H是ΔABC外接圆Γ垂心，过A,H圆ω与AB,AC交于D,E，AD=AE，…，证：QE,PD交点S在圆ω上

creasson

给三种表示：


第一种(%78)：
\[ s = tan \frac{A}{2}, \quad  t = tan \frac{B}{2}\]

第二种(%79):
\[ z = e^{i A}, \quad  w =  e^{i B}\]

第三种(%80):
将\(A,B,C\) 设为单位圆上的复数，\( |a| = |b| = |c| = 1\), 此为史勇常用的复数设法。

天山草

在原题中还可推出：

① A、O、S 三点共线。 ② AO 是角 PAQ 的角平分线。 ③  S 是三角形 APQ 的垂心。 ④ AP=AQ，SP=SQ，即三角形 APQ 和  SPQ  都是等腰三角形。

⑤ 以 AS 为直径的圆（圆心为 O3）交 AQ 于 D3，交 AP 于 E3，则该圆与圆 O 相内切于 A 点。D、D3、S、P 共线，E、E3、S、Q 共线。  

天山草

1. Clear["Global`*"]; (*设三角形ABC的外接圆为单位圆，C 点在正 x 轴上，如图一*)
   
2. 
   
3. \!\(\*OverscriptBox[\(o\), \(_\)]\) = o = 0;
   
4. \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1;
   
5. 
   
6. \!\(\*OverscriptBox[\(a\), \(_\)]\) = 1/a; b = t^2;
   
7. \!\(\*OverscriptBox[\(b\), \(_\)]\) = 1/b;
   
8. \!\(\*OverscriptBox[\(t\), \(_\)]\) = 1/t;
   
9. \!\(\*OverscriptBox[\(p\), \(_\)]\) = 1/p;
   
10. h = a + b + c(*当三角形外心在坐标原点时此式成立*);
    
11. \!\(\*OverscriptBox[\(h\), \(_\)]\) =
    
12. \!\(\*OverscriptBox[\(a\), \(_\)]\) +
    
13. \!\(\*OverscriptBox[\(b\), \(_\)]\) +
    
14. \!\(\*OverscriptBox[\(c\), \(_\)]\);
    
15. m = (a + h)/2;(* M是A、H的中点*)
    
16. \!\(\*OverscriptBox[\(m\), \(_\)]\) = (
    
17. \!\(\*OverscriptBox[\(a\), \(_\)]\) +
    
18. \!\(\*OverscriptBox[\(h\), \(_\)]\))/2;
    
19. k[a_, b_] := (a - b)/(
    
20. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
21. \!\(\*OverscriptBox[\(b\), \(_\)]\));
    
22. \!\(\*OverscriptBox[\(k\), \(_\)]\)[a_, b_] := 1/k[a, b];(*直线的复斜率*)
    
23. Jd[k1_, a1_, k2_, a2_] := -((k2 (a1 - k1
    
24. \!\(\*OverscriptBox[\(a1\), \(_\)]\)) - k1 (a2 - k2
    
25. \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
    
26. 
    
27. \!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k1_, a1_, k2_, a2_] := -((a1 - k1
    
28. \!\(\*OverscriptBox[\(a1\), \(_\)]\) - (a2 - k2
    
29. \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
    
30. (*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*)
    
31. FourPoint[a_, b_, c_, d_] := ((
    
32. \!\(\*OverscriptBox[\(c\), \(_\)]\) d - c
    
33. \!\(\*OverscriptBox[\(d\), \(_\)]\)) (a - b) - (
    
34. \!\(\*OverscriptBox[\(a\), \(_\)]\) b - a
    
35. \!\(\*OverscriptBox[\(b\), \(_\)]\)) (c - d))/((a - b) (
    
36. \!\(\*OverscriptBox[\(c\), \(_\)]\) -
    
37. \!\(\*OverscriptBox[\(d\), \(_\)]\)) - (
    
38. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
39. \!\(\*OverscriptBox[\(b\), \(_\)]\)) (c - d));
    
40. 
    
