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楼主: 蓉依山爸

[讨论] 20多年了,我无力解出来的一道高中奥数题!

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发表于 2014-1-3 10:21:35 | 显示全部楼层
这个四面体体积最大的问题  在百度贴吧有过,别人给定了数值,我用P是垂心,巧妙的放进三维坐标轴最后解4次方程,是个有根式解的4次方程,最后也可得到根式的体积。这里用字母代替了,相信最后的4次方程也有根式解,只是估计结果有几页长,这里就不算了
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 2014-3-19 10:25:09 | 显示全部楼层
怀疑表面积最大时候,那个点不是什么特殊点,暴力求6元高次函数极值软件也没有办法……

点评

6元高次函数极值?看样子空间的极值都不是那么好求的,就好比要求空间两圆圈上的最短距离,也是个高次方程!  发表于 2014-3-22 13:20
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 2014-3-20 19:09:04 | 显示全部楼层
如果将该题目类比到N维是否有解?
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 2014-5-28 11:30:42 | 显示全部楼层
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 2014-5-31 23:25:08 | 显示全部楼层
根据楼上提供的链接,出于网友kuing 精彩而初等的分析特将其解答过程转载如下(图片引用原文):

引理1:锐角\(\triangle ABC\) 中,\(H\)为其垂心,则\(AH\*HD=BH\*HE=CH\*HF=k\),

       \(\frac{k}{k+HA^2}+\frac{k}{k+HB^2}+\frac{k}{k+HC^2}=1\)

       注:1.由\(AFH \sim CDH\) 可以得到\(\frac{AH}{FH}=\frac{CH}{DH}\) 即\(AH\*HD=CH\*HF\),同理易得\(AH\*HD=BH\*HE\)

            2.\(\frac{S_{BHC}}{S_{ABC}}+\frac{S_{AHC}}{S_{ABC}}+\frac{S_{AHB}}{S_{ABC}}=1\)

              \(\frac{HD}{AH+HD}+\frac{HE}{BH+HE}+\frac{HF}{CH+HF}=1\)

             \(\frac{AH\*HD}{AH\*HD+HA^2}+\frac{BH\*HE}{BH\*HE+HB^2}+\frac{CH\*HF}{CH\*HF+HC^2}=1\)


201461.gif



在下面垂心四面体\(D-ABC\)的垂心\(H\)在四面体的内部,已知\(HA=a,HB=b,HC=c,HD=d\),体积为\(V\)

201462.gif

并记\(H\)在底面ABC的投影为\(E\),且记\(AE=x,BE=y,CE=z,HE=w\),由引理1,我们可以得到:

在\(\triangle ABC\) 中,设\(AE\*EF=k\),有  \(\frac{k}{k+x^2}+\frac{k}{k+y^2}+\frac{k}{k+z^2}=1\) ..........(1)

由于\(HE\perp \triangle ABC\) 易有\(x^2=a^2-w^2,y^2=b^2-w^2,z^2=c^2-w^2\)............................(2)

又\(\triangle AEH \sim \triangle DEF\), 有\(\frac{AE}{EH}=\frac{DE}{EF}\),即\(k=w(d+w)\).................(3)

将(2),(3)代入(1)并化简得到:\(3d^2w^4+2d(a^2+b^2+c^2+d^2)w^3+(a^2b^2+a^2+c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)w^2-a^2b^2c^2=0\).......(4)

设\(\triangle ABC\) 的面积为\(S\),其外接圆直径为D,则\(x=D\cos(A),y=D\cos(B),z=D\cos(C)\),代入\(\cos(A)^2+\cos(B)^2+\cos(C)^2+2\cos(A)\cos(B)\cos(C)=1\)

得到:\(D(x^2+y^2+z^2)+2xyz=D^3\).........................................................................................(5)

又\(a^2=(D\sin(A))^2=D^2-(D\cos(A))^2=D^2-x^2\),同理\(b^2=D^2-y^2\),\(c^2=D^2-z^2\).......(6)

代入\(16S^2=2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4\).............................................................(7)

得到:\(16S^2=3D^4-2(x^2+y^2+z^2)D^2+2(x^2y^2+x^2z^2+y^2z^2)-(x^4+y^4+z^4)\)..........(8)

将(2)代入(5)和(8)并消除\(D\)得到:

