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[讨论] 由三角形引导的一个几何变换——三坐标反演

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发表于 2019-4-20 12:46:43 来自手机 | 显示全部楼层
既然等价,必然都有,而且方程都相同,只是是切线构成的方程,需要转化一下。
或者换一个角度,既然对偶,我们可以对三坐标反演的任意一条不动曲线(比如不动三次曲线),求出任意一点的切线方程$ax+by+cz=0$,然后给出$a,b,c$满足的方程,那么就是对偶三坐标反演中不动曲线的方程
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-4-22 19:05:08 | 显示全部楼层
根据mathe 71#的计算方案:


三坐标反演的不动曲线为:

\(x(y^2+z^2)+y(x^2+z^2)+z(x^2+y^2)-\frac{21}{2}xyz=0\)

对上式分别关于\(x,y,z\)求导,设为

\(a_1x+b_1y+c_1z=0\)

则有:

\(4xy+4xz+2y^2-21yz+2z^2-2a_1=0\)

\(2x^2+4xy-21xz+4yz+2z^2-2b_1=0\)

\(2x^2-21xy+4xz+2y^2+4yz-2c_1=0\)

\(x+y+z-1=0\)

消元得:(已舍掉另一个因子)

\(165888a_1^4+2187648a_1^3b_1+2187648a_1^3c_1+7544178a_1^2b_1^2+16612452a_1^2b_1c_1+7544178a_1^2c_1^2+2187648a_1b_1^3+16612452a_1b_1^2c_1+16612452a_1b_1c_1^2+2187648a_1c_1^3+165888b_1^4+2187648b_1^3c_1+7544178b_1^2c_1^2+2187648b_1c_1^3+165888c_1^4-752796a_1^3-4298454a_1^2b_1-4298454a_1^2c_1-4298454a_1b_1^2+17951733a_1b_1c_1-4298454a_1c_1^2-752796b_1^3-4298454b_1^2c_1-4298454b_1c_1^2-752796c_1^3+293600a_1^2-5505950a_1b_1-5505950a_1c_1+293600b_1^2-5505950b_1c_1+293600c_1^2+1572500a_1+1572500b_1+1572500c_1-375000=0\)

再代入重心坐标公式

\(a_1=\frac{ya}{2s}\)

\(b_1= -\frac{a^2y+b^2y-c^2y-4as+4sx}{4as}\)

\(c_1=-\frac{a^2y-b^2y+c^2y-4sx}{4as}\)


化简得到:

