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[提问] 这个初中几何题如何求解?

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发表于 2018-3-10 19:06:47 | 显示全部楼层
本帖最后由 王守恩 于 2018-3-11 01:12 编辑
mathe 发表于 2018-3-5 19:39
对于一般情况,我们假设AA'分角A为u,v两个角,于是有
$tan(v)=tan(A-u)=\frac{tan(A)-tan(u)}{1+tan(A)tan ...


我还是在想:
1,此题先人应该有成熟的解法(先人比我们聪明多了)。
2,可以这样认为(?):任意一个三角形AEF都恰好对应另一个三角形ABC。
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
 楼主| 发表于 2018-3-21 19:42:20 | 显示全部楼层
  1. FullSimplify@Minimize[{Sqrt[b^2+(a-c)^2],(a-4)^2==a^2+d^2&&b^2+(c-4)^2==(b-d)^2+c^2&&b/3+c/4==1},{a,b,c,d}]
复制代码


我以为直接开放会降低求解速度,没想到也能快速求解出来

\[\left\{\frac{15}{\sqrt{24 \sqrt{6}+59}},\left\{a\to 2 \sqrt{6}-3,b\to 3 \left(\sqrt{6}-2\right),c\to -4 \left(\sqrt{6}-3\right),d\to 2 \left(\sqrt{6}-2\right)\right\}\right\}\]

点评

解析几何的办法最简单最直接,不费脑筋  发表于 2018-3-21 19:42
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
 楼主| 发表于 2018-3-22 10:03:22 | 显示全部楼层
本帖最后由 mathematica 于 2018-3-22 10:21 编辑
王守恩 发表于 2018-3-10 19:06
我还是在想:
1,此题先人应该有成熟的解法(先人比我们聪明多了)。
2,可以这样认为(?):任意一个三 ...


利用三次余弦定理,照样可以得到约束,然后求解最小值

  1. (*计算折纸的痕迹的最小长度*)
  2. Clear["Global`*"];(*Clear all variables*)
  3. FullSimplify@Minimize[
  4. {g,(*目标函数*)
  5. (*约束条件*)
  6. a+b==4&&
  7. c+d==3&&
  8. e+f==5&&
  9. b^2+c^2==a^2&&(*勾股定理或者余弦定理*)
  10. (a^2+f^2-g^2)/(2*a*f)==4/5&&(*余弦定理*)
  11. (d^2+e^2-f^2)/(2*d*e)==3/5&&(*余弦定理*)
  12. a>=0&&
  13. b>=0&&
  14. c>=0&&
  15. d>=0&&
  16. e>=0&&
  17. f>=0&&
  18. g>=0
  19. },
  20. {a,b,c,d,e,f,g}](*自变量*)
复制代码


求解结果

\[\left\{\frac{15}{\sqrt{24 \sqrt{6}+59}},\left\{a\to 7-2 \sqrt{6},b\to 2 \sqrt{6}-3,c\to 2 \left(\sqrt{6}-2\right),d\to 7-2 \sqrt{6},e\to -5 \left(\sqrt{6}-3\right),f\to 5 \left(\sqrt{6}-2\right),g\to \frac{15}{\sqrt{24 \sqrt{6}+59}}\right\}\right\}\]

变量含义见附件图片

  1. Clear["Global`*"];(*Clear all variables*)
  2. FullSimplify@Solve[
  3. a+b==4&&
  4. c+d==3&&
  5. e+f==5&&
  6. b^2+c^2==a^2&&(*勾股定理或者余弦定理*)
  7. (a^2+f^2-g^2)/(2*a*f)==4/5&&(*余弦定理*)
  8. (d^2+e^2-f^2)/(2*d*e)==3/5&&(*余弦定理*)
  9. a>=0&&
  10. b>=0&&
  11. c>=0&&
  12. d>=0&&
  13. e>=0&&
  14. f>=0&&
  15. g>=0
  16. ,
  17. {a,b,d,e,f,g}(*自变量*)
  18. ]
复制代码


求解出各个长度以c为变量
\[\left\{\left\{a\to \text{ConditionalExpression}\left[\frac{c^2}{8}+2,0<c<3\right],b\to \text{ConditionalExpression}\left[2-\frac{c^2}{8},0<c<3\right],d\to \text{ConditionalExpression}[3-c,0<c<3],e\to \text{ConditionalExpression}\left[-\frac{5 (c-8) (c+2)}{6 c+32},0<c<3\right],f\to \text{ConditionalExpression}\left[\frac{5 \left(c^2+16\right)}{6 c+32},0<c<3\right],g\to \text{ConditionalExpression}\left[\frac{3 \left(c^2+16\right)^{3/2}}{8 (3 c+16)},0<c<3\right]\right\}\right\}\]
02.png