\(256S^6+16(\sum_{a,b,c}^{}(a^4-10b^2c^2)+18w^2\sum_{a,b,c}^{}(a^2)-27w^4)S^4-8(\sum_{a,b,c}^{}(a^6b^2-4a^4b^4+a^2b^6+2a^2b^2c^4)+w^2\prod_{a,b,c}{}(a^2+b^2-2c^2))S^2+\prod_{a,b,c}^{}(a^2-b^2)^2=0\)…(9)

又\(S(d+w)=3V\).............................................................................................................................(10)

将(10)代入 (9) 消除\(S\)得到:

\(186624V^6+1296(d+w)^2(\sum_{a,b,c}^{}(a^4-10b^2c^2)+18w^2\sum_{a,b,c}^{}(a^2)-27w^4)V^4-72(d+w)^4(\sum_{a,b,c}^{}(a^6b^2-4a^4b^4+a^2b^6+2a^2b^2c^4)+w^2\prod_{a,b,c}^{}(a^2+b^2-2c^2))V^2+(d+w)^6\prod_{a,b,c}^{}(a^2-b^2)^2=0\)

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 2014-6-1 00:50:34 | 显示全部楼层
楼上最后一个等式写全了即:

\((b-c)^2(b+c)^2(a-c)^2(a+c)^2(a-b)^2(a+b)^2(d+w)^6-(72(a^6b^2+a^6c^2-2a^6w^2-4a^4b^4+2a^4b^2c^2+3a^4b^2w^2-4a^4c^4+3a^4c^2w^2+a^2b^6+2a^2b^4c^2+3a^2b^4w^2+2a^2b^2c^4-12a^2b^2c^2w^2+a^2c^6+3a^2c^4w^2+b^6c^2-2b^6w^2-4b^4c^4+3b^4c^2w^2+b^2c^6+3b^2c^4w^2-2c^6w^2))V^2(d+w)^4+(1296(a^4-10a^2b^2-10a^2c^2+18a^2w^2+b^4-10b^2c^2+18b^2w^2+c^4+18c^2w^2-27w^4))V^4(d+w)^2+186624V^6=0\)

最终计算结果为:

\(-(c-d)^2(c+d)^2(b-d)^2(b+d)^2(b-c)^2(b+c)^2(a-d)^2(a+d)^2(a-c)^2(a+c)^2(a-b)^2(a+b)^2+(72a^{10}b^6c^2+72a^{10}b^6d^2-288a^{10}b^4c^4+144a^{10}b^4c^2d^2-288a^{10}b^4d^4+72a^{10}b^2c^6+144a^{10}b^2c^4d^2+144a^{10}b^2c^2d^4+72a^{10}b^2d^6+72a^{10}c^6d^2-288a^{10}c^4d^4+72a^{10}c^2d^6-144a^8b^8c^2-144a^8b^8d^2+360a^8b^6c^4-288a^8b^6c^2d^2+360a^8b^6d^4+360a^8b^4c^6-144a^8b^4c^4d^2-144a^8b^4c^2d^4+360a^8b^4d^6-144a^8b^2c^8-288a^8b^2c^6d^2-144a^8b^2c^4d^4-288a^8b^2c^2d^6-144a^8b^2d^8-144a^8c^8d^2+360a^8c^6d^4+360a^8c^4d^6-144a^8c^2d^8+72a^6b^{10}c^2+72a^6b^{10}d^2+360a^6b^8c^4-288a^6b^8c^2d^2+360a^6b^8d^4-1296a^6b^6c^6+432a^6b^6c^4d^2+432a^6b^6c^2d^4-1296a^6b^6d^6+360a^6b^4c^8+432a^6b^4c^6d^2-432a^6b^4c^4d^4+432a^6b^4c^2d^6+360a^6b^4d^8+72a^6b^2c^{10}-288a^6b^2c^8d^2+432a^6b^2c^6d^4+432a^6b^2c^4d^6-288a^6b^2c^2d^8+72a^6b^2d^{10}+72a^6c^{10}d^2+360a^6c^8d^4-1296a^6c^6d^6+360a^6c^4d^8+72a^6c^2d^{10}-288a^4b^{10}c^4+144a^4b^{10}c^2d^2-288a^4b^{10}d^4+360a^4b^8c^6-144a^4b^8c^4d^2-144a^4b^8c^2d^4+360a^4b^8d^6+360a^4b^6c^8+432a^4b^6c^6d^2-432a^4b^6c^4d^4+432a^4b^6c^2d^6+360a^4b^6d^8-288a^4b^4c^{10}-144a^4b^4c^8d^2-432a^4b^4c^6d^4-432a^4b^4c^4d^6-144a^4b^4c^2d^8-288a^4b^4d^{10}+144a^4b^2c^{10}d^2-144a^4b^2c^8d^4+432a^4b^2c^6d^6-144a^4b^2c^4d^8+144a^4b^2c^2d^{10}-288a^4c^{10}d^4+360a^4c^8d^6+360a^4c^6d^8-288a^4c^4d^{10}+72a^2b^{10}c^6+144a^2b^{10}c^4d^2+144a^2b^{10}c^2d^4+72a^2b^{10}d^6-144a^2b^8c^8-288a^2b^8c^6d^2-144a^2b^8c^4d^4-288a^2b^8c^2d^6-144a^2b^8d^8+72a^2b^6c^{10}-288a^2b^6c^8d^2+432a^2b^6c^6d^4+432a^2b^6c^4d^6-288a^2b^6c^2d^8+72a^2b^6d^{10}+144a^2b^4c^{10}d^2-144a^2b^4c^8d^4+432a^2b^4c^6d^6-144a^2b^4c^4d^8+144a^2b^4c^2d^{10}+144a^2b^2c^{10}d^4-288a^2b^2c^8d^6-288a^2b^2c^6d^8+144a^2b^2c^4d^{10}+72a^2c^{10}d^6-144a^2c^8d^8+72a^2c^6d^{10}+72b^{10}c^6d^2-288b^{10}c^4d^4+72b^{10}c^2d^6-144b^8c^8d^2+360b^8c^6d^4+360b^8c^4d^6-144b^8c^2d^8+72b^6c^{10}d^2+360b^6c^8d^4-1296b^6c^6d^6+360b^6c^4d^8+72b^6c^2d^{10}-288b^4c^{10}d^4+360b^4c^8d^6+360b^4c^6d^8-288b^4c^4d^{10}+72b^2c^{10}d^6-144b^2c^8d^8+72b^2c^6d^{10})V^2+(-1296a^8b^4+12960a^8b^2c^2+12960a^8b^2d^2-1296a^8c^4+12960a^8c^2d^2-1296a^8d^4+2592a^6b^6-18144a^6b^4c^2-18144a^6b^4d^2-18144a^6b^2c^4-38880a^6b^2c^2d^2-18144a^6b^2d^4+2592a^6c^6-18144a^6c^4d^2-18144a^6c^2d^4+2592a^6d^6-1296a^4b^8-18144a^4b^6c^2-18144a^4b^6d^2+104976a^4b^4c^4+2592a^4b^4c^2d^2+104976a^4b^4d^4-18144a^4b^2c^6+2592a^4b^2c^4d^2+2592a^4b^2c^2d^4-18144a^4b^2d^6-1296a^4c^8-18144a^4c^6d^2+104976a^4c^4d^4-18144a^4c^2d^6-1296a^4d^8+12960a^2b^8c^2+12960a^2b^8d^2-18144a^2b^6c^4-38880a^2b^6c^2d^2-18144a^2b^6d^4-18144a^2b^4c^6+2592a^2b^4c^4d^2+2592a^2b^4c^2d^4-18144a^2b^4d^6+12960a^2b^2c^8-38880a^2b^2c^6d^2+2592a^2b^2c^4d^4-38880a^2b^2c^2d^6+12960a^2b^2d^8+12960a^2c^8d^2-18144a^2c^6d^4-18144a^2c^4d^6+12960a^2c^2d^8-1296b^8c^4+12960b^8c^2d^2-1296b^8d^4+2592b^6c^6-18144b^6c^4d^2-18144b^6c^2d^4+2592b^6d^6-1296b^4c^8-18144b^4c^6d^2+104976b^4c^4d^4-18144b^4c^2d^6-1296b^4d^8+12960b^2c^8d^2-18144b^2c^6d^4-18144b^2c^4d^6+12960b^2c^2d^8-1296c^8d^4+2592c^6d^6-1296c^4d^8)V^4+(-186624a^6+279936a^4b^2+279936a^4c^2+279936a^4d^2+279936a^2b^4-3639168a^2b^2c^2-3639168a^2b^2d^2+279936a^2c^4-3639168a^2c^2d^2+279936a^2d^4-186624b^6+279936b^4c^2+279936b^4d^2+279936b^2c^4-3639168b^2c^2d^2+279936b^2d^4-186624c^6+279936c^4d^2+279936c^2d^4-186624d^6)V^6+45349632V^8=0\)