\(452096a^{12}-3304560a^{11}x-1808384a^{10}b^2-1808384a^{10}c^2+5054889a^{10}x^2+16924698a^{10}y^2+13218240a^9b^2x+13218240a^9c^2x-3500658a^9x^3+10501974a^9xy^2+2712576a^8b^4+1808384a^8b^2c^2-20219556a^8b^2x^2-65938761a^8b^2y^2+2712576a^8c^4-20219556a^8c^2x^2-1760031a^8c^2y^2+1750329a^8x^4-10501974a^8x^2y^2+15752961a^8y^4-19827360a^7b^4x-13218240a^7b^2c^2x+14002632a^7b^2x^3+43174782a^7b^2xy^2-19827360a^7c^4x+14002632a^7c^2x^3-85182678a^7c^2xy^2-1808384a^6b^6+1808384a^6b^4c^2+30329334a^6b^4x^2+76048539a^6b^4y^2+1808384a^6b^2c^4+20219556a^6b^2c^2x^2-23739618a^6b^2c^2y^2-7001316a^6b^2x^4+21003948a^6b^2x^2y^2-1808384a^6c^6+30329334a^6c^4x^2-52308921a^6c^4y^2-7001316a^6c^2x^4+21003948a^6c^2x^2y^2+13218240a^5b^6x-13218240a^5b^4c^2x-21003948a^5b^4x^3-107353512a^5b^4xy^2-13218240a^5b^2c^4x-14002632a^5b^2c^2x^3-42007896a^5b^2c^2xy^2+13218240a^5c^6x-21003948a^5c^4x^3+149361408a^5c^4xy^2+452096a^4b^8-1808384a^4b^6c^2-20219556a^4b^6x^2-21979587a^4b^6y^2+2712576a^4b^4c^4+20219556a^4b^4c^2x^2+86158317a^4b^4c^2y^2+10501974a^4b^4x^4-21003948a^4b^4x^2y^2+10501974a^4b^4y^4-1808384a^4b^2c^6+20219556a^4b^2c^4x^2-106377873a^4b^2c^4y^2+7001316a^4b^2c^2x^4+42007896a^4b^2c^2x^2y^2-21003948a^4b^2c^2y^4+452096a^4c^8-20219556a^4c^6x^2+42199143a^4c^6y^2+10501974a^4c^4x^4-21003948a^4c^4x^2y^2+10501974a^4c^4y^4-3304560a^3b^8x+13218240a^3b^6c^2x+14002632a^3b^6x^3+43174782a^3b^6xy^2-19827360a^3b^4c^4x-14002632a^3b^4c^2x^3-171532242a^3b^4c^2xy^2+13218240a^3b^2c^6x-14002632a^3b^2c^4x^3+213540138a^3b^2c^4xy^2-3304560a^3c^8x+14002632a^3c^6x^3-85182678a^3c^6xy^2+5054889a^2b^8x^2-5054889a^2b^8y^2-20219556a^2b^6c^2x^2+20219556a^2b^6c^2y^2-7001316a^2b^6x^4+21003948a^2b^6x^2y^2+30329334a^2b^4c^4x^2-30329334a^2b^4c^4y^2+7001316a^2b^4c^2x^4-21003948a^2b^4c^2x^2y^2-20219556a^2b^2c^6x^2+20219556a^2b^2c^6y^2+7001316a^2b^2c^4x^4-21003948a^2b^2c^4x^2y^2+5054889a^2c^8x^2-5054889a^2c^8y^2-7001316a^2c^6x^4+21003948a^2c^6x^2y^2-3500658ab^8x^3+10501974ab^8xy^2+14002632ab^6c^2x^3-42007896ab^6c^2xy^2-21003948ab^4c^4x^3+63011844ab^4c^4xy^2+14002632ab^2c^6x^3-42007896ab^2c^6xy^2-3500658ac^8x^3+10501974ac^8xy^2+1750329b^8x^4-10501974b^8x^2y^2+1750329b^8y^4-7001316b^6c^2x^4+42007896b^6c^2x^2y^2-7001316b^6c^2y^4+10501974b^4c^4x^4-63011844b^4c^4x^2y^2+10501974b^4c^4y^4-7001316b^2c^6x^4+42007896b^2c^6x^2y^2-7001316b^2c^6y^4+1750329c^8x^4-10501974c^8x^2y^2+1750329c^8y^4+13218240a^9sy-128357460a^8sxy-13218240a^7b^2sy-39654720a^7c^2sy+128357460a^7sx^2y+72346932a^7sy^3+216275808a^6b^2sxy+297154032a^6c^2sxy-13218240a^5b^4sy-26436480a^5b^2c^2sy-214707024a^5b^2sx^2y-42007896a^5b^2sy^3+39654720a^5c^4sy-298722816a^5c^2sx^2y+42007896a^5c^2sy^3-47479236a^4b^4sxy+256714920a^4b^2c^2sxy-28005264a^4b^2sx^3y+84015792a^4b^2sxy^3-209235684a^4c^4sxy+28005264a^4c^2sx^3y-84015792a^4c^2sxy^3+13218240a^3b^6sy-39654720a^3b^4c^2sy+44341668a^3b^4sx^2y-128357460a^3b^4sy^3+39654720a^3b^2c^4sy-256714920a^3b^2c^2sx^2y+256714920a^3b^2c^2sy^3-13218240a^3c^6sy+212373252a^3c^4sx^2y-128357460a^3c^4sy^3-40439112a^2b^6sxy+121317336a^2b^4c^2sxy+56010528a^2b^4sx^3y-121317336a^2b^2c^4sxy+40439112a^2c^6sxy-56010528a^2c^4sx^3y+42007896ab^6sx^2y-14002632ab^6sy^3-126023688ab^4c^2sx^2y+42007896ab^4c^2sy^3+126023688ab^2c^4sx^2y-42007896ab^2c^4sy^3-42007896ac^6sx^2y+14002632ac^6sy^3-28005264b^6sx^3y+28005264b^6sxy^3+84015792b^4c^2sx^3y-84015792b^4c^2sxy^3-84015792b^2c^4sx^3y+84015792b^2c^4sxy^3+28005264c^6sx^3y-28005264c^6sxy^3=0\)