41. \!\(\*OverscriptBox[\(FourPoint\), \(_\)]\)[a_, b_, c_, d_] := -(((c
    
42. \!\(\*OverscriptBox[\(d\), \(_\)]\) -
    
43. \!\(\*OverscriptBox[\(c\), \(_\)]\) d) (
    
44. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
45. \!\(\*OverscriptBox[\(b\), \(_\)]\)) - ( a
    
46. \!\(\*OverscriptBox[\(b\), \(_\)]\) -
    
47. \!\(\*OverscriptBox[\(a\), \(_\)]\) b) (
    
48. \!\(\*OverscriptBox[\(c\), \(_\)]\) -
    
49. \!\(\*OverscriptBox[\(d\), \(_\)]\)))/((a - b) (
    
50. \!\(\*OverscriptBox[\(c\), \(_\)]\) -
    
51. \!\(\*OverscriptBox[\(d\), \(_\)]\)) - (
    
52. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
53. \!\(\*OverscriptBox[\(b\), \(_\)]\)) (c - d)));
    
54. (*过A、B点的直线与过C、D点的直线的交点*)
    
55. Duichengdian[a_, b_, p_] := (
    
56. \!\(\*OverscriptBox[\(a\), \(_\)]\) b - a
    
57. \!\(\*OverscriptBox[\(b\), \(_\)]\) +
    
58. \!\(\*OverscriptBox[\(p\), \(_\)]\) (a - b))/(
    
59. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
60. \!\(\*OverscriptBox[\(b\), \(_\)]\));(*P点关于直线 AB 的镜像*)
    
61. 
    
62. \!\(\*OverscriptBox[\(Duichengdian\), \(_\)]\)[a_, b_, p_] := (a
    
63. \!\(\*OverscriptBox[\(b\), \(_\)]\) -
    
64. \!\(\*OverscriptBox[\(a\), \(_\)]\) b + p (
    
65. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
66. \!\(\*OverscriptBox[\(b\), \(_\)]\)))/(a - b);
    
67. o1 = Simplify[ Jd[-k[a, h], m, k[a, t], a]];
    
68. 
    
69. \!\(\*OverscriptBox[\(o1\), \(_\)]\) = Simplify[
    
70. \!\(\*OverscriptBox[\(Jd\), \(_\)]\)[-k[a, h], m, k[a, t], a]];
    
71. Print["o1=", o1];
    
72. Duichengdian1[a_, k_, p_] := -k
    
73. \!\(\*OverscriptBox[\(a\), \(_\)]\) + k
    
74. \!\(\*OverscriptBox[\(p\), \(_\)]\) +
    
75.    a;   (* P 点关于直线的镜像, 直线过 A 点且复斜率为 k *)
    
76. 
    
77. \!\(\*OverscriptBox[\(Duichengdian1\), \(_\)]\)[a_, k_, p_] := (k
    
78. \!\(\*OverscriptBox[\(a\), \(_\)]\) - a + p)/k;
    
79. d = Simplify[Duichengdian1[o1, -(a - b)/(
    
80. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
81. \!\(\*OverscriptBox[\(b\), \(_\)]\)), a]] ;
    
82. 
    
83. \!\(\*OverscriptBox[\(d\), \(_\)]\) = Simplify[
    
84. \!\(\*OverscriptBox[\(Duichengdian1\), \(_\)]\)[o1, -(a - b)/(
    
85. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
86. \!\(\*OverscriptBox[\(b\), \(_\)]\)), a]];
    
87. Print["d = ", d];
    
88. e = Simplify[Duichengdian1[o1, -(a - c)/(
    
89. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
90. \!\(\*OverscriptBox[\(c\), \(_\)]\)), a]] ;
    
91. 
    
92. \!\(\*OverscriptBox[\(e\), \(_\)]\) = Simplify[
    
93. \!\(\*OverscriptBox[\(Duichengdian1\), \(_\)]\)[o1, -(a - c)/(
    
94. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
95. \!\(\*OverscriptBox[\(c\), \(_\)]\)), a]];
    
96. Print["e=", e];
    
97. x = Simplify[Duichengdian[o, o1, a]];
    
98. \!\(\*OverscriptBox[\(x\), \(_\)]\) = Simplify[
    
99. \!\(\*OverscriptBox[\(Duichengdian\), \(_\)]\)[o, o1, a]];
    
100. Print["x=", x];
     
101. q = Simplify[-(k[x, d]/x)];
     
102. \!\(\*OverscriptBox[\(q\), \(_\)]\) = Simplify[1/q]; p =
     
103. Simplify[-(k[x, e]/x)];
     
104. \!\(\*OverscriptBox[\(p\), \(_\)]\) = Simplify[1/p];
     