若做代换:

\(s_1 = a^2+b^2+c^2+d^2, s_2 = a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2, s_3 = a^2b^2c^2+a^2b^2d^2+a^2c^2d^2+b^2c^2d^2, s_4 = a^2b^2c^2d^2\)

可以将结果简化为:

\(45349632V^8+(-186624s_1^3+839808s_1s_2-5038848s_3)V^6+(15552s_1^3s_3-1296s_1^2s_2^2+7776s_1^2s_4-69984s_1s_2s_3+5184s_2^3-186624s_2s_4+209952s_3^2)V^4+(648s_1^3s_2s_4-432s_1^3s_3^2+72s_1^2s_2^2s_3-432s_1^2s_3s_4-2880s_1s_2^2s_4+1944s_1s_2s_3^2-288s_2^3s_3-6912s_1s_4^2+10368s_2s_3s_4-3888s_3^3)V^2+27s_1^4s_4^2-18s_1^3s_2s_3s_4+4s_1^3s_3^3+4s_1^2s_2^3s_4-s_1^2s_2^2s_3^2-144s_1^2s_2s_4^2+6s_1^2s_3^2s_4+80s_1s_2^2s_3s_4-18s_1s_2s_3^3-16s_2^4s_4+4s_2^3s_3^2+192s_1s_3s_4^2+128s_2^2s_4^2-144s_2s_3^2s_4+27s_3^4-256s_4^3=0\)

=======================================================

最终简化结果与54#简化结果是一致的!

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 2014-6-1 11:07:58 | 显示全部楼层
根据57#,58#  mathe的结论,猜想:

当 定长\(PA ,PB, PC, PD\) 构成的四面体各条棱边长最大时即\((AB+AC+BC+DA+DB+DC)_{\max}\),设\(\triangle ABC,\triangle DBC,\triangle DAC,\triangle DAB\)的内心分别为\(P_0,P_1,P_2,P_3\)

\(P\)点即为\(AP_1\cap BP_2 \cap CP_3  \cap DP_0\)  ?
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 2014-7-9 21:06:40 | 显示全部楼层
我感觉我解决了这个问题,P点不是特殊点,要想得到面积最值,只有依靠5个未知数的函数求极值了。我是拿两个长100,两个短1的TA  TB  TC TD 构成对称四面体求两个角度的方法。两个100的线夹角是2*0.7874489594,两个1线的夹角是2*0.9562693657.发现这个四面体没有什么特殊线或面的关系。大家可以用这个方法验证。如果求4个长度不一样的TA TB TC TD  估计软件能力有限。当然这个四面体会满足:每一个顶点到底面边的距离的增量,乘以各自的底边3个值会相等
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 2014-7-13 15:50:50 | 显示全部楼层
好了,发现这个最大体积最大表面积问题可以通过软件数值求解了。相信最长棱长也数值求解没有问题。设5个未知数,5个偏导数等于0,是可以解出来的。不要再研究满足什么条件或有根式解了,方程长度太恐怖了
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 2014-7-13 19:30:04 | 显示全部楼层
a := 3;
print(outputredirected...); # input placeholder
3
b := 4;
print(outputredirected...); # input placeholder
4
c := 5;
print(outputredirected...); # input placeholder
5
d := 6;
print(outputredirected...); # input placeholder
6
with(LinearAlgebra);
eq := <,>(x, y, sqrt(d^2-x^2-y^2)+h);
print(outputredirected...); # input placeholder
Vector[column](%id = 4424046202)
eq1 := <,>(sqrt(a^2-h^2), 0, 0);
print(outputredirected...); # input placeholder
Vector[column](%id = 4424106418)
eq2 := sqrt(b^2-h^2)*<,>(cos(alpha), sin(alpha), 0);
print(outputredirected...); # input placeholder
Vector[column](%id = 4424106490)
eq3 := sqrt(c^2-h^2)*<,>(cos(beta), sin(beta), 0);
print(outputredirected...); # input placeholder
Vector[column](%id = 4424107210)
v := sqrt(&x(eq2-eq1, eq3-eq1).&x(eq2-eq1, eq3-eq1))+sqrt(&x(eq-eq1, eq-eq2).&x(eq-eq1, eq-eq2))+sqrt(&x(eq-eq2, eq-eq3).&x(eq-eq2, eq-eq3))+sqrt(&x(eq-eq3, eq-eq1).&x(eq-eq3, eq-eq1));
print(outputredirected...); # input placeholder
/ // (1/2) (1/2)\
| ||/ 2 \ / 2 \ |
\conjugate\\\-h + 16/ cos(alpha) - \-h + 9/ /

