Geogebra有一个命令TriangleCenter(<Point>,<Point>,<Point>,<number>), 可以直接作出三角形的各种中心。
我把 number设为一个增量为1的滑动条，作图验证了number=1~100的情况，结果：
只有number=3时，O不必是相应的心，为任意点都行。
number=1,2,3,4分别为内心、重心、外心、垂心。
对于O为相应的心，除了number=1,2,3可以，貌似number=10也行，但不知这是个什么心。
number=10只是目测貌似，用IsInRegion()判定时，有显示False的点。
number=13十分接近，高度疑似，但应该不是，在三角形很钝的情况下有一点不易察觉的偏离。用IsInRegion()判定时，很容易显示False。
搜了下,number=10 是三角形的Spieker心, number=13是三角形的Fermat点
根据链接得知https://en.wikipedia.org/wiki/Encyclopedia_of_Triangle_Centers 得知,
三角形各中心百科全书（Encyclopedia of Triangle Centers，ETC）是一个在线的列表，收录了上万和三角形相关的三角形中心。
网站现正由伊凡斯维尔大学的数学教授克拉克·金伯林维护。截至2017年8月20日，此列表已一共收录有14143个三角形中心。 https://faculty.evansville.edu/ck6/encyclopedia/ETC.html
找到了一个合适的简化计算路径，证毕。
creasson 发表于 2021-5-26 00:22
找到了一个合适的简化计算路径，证毕。
楼上的cdf里的代码贴出来如下:
Clear["Global`*"];
(*设点*)
(*Z:=-(((-\-\ p+p u) (-\
\-\ p+p v))/(1+p+p u v));
PO = 1/2;*)
Z := -(((-I + u) (-I + v))/(1 + u v));
PO = 0;
PA = Z; PB = Z; PC = Z; PD = Z; PE =
Z; PF = Z;
(*AOD, BOE, COF三点共线*)
linst1 =
Factor[
ComplexExpand[
Im + PO * Conjugate + PD * Conjugate]]];
linst2 =
Factor[
ComplexExpand[
Im + PO * Conjugate + PE * Conjugate]]];
linst3 =
Factor[
ComplexExpand[
Im + PO * Conjugate + PF * Conjugate]]];
(*这三式不是独立的,任意两式可推出第三式，可解出e,f*)
efsolve =
Solve[{linst1, linst2, linst3} == 0, {e, f}] // Factor // Flatten;
(*设点G,H,I,J,K,L*)
(*Q[\_,\_]:=-((-1-2 p-p^2+p^2 \-\
\ p \-\ p^2 \)/(1+p+p \));*)
Q[\_, \_] := -((-1 + \ - I \)/(1 + \));
PG = Q; PH = Q; PI = Q; PJ = Q; PK =
Q; PL = Q;
(*KOH,LOI,GOJ共线*)
linst4 =
Factor[
ComplexExpand[
Im + PO * Conjugate + PH * Conjugate]]];
linst5 =
Factor[
ComplexExpand[
Im + PO * Conjugate + PI * Conjugate]]];
linst6 =
Factor[
ComplexExpand[
Im + PO * Conjugate + PJ * Conjugate]]];
(*可解出J2,K2,L2*)
JKLsolve =
Solve[{linst4, linst5, linst6} == 0, {J2, K2, L2}] // Factor //
Flatten;
(*角度相等*)
angst1 =
Factor]] //
Numerator;
angst2 =
Factor]] //
Numerator;
angst3 =
Factor]] //
Numerator;
angst4 =
Factor]] //
Numerator;
angst5 =
Factor]] //
Numerator;
angst6 =
Factor]] //
Numerator;
angsts =
FactorList[#][[-1]][] & /@ {angst1, angst2, angst3, angst4,
angst5, angst6};
(*代入前述关系化简*)
solved = Flatten[{efsolve, JKLsolve}];
angsts = Factor // Numerator;
angsts = FactorList[#][[-1]][] & /@ angsts;
(*GHIJKL六点共圆锥曲线*)
ReIms = Factor]];
ReIms = (ReIms /. solved) // Factor;
matrix = {#[]^2, #[]*#[], #[]^2, #[], #[], 1} & /@
ReIms;
conicst = Det // Factor // Numerator;
conicst = FactorList[[-1]][];
conicst = Factor // Numerator;
(*GroebnerBasis,{I2,H1,H2,J1,K1,L1}]*)
\