105. Print["p=", p];
     
106. Print["q=", q];
     
107. s = Simplify[FourPoint[q, e, p, d]];
     
108. \!\(\*OverscriptBox[\(s\), \(_\)]\) = Simplify[
     
109. \!\(\*OverscriptBox[\(FourPoint\), \(_\)]\)[q, e, p, d]];
     
110. Print["s=", s];
     
111. Print["SE的复斜率/SD的复斜率 = ", Simplify[k[s, e]/k[s, d]]]
     
112. Print["AE的复斜率/AD的复斜率 = ", Simplify[k[a, e]/k[a, d]]]
     
113. If[Simplify[k[s, e]/k[s, d]] == Simplify[k[a, e]/k[a, d]],
     
114. Print["由于上述两个复斜率的比值相等，因此 ADSE 四点共圆。"]]
     
115. Print["OA的复斜率 = ", Simplify[k[o, a]], "     OS的复斜率 = ",
     
116.   Simplify[k[o, s]]];
     
117. If[Simplify[k[o, a]/k[o, s]] == 1,
     
118. Print["因为OA的复斜率与OS的复斜率相等，故 A、O、S 三点共线"]]
     
119. Print["AS的复斜率 = ", Simplify[k[a, s]], "     PQ的复斜率 = ",
     
120.   Simplify[k[p, q]]];
     
121. If[Simplify[k[a, s]/k[p, q]] == -1,
     
122. Print["因为AS的复斜率与PQ的复斜率是相反数，故这两条直线垂直"]]
     
123. Print["PS的复斜率 = ", Simplify[k[p, s]], "     AQ的复斜率 = ",
     
124.   Simplify[k[a, q]]];
     
125. If[Simplify[k[p, s]/k[a, q]] == -1,
     
126. Print["因为PS的复斜率与AQ的复斜率是相反数，故这两条直线垂直"]]
     
127. Print["QS的复斜率 = ", Simplify[k[q, s]], "     AP的复斜率 = ",
     
128.   Simplify[k[a, p]]];
     
129. If[Simplify[k[q, s]/k[a, p]] == -1,
     
130. Print["因为QS的复斜率与AP的复斜率是相反数，故这两条直线垂直"]]
     
131. S2[a_, b_] := (a - b) (
     
132. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
     
133. \!\(\*OverscriptBox[\(b\), \(_\)]\)); (* A点到B点长度的平方 *)
     
134. Print["AP = ", Simplify[S2[a, p]], "    AQ = ", Simplify[S2[a, q]]];
     
135. If[Simplify[S2[a, p]/S2[a, q]] == 1, Print["因为 AP = AQ, 故二者长度相等"]]
     
136. Print["SP = ", Simplify[S2[s, p]], "    SQ = ", Simplify[S2[s, q]]];
     
137. If[Simplify[S2[s, p]/S2[s, q]] == 1, Print["因为 SP = SQ, 故二者长度相等"]]
     
138. o2 = Simplify[ Jd[-k[a, b], m, k[a, t], a]];  
     
139. \!\(\*OverscriptBox[\(o2\), \(_\)]\) = Simplify[
     
140. \!\(\*OverscriptBox[\(Jd\), \(_\)]\)[-k[a, b], m, k[a, t], a]];
     
141. Print["o2=", o2];
     
142. d2 = Simplify[Duichengdian1[o2, -(a - b)/(
     
143. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
     
144. \!\(\*OverscriptBox[\(b\), \(_\)]\)), a]] ;
     
145. 
     
146. \!\(\*OverscriptBox[\(d2\), \(_\)]\) = Simplify[
     
147. \!\(\*OverscriptBox[\(Duichengdian1\), \(_\)]\)[o2, -(a - b)/(
     
148. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
     
149. \!\(\*OverscriptBox[\(b\), \(_\)]\)), a]];
     
150. Print["d2 = ", d2];
     
151. e2 = Simplify[Duichengdian1[o2, -(a - c)/(
     
152. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
     
153. \!\(\*OverscriptBox[\(c\), \(_\)]\)), a]] ;
     
154. 
     
