[\(o\), \(_\)]\) = 1/( 1 - u^2 v^2);(*三角形 ABC 的外心坐标*)
f = \[Lambda] b; \!\(\*OverscriptBox[\(f\), \(_\)]\) = \[Lambda] \!\(\*OverscriptBox[\(b\), \(_\)]\); e = \[Mu] c;
\!\(\*OverscriptBox[\(e\), \(_\)]\) = \[Mu] \!\(\*OverscriptBox[\(c\), \(_\)]\); d = \[Nu] b + (1 - \[Nu]) c;
\!\(\*OverscriptBox[\(d\), \(_\)]\) = \[Nu] \!\(\*OverscriptBox[\(b\), \(_\)]\) + (1 - \[Nu]) \!\(\*OverscriptBox[\(c\), \(_\)]\);
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));(*由三角形顶点求其外心坐标*)
\!\(\*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)));
oa = Simplify@WX[a, e, f]; \!\(\*OverscriptBox[\(oa\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(WX\), \(_\)]\)[a, e, f];(*圆AEF的圆心坐标*)
ob = Simplify@WX[b, d, f]; \!\(\*OverscriptBox[\(ob\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(WX\), \(_\)]\)[b, d, f];(*圆BDF的圆心坐标*)
oc = Simplify@WX[c, d, e]; \!\(\*OverscriptBox[\(oc\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(WX\), \(_\)]\)[c, d, e];(*圆CDE的圆心坐标*)
k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
Simplify@Solve[{(oc - c) (\!\(\*OverscriptBox[\(oc\), \(_\)]\) - \!\(\*OverscriptBox[\(c\), \(_\)]\)) == (oa - a) (\!\(\*OverscriptBox[\(oa\), \(_\)]\) - \!\(\*OverscriptBox[\(a\), \(_\)]\)) == (ob - b) (\!\(\*OverscriptBox[\(ob\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\))}, {\[Mu], \[Nu]}];(*由三个小圆半径相等来求\[Mu]、\[Nu]与\[Lambda]的关系*)
\[Mu] = -((\[Lambda] + \[Lambda] (-u^2) v^4 + (\[Lambda] - 1) (u^2 - 1) v^2)/((v^2 + 1) (u^2 v^2 - 1))); \[Nu] = (\[Lambda] + u^2 (\[Lambda] - (\[Lambda] - 1) v^2) - \[Lambda] v^2 - 1)/((u^2 - 1) (v^2 + 1));
e = \[Mu] c; \!\(\*OverscriptBox[\(e\), \(_\)]\) = \[Mu] \!\(\*OverscriptBox[\(c\), \(_\)]\); d = \[Nu] b + (1 - \[Nu]) c;
\!\(\*OverscriptBox[\(d\), \(_\)]\) = \[Nu] \!\(\*OverscriptBox[\(b\), \(_\)]\) + (1 - \[Nu]) \!\(\*OverscriptBox[\(c\), \(_\)]\);(*将 \[Mu]、\[Nu] 与 \[Lambda] 的关系代回 E、D 点的坐标，消去 \[Mu]、\[Nu] 参数 *)
Print["AEFO 是否四点共圆：", Simplify[k[e, f]/k[e, o] == k[a, f]/k[a, o]]];(*如果圆周角\[Angle]OEF=\[Angle]OAF，则四点共圆*)
Print["CDEO 是否四点共圆：", Simplify[k[d, e]/k[d, o] == k[c, e]/k[c, o]]];(*如果圆周角\[Angle]ODE=\[Angle]OCE，则四点共圆*)
Print["BDFO 是否四点共圆：", Simplify[k[b, o]/k[b, f] == k[d, o]/k[d, f]]];(*如果圆周角\[Angle]FBO=\[Angle]FDO，则四点共圆*)
Print["因此，只要三个小圆大小相同，它们一定都通过三角形ABC的外心。"];

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