(*各变量的幂次*)
Exponent[#, {G1, G2, H1, H2, I1, I2, J1, J2, K1, K2, L1,
L2}] & /@ angsts
(*消元求解*)
(*消去L1*)
angst5 =
SubresultantPolynomialRemainders], angsts[], L1][[-1]];
angst5 = FactorList[[-1]][];
(*消去K1*)
angst4 =
SubresultantPolynomialRemainders], K1][[-1]];
angst4 = FactorList[[-1]][];
(*消去J1*)
angst3 =
SubresultantPolynomialRemainders], J1][[-1]];
angst3 = FactorList[[-1]][];
(*消去H2*)
angst2 =
SubresultantPolynomialRemainders], H2][[-1]];
angst2 = FactorList[[-1]][];
angst1 =
SubresultantPolynomialRemainders], angsts[], H2][[-1]];
angst1 = FactorList[[-1]][];
(*消去H1, 有两个有效因式, 取最后一个因式即可*)
solve1 = SubresultantPolynomialRemainders[[-1]];
solve1 = FactorList[[-1]][];(*G1,G2,I1,I2*)
(*回代消元*)
solve2 = SubresultantPolynomialRemainders[[-1]];
solve2 = FactorList[[-1]][];(*G1,G2,I1,H1*)
solve3 = SubresultantPolynomialRemainders[[-1]];
solve3 = FactorList[[-1]][];
solve3 = SubresultantPolynomialRemainders[[-1]];
solve3 =
FactorList[[-1]][[
1]];(*G1,G2,I1,H2*)
(*solve4= \
SubresultantPolynomialRemainders], solve1, I2][[-1]];
solve4 = FactorList[[-1]][];(*G1,G2,I1,J1*)
solve5 = \
SubresultantPolynomialRemainders],solve2, H1][[-1]];
solve5 = FactorList[[-1]][];
solve5 = SubresultantPolynomialRemainders[[-1]];
solve5 = FactorList[[-1]][];
solve5 = SubresultantPolynomialRemainders[[-1]];
solve5 = FactorList[[-1]][];(*G1,G2,I1,K1*)
solve6= \
SubresultantPolynomialRemainders],solve1, I2][[-1]];
solve6= FactorList[[-1]][];(*G1,G2,I1,L1*)*)
(*共圆锥条件检验*)
check = SubresultantPolynomialRemainders],
I2][[-1]];
check =FactorList[[-1]][];
check = SubresultantPolynomialRemainders[[-1]];
check =FactorList[[-1]][];
PolynomialGCD
(*存在公因式则表明conicst在以上条件等式下是等于0的*) hujunhua 发表于 2021-5-23 17:35
上面的作图命令得一行行地拷贝，要是Geogebra能够整段地拷贝代码就好了。
这是一个很好的问题, 我查了下文档.文档说执行Geogebra命令的方式有四种. https://wiki.geogebra.org/en/Scripting
其中直接执行文件这种好像得是JavaScript脚本.
纯粹的GGBScript 命令,得通过交互控件触发.
都不完美,
没办法,我随便新建了一个Button 控件,然后设置 Button被点击的事件的脚本(On Click),黏贴如下内容,然后退出设置的状态, 点击按钮,都执行成功了.
A=(2,3);
B=(4,-4);
C=(-1,1);
sn=Slider(1,100,1);
P=TriangleCenter(A,B,C,sn);
a=Segment(B,C);
b=Segment(C,A);
c=Segment(A,B);
D=Intersect(a,Line(A,P));
E=Intersect(b,Line(B,P));
F=Intersect(c,Line(C,P));
G=TriangleCenter(P,A,E,sn);
H=TriangleCenter(P,A,F,sn);
I=TriangleCenter(P,B,F,sn);
J=TriangleCenter(P,B,D,sn);
K=TriangleCenter(P,C,D,sn);
L=TriangleCenter(P,C,E,sn);
conic5=Conic(G,H,I,J,K);
tf=IsInRegion(L,conic5);
wayne 发表于 2021-5-26 09:05
这是一个很好的问题, 我查了下文档.文档说执行Geogebra命令的方式有四种. https://wiki.geogebra.org/en/ ...