再取\(a=5,b=4,c=3,s=6\)得到

\(580717154304x^4+677503346688x^3y+4077118354176x^2y^2+2263049373312xy^3+6479325884304y^4-5807171543040x^3-60531697820160x^2y-52731671690880xy^2+23537864286720y^3+41927271321600x^2+301709688537600xy-67890352521600y^2-137046712320000x-182728949760000y+93746626560000=0\)

画图:

00.GIF

以上结果为4次曲线(绿线),红线为不动直线,黑线为已知三角形,显然绿线不满足我们的要求,以上计算不知是哪一步搞错了!!!
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-4-22 22:45:38 来自手机 | 显示全部楼层
我们根据求导得出a_1等以后,对应的x,y,z要满足原三次曲线方程,由此消去x,y,z可以得出关于a,b,c的方程,再分别将a,b,c改写为x,y,z即可

点评

你说的应该是不动曲线吧,应该和部分不动直线相切才对  发表于 2019-4-23 13:13
切线方程系数a,b,c不一定满足a+b+c=1? 吧,能给出不动曲线上的几个不动点吗?  发表于 2019-4-23 12:33
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-4-22 22:51:00 来自手机 | 显示全部楼层
不过从另外一个角度分析,我们凭什么认为三次曲线的切线比如是三次的呢?

点评

现在的问题是如何验证最终答案是否正确?例如可取特殊点代入验算~  发表于 2019-4-22 23:00
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-4-23 20:57:59 | 显示全部楼层
由于楼上72#中默认了\(a_1+b_1+c_1=1\),因此出现了不变曲线偏移(即与不变直线相交而不是相切)

因为我们可以重新统一乘以一个参数\(k\)

\(a_1=\frac{yak}{2s}\)

\(b_1=-\frac{(a^2y+b^2y-c^2y-4as+4sx)k}{4as}\)

\(c_1=-\frac{(a^2y-b^2y+c^2y-4sx)k}{4as}\)

为了求得\(k\)值,我们取特殊正三角形\(a=1,b=1,c=1,s=\frac{\sqrt{3}}{4},x=\frac{1}{4},y=\frac{\sqrt{3}}{4}\)

得到:\((121k+150)(k-2)(-4+9k)^2=0\)

经检验\(k=\frac{4}{9}\)时满足不变直线与不变曲线相切

因此:我们得到最终的不变曲线(4次,重心坐标方程)

\(524288x_1^4+6914048x_1^3y_1+6914048x_1^3z_1+23843328x_1^2y_1^2+52503552x_1^2y_1z_1+23843328x_1^2z_1^2+6914048x_1y_1^3+52503552x_1y_1^2z_1+52503552x_1y_1z_1^2+6914048x_1z_1^3+524288y_1^4+6914048y_1^3z_1+23843328y_1^2z_1^2+6914048y_1z_1^3+524288z_1^4-5353216x_1^3-30566784x_1^2y_1-30566784x_1^2z_1-30566784x_1y_1^2+127656768x_1y_1z_1-30566784x_1z_1^2-5353216y_1^3-30566784y_1^2z_1-30566784y_1z_1^2-5353216z_1^3+4697600x_1^2-88095200x_1y_1-88095200x_1z_1+4697600y_1^2-88095200y_1z_1+4697600z_1^2+56610000x_1+56610000y_1+56610000z_1-30375000\)

代入重心坐标公式得到:

\(388962a^8y^4+259308a^4b^4y^4-518616a^4b^2c^2y^4+259308a^4c^4y^4+43218b^8y^4-172872b^6c^2y^4+259308b^4c^4y^4-172872b^2c^6y^4+43218c^8y^4+5315814a^7sy^3-1037232a^5b^2sy^3+1037232a^5c^2sy^3+2074464a^4b^2sxy^3-2074464a^4c^2sxy^3-6698790a^3b^4sy^3+13397580a^3b^2c^2sy^3-6698790a^3c^4sy^3-345744ab^6sy^3+1037232ab^4c^2sy^3-1037232ab^2c^4sy^3+345744ac^6sy^3+691488b^6sxy^3-2074464b^4c^2sxy^3+2074464b^2c^4sxy^3-691488c^6sxy^3-4624326a^6s^2y^2-4148928a^5s^2xy^2+26795160a^4b^2s^2y^2-26795160a^4c^2s^2y^2+4148928a^4s^2x^2y^2-53590320a^3b^2s^2xy^2+53590320a^3c^2s^2xy^2+7390278a^2b^4s^2y^2-14780556a^2b^2c^2s^2y^2+7390278a^2c^4s^2y^2-4148928ab^4s^2xy^2+8297856ab^2c^2s^2xy^2-4148928ac^4s^2xy^2+4148928b^4s^2x^2y^2-8297856b^2c^2s^2x^2y^2+4148928c^4s^2x^2y^2-26795160a^5s^3y+107180640a^4s^3xy-26795160a^3b^2s^3y+26795160a^3c^2s^3y-107180640a^3s^3x^2y+59122224a^2b^2s^3xy-59122224a^2c^2s^3xy-16595712ab^2s^3x^2y+16595712ac^2s^3x^2y+11063808b^2s^3x^3y-11063808c^2s^3x^3y+26103672a^4s^4-107180640a^3s^4x+118244448a^2s^4x^2-22127616as^4x^3+11063808s^4x^4=0\)