点评

mathematica求解错误.有的表达式的范围可以是闭区间  发表于 2018-3-22 10:23
不添加任何辅助线  发表于 2018-3-22 10:05
一下子把所有的长度一网打尽!mathematica万岁!  发表于 2018-3-22 10:05
很奇怪,mathematica究竟如何求解这个最优化问题的解的呢?具体操作是?  发表于 2018-3-22 10:04
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
 楼主| 发表于 2018-3-22 16:50:28 | 显示全部楼层
本帖最后由 mathematica 于 2018-3-22 16:51 编辑
mathematica 发表于 2018-3-22 10:03
利用三次余弦定理,照样可以得到约束,然后求解最小值

  1. Clear["Global`*"];(*Clear all variables*)
  2. FullSimplify@Solve[
  3. a+b==4&&
  4. c+d==3&&
  5. e+f==5&&
  6. b^2+c^2==a^2&&(*勾股定理或者余弦定理*)
  7. (a^2+f^2-g^2)/(2*a*f)==4/5&&(*余弦定理*)
  8. (d^2+e^2-f^2)/(2*d*e)==3/5&&(*余弦定理*)
  9. a>=0&&
  10. b>=0&&
  11. c>=0&&
  12. d>=0&&
  13. e>=0&&
  14. f>=0&&
  15. g>=0
  16. ,
  17. {a,b,c,d,e,f}(*自变量*)
  18. ]
复制代码

查看共同定义域的最小值就可以了
\[\left\{\left\{a\to \text{ConditionalExpression}\left[\frac{1}{8} \left(\left(\text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,1\right]-6\right) \text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,1\right]+25\right),\frac{3}{2}<g<\frac{15}{8}\lor \frac{15}{\sqrt{24 \sqrt{6}+59}}<g<\frac{3}{2}\right],b\to \text{ConditionalExpression}\left[-\frac{1}{8} \left(\text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,1\right]-7\right) \left(\text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,1\right]+1\right),\frac{3}{2}<g<\frac{15}{8}\lor \frac{15}{\sqrt{24 \sqrt{6}+59}}<g<\frac{3}{2}\right],c\to \text{ConditionalExpression}\left[3-\text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,1\right],\frac{3}{2}<g<\frac{15}{8}\lor \frac{15}{\sqrt{24 \sqrt{6}+59}}<g<\frac{3}{2}\right],d\to \text{ConditionalExpression}\left[\text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,1\right],\frac{3}{2}<g<\frac{15}{8}\lor \frac{15}{\sqrt{24 \sqrt{6}+59}}<g<\frac{3}{2}\right],e\to \text{ConditionalExpression}\left[\frac{5 \left(\text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,1\right]^2-25\right)}{6 \text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,1\right]-50},\frac{3}{2}<g<\frac{15}{8}\lor \frac{15}{\sqrt{24 \sqrt{6}+59}}<g<\frac{3}{2}\right],f\to \text{ConditionalExpression}\left[\frac{5}{18} \left(-3 \text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,1\right]+\frac{400}{25-3 \text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,1\right]}-7\right),\frac{3}{2}<g<\frac{15}{8}\lor \frac{15}{\sqrt{24 \sqrt{6}+59}}<g<\frac{3}{2}\right]\right\},\left\{a\to \text{ConditionalExpression}\left[\frac{1}{8} \left(\left(\text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,2\right]-6\right) \text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,2\right]+25\right),\frac{15}{\sqrt{24 \sqrt{6}+59}}<g<\frac{3}{2}\right],b\to \text{ConditionalExpression}\left[-\frac{1}{8} \left(\text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,2\right]-7\right) \left(\text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,2\right]+1\right),\frac{15}{\sqrt{24 \sqrt{6}+59}}<g<\frac{3}{2}\right],c\to \text{ConditionalExpression}\left[3-\text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,2\right],\frac{15}{\sqrt{24 \sqrt{6}+59}}<g<\frac{3}{2}\right],d\to \text{ConditionalExpression}\left[\text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,2\right],\frac{15}{\sqrt{24 \sqrt{6}+59}}<g<\frac{3}{2}\right],e\to \text{ConditionalExpression}\left[\frac{5 \left(\text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,2\right]^2-25\right)}{6 \text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,2\right]-50},\frac{15}{\sqrt{24 \sqrt{6}+59}}<g<\frac{3}{2}\right],f\to \text{ConditionalExpression}\left[\frac{5}{18} \left(-3 \text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,2\right]+\frac{400}{25-3 \text{Root}\left[9 \text{$\#$1}^6-162 \text{$\#$1}^5+1647 \text{$\#$1}^4-10044 \text{$\#$1}^3+\left(41175-576 g^2\right) \text{$\#$1}^2+\left(9600 g^2-101250\right) \text{$\#$1}-40000 g^2+140625\&,2\right]}-7\right),\frac{15}{\sqrt{24 \sqrt{6}+59}}<g<\frac{3}{2}\right]\right\}\right\}\]


可知道结果是
\[\frac{15}{\sqrt{24 \sqrt{6}+59}}\]

点评

万岁的mathematica就是不会把\(\frac{15}{\sqrt{59+24 \sqrt{6}}}\)简化成 \(12 \sqrt{2}-9 \sqrt{3}\)。  发表于 2018-3-22 20:28
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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