在线试了一下，果然可以。
这就不错了。
@mathe可以试一下了，看看number=10,13是不是真的行。 fermat点应该是退化情况 对于三角形\(ABC\)的内心\(I\)将原三角形分成6个三角形\(AIE,AIF,BIF,BID,BID,CID,CIE\)，(其中\(D,E,F\)分别为\(AI,BI,CI\)延长线交对边的点）且这6个小三角形的内心分别设为\(I1,I2,I3,I4,I5,I6\)
为了便于分析与计算，我们将\(B\)点设为原点\(O,C\)点位于\(X\)正半轴上，并设
\(A,B,C，D,E,F,I,I1,I2,I3,I4,I5,I6\)
\(AB=c,AC=b,BC=a,AD=t1,BE=t2,CF=t3,BD=a11,DC=a12,CE=b11,AE=b12,AF=c11,BF=c12\)
我们可以得到：
\(x1=\frac{a^2-b^2+c^2}{2a},y1=\frac{2s}{a},x2=0,y2=0,x3=a,y3=0,x0=\frac{a-b+c}{2},y0=\frac{2s}{a+b+c},x11=\frac{ac}{b+c},y11=0,x12=\frac{(a+c)^2-b^2}{2(a+c)},y12=\frac{2s}{a+c}
x13=\frac{a^2-b^2+c^2}{2(a+b)},y13=\frac{2s}{a+b}\)
\(a11=\frac{ac}{b+c},a12=\frac{ab}{b+c},b11=\frac{ab}{a+c},b12=\frac{bc}{a+c},c11=\frac{bc}{a+b},c12=\frac{ac}{a+b},a1=\frac{t1(b+c)}{a+b+c},a2=\frac{at1}{a+b+c},b1=\frac{(a+c)t2}{a+b+c}，
b2=\frac{bt2}{a+b+c},c1=\frac{(a+b)t3}{a+b+c},c2=\frac{ct3}{a+b+c}\)
\(x21=\frac{a1x12 + b12x0 + b2x1}{b2 + b12 + a1}, y21=\frac{a1y12 + b12y0 + b2y1}{b2 + b12 + a1}, x22=\frac{a1x13 + c11x0 + c2x1}{c2 + c11 + a1}, y22=\frac{a1y13 + c11y0 + c2y1}{c2 + c11 + a1}, x23=\frac{b1x13 + c12x0 + c2x2}{c2 + c12 + b1}, y23=\frac{b1y13 + c12y0 + c2y2}{c2 + c12 + b1}
, x24=\frac{a11x0 + a2x2 + b1x11}{a2 + a11 + b1}, y24 =\frac{a11y0 + a2y2 + b1y11}{a2 + a11 + b1}, x25=\frac{a12x0 + a2x3 + c1x11}{a2 + a12 + c1}, y25=\frac{a12y0 + a2y3 + c1y11}{a2 + a12 + c1}, x26=\frac{b11x0 + b2x3 + c1x12}{b2 + b11 + c1}, y26=\frac{b11y0 + b2y3 + c1y12}{b2 + b11 + c1}\)
\(t1=\frac{\sqrt{bc(a+b+c)(-a+b+c)}}{b+c},t2=\frac{\sqrt{ac(a+b+c)(a-b+c)}}{a+c},t3=\frac{\sqrt{ab(a+b+c)(a+b-c)}}{a+b},s=\frac{\sqrt{2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4}}{4}\)
\(I1,I2,I3,I4,I5,I6\)选较简单的后5个点坐标构成椭圆曲线
\[\det{\left(
\begin{array}{cccccc}
x^2& xy& y^2& x&y,&1\\
x22^2&x22y22&y22^2&x22&y22&1\\
x23^2&x23y23&y23^2&x23&y23&1\\
x24^2&x24y24&y24^2&x24&y24&1\\
x25^2&x25y25&y25^2&x25&y25&1\\
x26^2&x26y26&y26^2&x26&y26&1\\
\end{array}
\right) }=0\]
例如：取\({a=5,b=4,c=3}\)
得到
\(a = 5, a1=\sqrt{2}, a11 = \frac{15}{7}, a12 =\frac{20}{7}, a2 = \frac{5\sqrt{2}}{7}, b = 4, b1 = \sqrt{5}, b11=\frac{5}{2}, b12=\frac{3}{2}, b2=\frac{\sqrt{5}}{2}, c = 3, c1 =\sqrt{10}, c11=\frac{4}{3}, c12=\frac{5}{3}, c2 =\sqrt{10}{3}, s = 6, t1=\frac{12\sqrt{2}}{7}, t2=\frac{3\sqrt{5}}{2}, t3=\frac{4\sqrt{10}}{3}, x0 = 2, x1 =\frac{9}{5}, x11=\frac{15}{7}, x12 = 3, x13 = 1, x2 = 0, x3 = 5, y0 = 1, y1=\frac{12}{5}, y11 = 0, y12 =\frac{3}{2}, y13 =\frac{4}{3}, y2 = 0, y3 = 0\)