以上因形如三叶草,称其为"mathe 三叶草不变曲线"

例如:以下红色直线为不变直线,绿色为相应的mathe 三叶草不变曲线

取\(a=5,b=4,c=3,s=6\)

5431.GIF

取\(a=1,b=1,c=1,s=\frac{\sqrt{3}}{4}\)

1111.GIF

点评

很漂亮的图,只是我有点担心,对于不同的$a_1,b_1,c_1$,其对应的k可能是不同的。  发表于 2019-4-23 22:15
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-4-24 20:10:06 | 显示全部楼层
我们仿照楼上75#的计算方案:

我们对圆锥曲线

\(x^2x_0^2+y^2y_0^2+z^2z_0^2-2xx_0yy_0-2xx_0zz_0-2yy_0zz_0=0\)

在不变曲线上有\(x=x_0,y=y_0,z=z_0\)

即有不变曲线的切线方程:

\((4x_0^3-4x_0y_0^2-4x_0z_0^2)x+(-4x_0^2y_0+4y_0^3-4y_0z_0^2)y+(-4x_0^2z_0-4y_0^2z_0+4z_0^3)z=0\)

我们设切线方程的系数:

\(a_1=4x_0^3-4x_0y_0^2-4x_0z_0^2\)

\(b_1=-4x_0^2y_0+4y_0^3-4y_0z_0^2\)

\(c_1=-4x_0^2z_0-4y_0^2z_0+4z_0^3\)

\(x_0+y_0+z_0=1\)

消元得到:

\(a_1^4b_1^2-2a_1^4b_1c_1+a_1^4c_1^2-2a_1^3b_1^3+2a_1^3b_1^2c_1+2a_1^3b_1c_1^2-2a_1^3c_1^3+a_1^2b_1^4+2a_1^2b_1^3c_1-6a_1^2b_1^2c_1^2+2a_1^2b_1c_1^3+a_1^2c_1^4-2a_1b_1^4c_1+2a_1b_1^3c_1^2+2a_1b_1^2c_1^3-2a_1b_1c_1^4+b_1^4c_1^2-2b_1^3c_1^3+b_1^2c_1^4+8a_1^5-4a_1^4b_1-4a_1^4c_1-8a_1^3b_1^2-8a_1^3c_1^2-8a_1^2b_1^3+16a_1^2b_1^2c_1+16a_1^2b_1c_1^2-8a_1^2c_1^3-4a_1b_1^4+16a_1b_1^2c_1^2-4a_1c_1^4+8b_1^5-4b_1^4c_1-8b_1^3c_1^2-8b_1^2c_1^3-4b_1c_1^4+8c_1^5-32a_1^4-72a_1^3b_1-72a_1^3c_1-112a_1^2b_1^2-288a_1^2b_1c_1-112a_1^2c_1^2-72a_1b_1^3-288a_1b_1^2c_1-288a_1b_1c_1^2-72a_1c_1^3-32b_1^4-72b_1^3c_1-112b_1^2c_1^2-72b_1c_1^3-32c_1^4-32a_1^2b_1-32a_1^2c_1-32a_1b_1^2-64a_1b_1c_1-32a_1c_1^2-32b_1^2c_1-32b_1c_1^2=0\)

依楼上的计算我们有:

\(a_1=\frac{yat}{2s}\)

\(b_1=-\frac{(a^2y+b^2y-c^2y-4as+4sx)t}{4as}\)

\(c_1=-\frac{(a^2y-b^2y+c^2y-4sx)t}{4as}\)