155. \!\(\*OverscriptBox[\(e2\), \(_\)]\) = Simplify[
     
156. \!\(\*OverscriptBox[\(Duichengdian1\), \(_\)]\)[o2, -(a - c)/(
     
157. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
     
158. \!\(\*OverscriptBox[\(c\), \(_\)]\)), a]];
     
159. Print["e2 = ", e2];
     
160. x2 = Simplify[Duichengdian[o, o2, a]];
     
161. \!\(\*OverscriptBox[\(x2\), \(_\)]\) = Simplify[
     
162. \!\(\*OverscriptBox[\(Duichengdian\), \(_\)]\)[o, o2, a]];
     
163. Print["x2=", x2];
     
164. Print["X2D2 的复斜率 = ", Simplify[k[x2, d2]], "，    D2Q 的复斜率 = ",
     
165.   Simplify[k[d2, q]]];
     
166. If[Simplify[k[x2, d2]/ k[d2, q]] == 1,
     
167. Print["由于X2D2的复斜率等于D2Q的复斜率，因此 X2、D2、Q 三点共线。"]]
     
168. Print["X2E2 的复斜率 = ", Simplify[k[x2, e2]], "，    E2P 的复斜率 = ",
     
169.   Simplify[k[e2, p]]];
     
170. If[Simplify[k[x2, e2]/ k[e2, p]] == 1,
     
171. Print["由于X2E2的复斜率等于E2P的复斜率，因此 X2、E2、P 三点共线。"]]
     
172. o3 = (a + s)/2;
     
173. \!\(\*OverscriptBox[\(o3\), \(_\)]\) = (
     
174. \!\(\*OverscriptBox[\(a\), \(_\)]\) +
     
175. \!\(\*OverscriptBox[\(s\), \(_\)]\))/2;
     
176. d3 = Simplify[Duichengdian1[o3, -(a - q)/(
     
177. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
     
178. \!\(\*OverscriptBox[\(q\), \(_\)]\)), a]] ;
     
179. 
     
180. \!\(\*OverscriptBox[\(d3\), \(_\)]\) = Simplify[
     
181. \!\(\*OverscriptBox[\(Duichengdian1\), \(_\)]\)[o3, -(a - q)/(
     
182. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
     
183. \!\(\*OverscriptBox[\(q\), \(_\)]\)), a]];
     
184. Print["d3 = ", d3];
     
185. e3 = Simplify[Duichengdian1[o3, -(a - p)/(
     
186. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
     
187. \!\(\*OverscriptBox[\(p\), \(_\)]\)), a]] ;
     
188. 
     
189. \!\(\*OverscriptBox[\(e3\), \(_\)]\) = Simplify[
     
190. \!\(\*OverscriptBox[\(Duichengdian1\), \(_\)]\)[o3, -(a - p)/(
     
191. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
     
192. \!\(\*OverscriptBox[\(p\), \(_\)]\)), a]];
     
193. Print["e3 = ", e3];
     
194. Print["DD3 的复斜率 = ", Simplify[k[d, d3]], "，    D3S 的复斜率 = ",
     
195.   Simplify[k[d3, s]], "，    SP 的复斜率 = ", Simplify[k[s, p]]];
     
196. If[Simplify[k[d, d3]/ k[d3, s]] == 1 &&
     
197.   Simplify[k[d3, s]/ k[s, p]] == 1,
     
198. Print["由于DD3、D3S、SP 的复斜率都相等，因此 D、D3、S、P 四点共线。"]]
     
199. Print["EE3 的复斜率 = ", Simplify[k[e, e3]], "，    E3S 的复斜率 = ",
     
200.   Simplify[k[e3, s]], "，    SQ 的复斜率 = ", Simplify[k[s, q]]];
     
201. If[Simplify[k[e, e3]/ k[e3, s]] == 1 &&
     
202.   Simplify[k[e3, s]/ k[s, q]] == 1,
     
203. Print["由于EE3、E3S、SQ 的复斜率都相等，因此 E、E3、S、Q 四点共线。"]]
     
204. 
     

复制代码

运 行 结 果：




上面这个程序是 2021-12-10 改进以后的，运行结果与原先的完全相同。改进之处是求 d、e、d2、e2、d3、e3 这些点的坐标的方法不用解方程了，因而程序更加简练。

denglongshan

假设A在原点，角BAC平分线与实轴重合，AB的长度等于单位长度