(1/2) (1/2) /
/ 2 \ / 2 \ |
\-h + 25/ sin(beta) - \-h + 16/ sin(alpha) \

(1/2) (1/2)\\ // (1/2)
/ 2 \ / 2 \ || ||/ 2 \
\-h + 25/ cos(beta) - \-h + 9/ // \\\-h + 16/

(1/2)\ (1/2)
/ 2 \ | / 2 \
cos(alpha) - \-h + 9/ / \-h + 25/ sin(beta) -

(1/2) / (1/2)
/ 2 \ |/ 2 \
\-h + 16/ sin(alpha) \\-h + 25/ cos(beta)

(1/2)\\\ / / / (1/2)
/ 2 \ ||| | | |/ 2 2 \
- \-h + 9/ ///^(1/2) + \conjugate\y \\-x - y + 36/

\
|
+ h/

/ (1/2) \ / (1/2) \\
|/ 2 2 \ | | / 2 \ ||
- \\-x - y + 36/ + h/ \y - \-h + 16/ sin(alpha)//

/ / (1/2) \
| |/ 2 2 \ |
\y \\-x - y + 36/ + h/

/ (1/2) \ / (1/2) \\
|/ 2 2 \ | | / 2 \ ||
- \\-x - y + 36/ + h/ \y - \-h + 16/ sin(alpha)// +

/ / (1/2)\ / (1/2) \
| | / 2 \ | |/ 2 2 \ |
conjugate\-\x - \-h + 9/ / \\-x - y + 36/ + h/

/ (1/2) \ / (1/2) \\
|/ 2 2 \ | | / 2 \ ||
+ \\-x - y + 36/ + h/ \x - \-h + 16/ cos(alpha)//

/ / (1/2)\ / (1/2) \
| | / 2 \ | |/ 2 2 \ |
\-\x - \-h + 9/ / \\-x - y + 36/ + h/

/ (1/2) \ / (1/2) \\
|/ 2 2 \ | | / 2 \ ||
+ \\-x - y + 36/ + h/ \x - \-h + 16/ cos(alpha)// +

// (1/2)\ / (1/2) \
|| / 2 \ | | / 2 \ |
conjugate\\x - \-h + 9/ / \y - \-h + 16/ sin(alpha)/

/ (1/2) \\ // (1/2)\ /
| / 2 \ || || / 2 \ | |
- y \x - \-h + 16/ cos(alpha)// \\x - \-h + 9/ / \y

(1/2) \
/ 2 \ |
- \-h + 16/ sin(alpha)/

/ (1/2) \\\ / //
| / 2 \ ||| | ||
- y \x - \-h + 16/ cos(alpha)///^(1/2) + \conjugate\\

(1/2) \ / (1/2) \
/ 2 2 \ | | / 2 \ |
\-x - y + 36/ + h/ \y - \-h + 16/ sin(alpha)/

/ (1/2) \ / (1/2) \\
|/ 2 2 \ | | / 2 \ ||
- \\-x - y + 36/ + h/ \y - \-h + 25/ sin(beta)//

// (1/2) \ / (1/2) \
||/ 2 2 \ | | / 2 \ |
\\\-x - y + 36/ + h/ \y - \-h + 16/ sin(alpha)/

/ (1/2) \ / (1/2) \\
|/ 2 2 \ | | / 2 \ ||
- \\-x - y + 36/ + h/ \y - \-h + 25/ sin(beta)// +

/
|
conjugate\
/ (1/2) \ / (1/2) \
|/ 2 2 \ | | / 2 \ |
-\\-x - y + 36/ + h/ \x - \-h + 16/ cos(alpha)/

/ (1/2) \ / (1/2) \\
|/ 2 2 \ | | / 2 \ ||
+ \\-x - y + 36/ + h/ \x - \-h + 25/ cos(beta)//