为了求得\(t\)值,我们取特殊正三角形及对偶不动曲线的三个不动点

\(a=1,b=1,c=1,s=\frac{\sqrt{3}}{4},x=\frac{(9a^2-b^2+c^2)}{18a},y=\frac{2s}{9a}\)

\(a=1,b=1,c=1,s=\frac{\sqrt{3}}{4},x=\frac{2(3a^2-b^2+c^2)}{9a},y=\frac{8s}{9a}\)

\(a=1,b=1,c=1,s=\frac{\sqrt{3}}{4},x=\frac{3a^2-2b^2+2c^2}{9a},y=\frac{8s}{9a}\)

均算得

\(t^2+228t+96=0\)

取\(t=-114+10\sqrt{129}\)

然后算得不动曲线的方程(重心坐标)

\(-32(y_1+z_1))(x_1+z_1)(x_1+y_1)+(-32x_1^4-72x_1^3y_1-72x_1^3z_1-112x_1^2y_1^2-288x_1^2y_1z_1-112x_1^2z_1^2-72x_1y_1^3-288x_1y_1^2z_1-288x_1y_1z_1^2-72x_1z_1^3-32y_1^4-72y_1^3z_1-112y_1^2z_1^2-72y_1z_1^3-32z_1^4)t+(8x_1^5-4x_1^4y_1-4x_1^4z_1-8x_1^3y_1^2-8x_1^3z_1^2-8x_1^2y_1^3+16x_1^2y_1^2z_1+16x_1^2y_1z_1^2-8x_1^2z_1^3-4x_1y_1^4+16x_1y_1^2z_1^2-4x_1z_1^4+8y_1^5-4y_1^4z_1-8y_1^3z_1^2-8y_1^2z_1^3-4y_1z_1^4+8z_1^5)t^2+(y_1-z_1)^2(x_1-z_1)^2(x_1-y_1)^2t^3=0\)

代入重心坐标反演公式得:(神奇了得到了胡子不变曲线的高级版,6次方程)

\(1024a^4s^3(ay-2s)(a^2y+b^2y-c^2y+4sx)(a^2y-b^2y+c^2y+4as-4sx)-128a^2s^2(9a^8y^4+6a^4b^4y^4-12a^4b^2c^2y^4+6a^4c^4y^4+b^8y^4-4b^6c^2y^4+6b^4c^4y^4-4b^2c^6y^4+c^8y^4+6a^7sy^3-24a^5b^2sy^3+24a^5c^2sy^3+48a^4b^2sxy^3-48a^4c^2sxy^3-38a^3b^4sy^3+76a^3b^2c^2sy^3-38a^3c^4sy^3-8ab^6sy^3+24ab^4c^2sy^3-24ab^2c^4sy^3+8ac^6sy^3+16b^6sxy^3-48b^4c^2sxy^3+48b^2c^4sxy^3-16c^6sxy^3-20a^6s^2y^2-96a^5s^2xy^2+152a^4b^2s^2y^2-152a^4c^2s^2y^2+96a^4s^2x^2y^2-304a^3b^2s^2xy^2+304a^3c^2s^2xy^2+44a^2b^4s^2y^2-88a^2b^2c^2s^2y^2+44a^2c^4s^2y^2-96ab^4s^2xy^2+192ab^2c^2s^2xy^2-96ac^4s^2xy^2+96b^4s^2x^2y^2-192b^2c^2s^2x^2y^2+96c^4s^2x^2y^2-112a^5s^3y+608a^4s^3xy-112a^3b^2s^3y+112a^3c^2s^3y-608a^3s^3x^2y+352a^2b^2s^3xy-352a^2c^2s^3xy-384ab^2s^3x^2y+384ac^2s^3x^2y+256b^2s^3x^3y-256c^2s^3x^3y+256a^4s^4-448a^3s^4x+704a^2s^4x^2-512as^4x^3+256s^4x^4)t+8a^2s(27a^9y^5-18a^5b^4y^5+36a^5b^2c^2y^5-18a^5c^4y^5-9ab^8y^5+36ab^6c^2y^5-54ab^4c^4y^5+36ab^2c^6y^5-9ac^8y^5+54a^8sy^4+72a^6b^2sy^4-72a^6c^2sy^4-144a^5b^2sxy^4+144a^5c^2sxy^4+132a^4b^4sy^4-264a^4b^2c^2sy^4+132a^4c^4sy^4+72a^2b^6sy^4-216a^2b^4c^2sy^4+216a^2b^2c^4sy^4-72a^2c^6sy^4-144ab^6sxy^4+432ab^4c^2sxy^4-432ab^2c^4sxy^4+144ac^6sxy^4+22b^8sy^4-88b^6c^2sy^4+132b^4c^4sy^4-88b^2c^6sy^4+22c^8sy^4-144a^7s^2y^3+288a^6s^2xy^3-528a^5b^2s^2y^3+528a^5c^2s^2y^3-288a^5s^2x^2y^3+1056a^4b^2s^2xy^3-1056a^4c^2s^2xy^3-560a^3b^4s^2y^3+1120a^3b^2c^2s^2y^3-560a^3c^4s^2y^3+864a^2b^4s^2xy^3-1728a^2b^2c^2s^2xy^3+864a^2c^4s^2xy^3-176ab^6s^2y^3+528ab^4c^2s^2y^3-864ab^4s^2x^2y^3-528ab^2c^4s^2y^3+1728ab^2c^2s^2x^2y^3+176ac^6s^2y^3-864ac^4s^2x^2y^3+352b^6s^2xy^3-1056b^4c^2s^2xy^3+1056b^2c^4s^2xy^3-352c^6s^2xy^3+448a^6s^3y^2-2112a^5s^3xy^2+1664a^4b^2s^3y^2-1664a^4c^2s^3y^2+2112a^4s^3x^2y^2-4480a^3b^2s^3xy^2+4480a^3c^2s^3xy^2+704a^2b^4s^3y^2-1408a^2b^2c^2s^3y^2+3456a^2b^2s^3x^2y^2+704a^2c^4s^3y^2-3456a^2c^2s^3x^2y^2-2112ab^4s^3xy^2+4224ab^2c^2s^3xy^2-2304ab^2s^3x^3y^2-2112ac^4s^3xy^2+2304ac^2s^3x^3y^2+2112b^4s^3x^2y^2-4224b^2c^2s^3x^2y^2+2112c^4s^3x^2y^2-1408a^5s^4y+6656a^4s^4xy-1408a^3b^2s^4y+1408a^3c^2s^4y-8960a^3s^4x^2y+5632a^2b^2s^4xy-5632a^2c^2s^4xy+4608a^2s^4x^3y-8448ab^2s^4x^2y+8448ac^2s^4x^2y-2304as^4x^4y+5632b^2s^4x^3y-5632c^2s^4x^3y+1024a^4s^5-5632a^3s^5x+11264a^2s^5x^2-11264as^5x^3+5632s^5x^4)t^2=0\)