\(x21 =\frac{2880 + 2880\sqrt{2}+ 864\sqrt{5}}{10(144 + 96\sqrt{2}+ 48\sqrt{5})}, x22 =\frac{2880 + 1080\sqrt{2}+ 648\sqrt{10}}{10(144 + 108\sqrt{2}+ 36\sqrt{10})}, x23 =\frac{720 + 216\sqrt{5}}{2(180 + 108\sqrt{5}+ 36\sqrt{10})}, x24 = \frac{15(48 + 24\sqrt{5})}{2(180 + 60\sqrt{2}+ 84\sqrt{5})}, x25 =\frac{5(192 + 120\sqrt{2}+ 72\sqrt{10})}{2(240 + 60\sqrt{2}+ 84\sqrt{10})}, x26 =\frac{960 + 576\sqrt{10}+ 480\sqrt{5}}{2(240 + 96\sqrt{10} + 48\sqrt{5})}, y21 = \frac{12(60 + 60\sqrt{2}+ 48\sqrt{5}}{5(144 + 96\sqrt{2}+ 48\sqrt{5})}, y22 =\frac{12(60 + 60\sqrt{2}+ 36\sqrt{10})}{5(144 + 108\sqrt{2}+ 36\sqrt{10})}, y23 = \frac{12(15 + 12\sqrt{5})}{180 + 108\sqrt{5} + 36\sqrt{10}}, y24 =\frac{180}{180 + 60\sqrt{2}+ 84\sqrt{5}}, y25 = \frac{240}{240 + 60\sqrt{2}+ 84\sqrt{10}}, y26 =\frac{12(20 + 12\sqrt{10})}{240 + 96\sqrt{10}+ 48\sqrt{5}}\)
椭圆曲线方程:
\((465865600000\sqrt{10}+ 657851200000\sqrt{5}+ 1038608000000\sqrt{2}+ 1477336000000)x^2 + ((43001600000\sqrt{10}+ 97088000000\sqrt{2}+ 63683200000\sqrt{5}+ 136976000000)y - 1952544000000\sqrt{10} - 4351680000000\sqrt{2}- 2757312000000\sqrt{5} - 6189600000000)x + (1024374400000\sqrt{10}+ 2294192000000\sqrt{2}+ 1448708800000\sqrt{5}+ 3245704000000)y^2 + (-2063712000000\sqrt{10}- 4633440000000\sqrt{2}- 2924256000000\sqrt{5}- 6549600000000)y + 8231400000000 + 5799600000000\sqrt{2}+ 2597040000000\sqrt{10} + 3668760000000\sqrt{5}=0\)
若消元可以得到：（一般表达式见附件，只含\(a,b,c,x,y\))
F1: \(121186279x^8 - 168512396x^7y - 108228692x^6y^2 - 65161548x^5y^3 - 402431070x^4y^4 + 131013868x^3y^5 - 82188452x^2y^6 + 7291116xy^7 + 21119y^8 - 2264732500x^7 + 2182105400x^6y + 1544825300x^5y^2 - 917671000x^4y^3 + 3749646500x^3y^4 - 1667779000x^2y^5 + 425714300xy^6 - 16143400y^7 + 16821693700x^6 - 9733579100x^5y - 8472300500x^4y^2 + 10502577000x^3y^3 - 14462454500x^2y^4 + 5397472900xy^5 - 542302700y^6 - 63619801500x^5 + 17096885000x^4y + 23690115000x^3y^2 - 31225170000x^2y^3 + 25547692500xy^4 - 5204323000y^5 + 127737391250x^4 - 2691082500x^3y - 22808947500x^2y^2 + 31330282500xy^3 - 16654783750y^4 - 118412287500x^3 - 16203975000x^2y - 28204387500xy^2 - 4653675000y^3 + 3679312500x^2 - 16931812500xy + 51136312500y^2 + 68588437500x + 39639375000y - 30406640625=0\)
F2: \(41659039x^8 - 69730684x^7y + 286833508x^6y^2 - 37209612x^5y^3 + 386084730x^4y^4 + 219518012x^3y^5 - 146455292x^2y^6 - 15692916xy^7 + 11868239y^8 - 601730380x^7 - 72003080x^6y - 3572639620x^5y^2 - 2078162600x^4y^3 - 4549171300x^3y^4 - 59095640x^2y^5 + 610175060xy^6 + 7908040y^7 + 4063741900x^6 + 7154965700x^5y + 22172801500x^4y^2 + 21578181000x^3y^3 + 13495178500x^2y^4 - 2238922300xy^5 - 889337900y^6 - 17105332500x^5 - 48172625000x^4y - 83709675000x^3y^2 - 67073850000x^2y^3 - 11779462500xy^4 + 3633115000y^5 + 48826666250x^4 + 150796807500x^3y + 176933227500x^2y^2 + 77586142500xy^3 + 688216250y^4 - 94309912500x^3 - 252919275000x^2y - 180009262500xy^2 - 29643825000y^3 + 116389687500x^2 + 215322187500xy + 68470312500y^2 - 81042187500x - 72309375000y + 23762109375=0\)
画图得到：
绿色曲线为F2,其中有部分与红色椭圆曲线重合
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