/ / (1/2) \ / (1/2) \
| |/ 2 2 \ | | / 2 \ |
\-\\-x - y + 36/ + h/ \x - \-h + 16/ cos(alpha)/

/ (1/2) \ / (1/2) \\
|/ 2 2 \ | | / 2 \ ||
+ \\-x - y + 36/ + h/ \x - \-h + 25/ cos(beta)// +

// (1/2) \ /
|| / 2 \ | |
conjugate\\x - \-h + 16/ cos(alpha)/ \y

(1/2) \
/ 2 \ |
- \-h + 25/ sin(beta)/

/ (1/2) \ /
| / 2 \ | |
- \y - \-h + 16/ sin(alpha)/ \x

(1/2) \\ // (1/2)
/ 2 \ || || / 2 \
- \-h + 25/ cos(beta)// \\x - \-h + 16/ cos(alpha)

\ / (1/2) \
| | / 2 \ |
/ \y - \-h + 25/ sin(beta)/

/ (1/2) \ /
| / 2 \ | |
- \y - \-h + 16/ sin(alpha)/ \x

(1/2) \\\ / //
/ 2 \ ||| | ||
- \-h + 25/ cos(beta)///^(1/2) + \conjugate\\

(1/2) \ / (1/2) \
/ 2 2 \ | | / 2 \ |
\-x - y + 36/ + h/ \y - \-h + 25/ sin(beta)/

/ (1/2) \\ // (1/2) \ /
|/ 2 2 \ || ||/ 2 2 \ | |
- y \\-x - y + 36/ + h// \\\-x - y + 36/ + h/ \y

(1/2) \ / (1/2) \\
/ 2 \ | |/ 2 2 \ ||
- \-h + 25/ sin(beta)/ - y \\-x - y + 36/ + h// +

/
|
conjugate\
/ (1/2) \ / (1/2) \
|/ 2 2 \ | | / 2 \ |
-\\-x - y + 36/ + h/ \x - \-h + 25/ cos(beta)/

/ (1/2)\ / (1/2) \\ /
| / 2 \ | |/ 2 2 \ || |
+ \x - \-h + 9/ / \\-x - y + 36/ + h// \
/ (1/2) \ / (1/2) \
|/ 2 2 \ | | / 2 \ |
-\\-x - y + 36/ + h/ \x - \-h + 25/ cos(beta)/

/ (1/2)\ / (1/2) \\ /
| / 2 \ | |/ 2 2 \ || |
+ \x - \-h + 9/ / \\-x - y + 36/ + h// + conjugate\

/ (1/2) \
| / 2 \ |
\x - \-h + 25/ cos(beta)/ y

/ (1/2) \ / (1/2)\\ //
| / 2 \ | | / 2 \ || ||
- \y - \-h + 25/ sin(beta)/ \x - \-h + 9/ // \\x

(1/2) \
/ 2 \ |
- \-h + 25/ cos(beta)/ y

/ (1/2) \ / (1/2)\\\
| / 2 \ | | / 2 \ |||
- \y - \-h + 25/ sin(beta)/ \x - \-h + 9/ ///^(1/2)
fsolve({diff(v, h) = 0., diff(v, x) = 0., diff(v, y) = 0., diff(v, alpha) = 0., diff(v, beta) = 0.}, {h, x, y, alpha, beta});
print(outputredirected...); # input placeholder
{h = 1.150423027, x = 0.3116699162, y = 0.03559146687,

alpha = 2.147085487, beta = -2.096605483}
h := 1.150423027;
print(outputredirected...); # input placeholder
1.150423027
x := .3116699162;
print(outputredirected...); # input placeholder
0.3116699162
y := 0.3559146687e-1;
print(outputredirected...); # input placeholder
0.03559146687
alpha := 2.147085487;
print(outputredirected...); # input placeholder
2.147085487
beta := -2.096605483;
print(outputredirected...); # input placeholder
-2.096605483
evalf((1/2)*v);
print(outputredirected...); # input placeholder
92.41717020

点评

maple里面的 已经复制在下面latex转换了的 还是这样  发表于 2014-7-22 21:27
发上来的是一堆乱码,你要学会用Latex编写相关的过程或者结论,否则没有人愿意看,也看不明白啊!  发表于 2014-7-22 21:19
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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