我们取(下面为了输入方便,我们设\(\sqrt{3}=\alpha,\sqrt{129}=\beta\)

为了作对比:我们把交点变换(对偶变换)的不变曲线(金色),和对称变换的不变曲线(红色)放在一起,mathe圆锥曲线变换(简称mathe变换)的不动曲线(蓝色),mathe不动直线(红色点线)

\(a=1,b=1,c=1,s=\frac{\sqrt{3}}{4}\)

mathe 变换的不变曲线:

\(((2365740x^4y-6984360x^2y^3-10260y^5-4731480x^3y+6984360xy^3+3038970x^2y-1740990y^3-673230xy+18915y)\beta-26869644x^4y+79327080x^2y^3+116532y^5+53739288x^3y-79327080xy^3-34516110x^2y+19773850y^3+7646466xy-214845y)\alpha+(1167480x^6-7004880x^4y^2+10507320x^2y^4-3502440x^5+14009760x^3y^2-10507320xy^4+3737700x^4-5366880x^2y^2+2611260y^4-1638000x^3-1638000xy^2+178515x^2+851745y^2+56745x-10440)\beta-13260024x^6+79560144x^4y^2-119340216y^4x^2+39780072x^5-159120288x^3y^2+119340216xy^4-42452100x^4+60956064x^2y^2-29658204y^4+18604080x^3+18604080xy^2-2027493x^2-9673959y^2-644535x+118584=0\)

交点变换与对称变换曲线一样:

\((x^2-8y^2-x)\alpha-27x^2y+9y^3+27xy-y=0\)

1112.GIF


\(a=5,b=4,c=3,s=6\)

mathe变换不变曲线:

\(-1461343106349y^6-320786500608x^5y+4085943795360x^3y^3-9844522488108xy^5+(16139243520x^6+28243676160x^5y-294625252800x^4y^2-359747287200x^3y^3+1380016815300x^2y^4+866761863660xy^5+128664084605y^6-242088652800x^5+157953888000x^4y+3542419008000x^3y^2-1951950636000x^2y^3-9714047431500xy^4-2604455600400y^5+1291749120000x^4-3602957760000x^3y-9849658320000x^2y^2+19856464560000xy^3+16093349070000y^4-2830464000000x^3+13933782000000x^2y-363352500000xy^2-31964886000000y^3+1542369600000x^2-17277494400000xy+18784850400000y^2+2451384000000x+3268512000000y-2255040000000)\beta+2749598576640x^5-183306571776x^6-40234146854400x^3y^2+110330370758700xy^4-15673978219140y^4x^2+3346299656640x^4y^2+196236017280000xy-225526064400000xy^3+22169896408800x^2y^3-158257625040000x^2y+40921744320000x^3y-1794011846400x^4y+111870648720000x^2y^2+4126602780000xy^2+29580931397520y^5+363051337680000y^3-213354384480000y^2-14671445760000x^4+32147850240000x^3-182785274730000y^4-17517539520000x^2-27843912000000x-37125216000000y+25614144000000=0\)

交点变换的不变曲线:

\(193536x^3-4810752x^2y-2419200xy^2+4147200y^3+276480x^2+24053760xy-15552000y^2-6220800x=0\)

对称变换的不变曲线:

\(-77760x^2y-45360xy^2+77760y^3+13824x^2+396864xy-245424y^2-69120x-92160y=0\)

5433.GIF
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-4-25 05:42:36 | 显示全部楼层
关于不动曲线\(x(y^2+z^2)+y(x^2+z^2)+z(x^2+y^2)-\frac{21}{2}xyz=0\)对应的切线方程我们可以尝试如下计算,
记$p=x+y+z,q=xy+yz+zx,r=xyz$,于是不动曲线方程变成$2pq=27r$
并且记
$u=4xy+4xz+2y^2−21yz+2z^2,v=4yz+4yx+2z^2-21zx+2x^2,w=4zx+4zy+2x^2-21xy+2y^2$
于是经过计算可以得到
$s_1=u+v+w=4p^2-21q$
$s_2=u^2+v^2+w^2=8p^4-184p^2q+633q^2$
$s_3=u^3+v^3+w^3=16p^6-7468/9 p^4q+6948p^2q^2-15609q^3$
然后分别消去q,p可以得到
$-28472283s_1^6 + 124496946s_2s_1^4 - 78633828s_3s_1^3 - 112826763s_2^2s_1^2 + 132453468s_3s_2s_1 + (-2381400s_2^3 - 34720812s_3^2)=0$
最后分别将$s_1,s_2,s_3$的表达式代入可以得出一条关于$u,v,w$的6次齐次方程
$-84672*u^6 + (-1947456*v - 1947456*w)*u^5 + (-4678128*v^2 - 57661632*w*v - 4678128*w^2)*u^4 + (13420512*v^3 - 271140912*w*v^2 - 271140912*w^2*v + 13420512*w^3)*u^3 + (-4678128*v^4 - 271140912*w*v^3 - 1012809420*w^2*v^2 - 271140912*w^3*v - 4678128*w^4)*u^2 + (-1947456*v^5 - 57661632*w*v^4 - 271140912*w^2*v^3 - 271140912*w^3*v^2 - 57661632*w^4*v - 1947456*w^5)*u + (-84672*v^6 - 1947456*w*v^5 - 4678128*w^2*v^4 + 13420512*w^3*v^3 - 4678128*w^4*v^2 - 1947456*w^5*v - 84672*w^6)=0$
不知道是否有计算错误
然后试着将星空关于$u,v,w$的重心坐标公式代入(取$a=b=c=1,s=sqrt(3)/4$)
结果得到如下图一个不够对称的图,感觉应该哪里算错了?这么对称的原始方程应该得到一条三个方向对称的曲线才对。
2m.png

点评

经检验,你的结果是对的  发表于 2019-4-25 19:19
你也计算一个版本看看,我也担心算错了  发表于 2019-4-25 11:48
嗯,这一步没问题了.但是关于s1,s2,s3的代数方程好像有问题?  发表于 2019-4-25 09:57
再把$r=2/27 pq$代入呢?  发表于 2019-4-25 09:14
s2,s3的表达式有误~s2=8p^4-100p^2q-1134pr+633q^2,s3=16p^6-300p^4q-7152p^3r+3726p^2q^2+47871pqr-15609q^3-59049r^2  发表于 2019-4-25 08:24
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-4-25 06:08:51 | 显示全部楼层
2m.png
试着把三角形(蓝色),原始不动曲线(绿色),和对偶不变曲线(红色)画在了一起。
感觉有点奇怪,怎么原始不动曲线也有点不对称,是不是参数$a=1,b=1,c=1,s={sqrt{3}}/4$
或者对应变换公式
subst(%,u,a*y/(2*s))
subst(%,v,-(a^2*y+b^2*y-c^2*y-4*a*s+4*s*x)/(2*a*s))
subst(%,w,-(a^2*y-b^2*y+c^2*y-4*s*x)/(4*a*s))
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-4-25 06:36:23 | 显示全部楼层
2m.png
找到问题了,星空前面贴的重心公式错了一个数字,应该是
subst(%,u,a*y/(2*s))
subst(%,v,-(a^2*y+b^2*y-c^2*y-4*a*s+4*s*x)/(4*a*s))
subst(%,w,-(a^2*y-b^2*y+c^2*y-4*s*x)/(4*a*s))

得出一条六次曲线
$-200704.0000*y^6 + ( (4616192.000*y - 3997739.541))*y^5 + (85876224.00*x^2 + ( - 85876224.00)*x + (-39714304.00*y^2 + 68787192.32*y - 8316672.000))*y^4 + ( (-553604352.0*y + 479435432.5)*x^2 + ( 553604352.0*y - 479435432.5)*x + (152723200.0*y^3 - 396786512.8*y^2 + 205226112.0*y + 20662229.91))*y^3 + (-639845136.0*x^4 + ( 1279690272)*x^3 + (875325024.0*y^2 - 1516107415*y - 303273936.0)*x^2 + ( - 875325024.0*y^2 + 1516107415*y - 336571200.0)*x + (-231770000.0*y^4 + 802874831.3*y^3 - 824133744.0*y^2 + 223129269.8*y - 6237504.000))*y^2 + ( (143857728.0*y - 124584447.0)*x^4 + ( - 287715456.0*y + 249168894.0)*x^3 + (-181104000.0*y^3 + 470521994.2*y^2 - 191697408.0*y - 69246171.92)*x^2 + ( 181104000.0*y^3 - 470521994.2*y^2 + 335555136.0*y - 55338275.06)*x + (49000000.00*y^5 - 212176223.9*y^4 + 322224000.0*y^3 - 200633837.3*y^2 + 44932608.00*y - 2248728.492))*y + (-19051200.00*x^6 + ( 57153600.00)*x^5 + (22184064.00*y^2 - 38423925.97*y - 54803952.00)*x^4 + ( - 44368128.00*y^2 + 76847851.93*y + 14351904.00)*x^3 + (-5880000.000*y^4 + 20368917.50*y^3 + 6816096.000*y^2 - 42359200.83*y + 3789072.000)*x^2 + ( 5880000.000*y^4 - 20368917.50*y^3 + 15367968.00*y^2 + 3935274.860*y - 1439424.000)*x + (  - 1470000.000*y^4 + 5092229.374*y^3 - 5228496.000*y^2 + 1417676.658*y - 84672.00000))=0$

点评

嗯,贴错了,已修正.我们得到的结果不一样啊?  发表于 2019-4-25 07:39
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-4-25 17:31:04 | 显示全部楼层
我重新计算的确有点不同,是不是
$336265625*s_1^6 - 25063528750*s_2*s_1^4 - 136076467500*s_3*s_1^3 - 220095474375*s_2^2*s_1^2 - 42330667500*s_3*s_2*s_1 + (1719637051392*s_2^3 - 1734797020284*s_3^2)=0$
然后在正三角形情况为
$13.75709641*x^6 + (- 41.27128923)*x^5 + (20.45372499*y^2 + 12.52236448*y + 45.15939477)*x^4 + (- 40.90744998*y^2 - 25.04472896*y - 21.53330747)*x^3 + (55.14966540*y^4 - 47.46046292*y^3 + 41.98703246*y^2 + 20.66981369*y + 2.246754023)*x^2 + (- 55.14966540*y^4 + 47.46046292*y^3 - 21.53330747*y^2 - 8.147449214*y + 1.641351508)*x + (11.44403372*y^6 - 27.93209054*y^5 + 38.68852975*y^4 - 23.46628572*y^3 + 6.950686020*y^2 + 0.9476347352*y - 0.4383898414)=0$
得到图
2m.png
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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