一道初中竞赛题(平面几何)

mathematica

1. Clear["Global`*"];
   
2. ans=FullSimplify@Solve[{
   
3.     AC^2+1^2==AP^2,(*勾股定理*)
   
4.     AC^2+(1+3)^2==AB^2,(*勾股定理*)
   
5.     cosAPC==1/AP,(*余弦定义*)
   
6.     cosB==4/AB,(*余弦定义*)
   
7.     cosAPC==4*cosB^3-3*cosB,(*三倍角公式*)
   
8.     AC>=0&&AP>=0&&AB>=0(*限制变量范围*)
   
9. },{AC,AP,AB,cosAPC,cosB}];
   
10. aaa=Grid[ans]
    

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求解结果:
\[\begin{array}{ccccc}
\text{AC}\to 0 & \text{AP}\to 1 & \text{AB}\to 4 & \text{cosAPC}\to 1 & \text{cosB}\to 1 \\
\text{AC}\to \frac{4}{\sqrt{11}} & \text{AP}\to 3 \sqrt{\frac{3}{11}} & \text{AB}\to 8 \sqrt{\frac{3}{11}} & \text{cosAPC}\to \frac{\sqrt{\frac{11}{3}}}{3} & \text{cosB}\to \frac{\sqrt{\frac{11}{3}}}{2} \\
\end{array}\]

1. Clear["Global`*"];
   
2. ans=FullSimplify@Solve[{
   
3.     AC^2+1^2==AP^2,(*勾股定理*)
   
4.     AC^2+(1+3)^2==AB^2,(*勾股定理*)
   
5.     cosAPC==1/AP,(*余弦定义*)
   
6.     cosB==4/AB,(*余弦定义*)
   
7.     cosAPC==4*cosB^3-3*cosB(*三倍角公式*)
   
8. },{AC,AP,AB,cosAPC,cosB}];
   
9. aaa=Grid[ans,Alignment->Left]
   

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\[\begin{array}{lllll}
\text{AC}\to 0 & \text{AP}\to -1 & \text{AB}\to -4 & \text{cosAPC}\to -1 & \text{cosB}\to -1 \\
\text{AC}\to 0 & \text{AP}\to 1 & \text{AB}\to 4 & \text{cosAPC}\to 1 & \text{cosB}\to 1 \\
\text{AC}\to -4 \sqrt{\frac{7}{13}} & \text{AP}\to 5 \sqrt{\frac{5}{13}} & \text{AB}\to -8 \sqrt{\frac{5}{13}} & \text{cosAPC}\to \frac{\sqrt{\frac{13}{5}}}{5} & \text{cosB}\to -\frac{\sqrt{\frac{13}{5}}}{2} \\
\text{AC}\to 4 \sqrt{\frac{7}{13}} & \text{AP}\to 5 \sqrt{\frac{5}{13}} & \text{AB}\to -8 \sqrt{\frac{5}{13}} & \text{cosAPC}\to \frac{\sqrt{\frac{13}{5}}}{5} & \text{cosB}\to -\frac{\sqrt{\frac{13}{5}}}{2} \\
\text{AC}\to -4 \sqrt{\frac{7}{13}} & \text{AP}\to -5 \sqrt{\frac{5}{13}} & \text{AB}\to 8 \sqrt{\frac{5}{13}} & \text{cosAPC}\to -\frac{\sqrt{\frac{13}{5}}}{5} & \text{cosB}\to \frac{\sqrt{\frac{13}{5}}}{2} \\
\text{AC}\to 4 \sqrt{\frac{7}{13}} & \text{AP}\to -5 \sqrt{\frac{5}{13}} & \text{AB}\to 8 \sqrt{\frac{5}{13}} & \text{cosAPC}\to -\frac{\sqrt{\frac{13}{5}}}{5} & \text{cosB}\to \frac{\sqrt{\frac{13}{5}}}{2} \\
\text{AC}\to -\frac{4}{\sqrt{11}} & \text{AP}\to -3 \sqrt{\frac{3}{11}} & \text{AB}\to -8 \sqrt{\frac{3}{11}} & \text{cosAPC}\to -\frac{\sqrt{\frac{11}{3}}}{3} & \text{cosB}\to -\frac{\sqrt{\frac{11}{3}}}{2} \\
\text{AC}\to \frac{4}{\sqrt{11}} & \text{AP}\to -3 \sqrt{\frac{3}{11}} & \text{AB}\to -8 \sqrt{\frac{3}{11}} & \text{cosAPC}\to -\frac{\sqrt{\frac{11}{3}}}{3} & \text{cosB}\to -\frac{\sqrt{\frac{11}{3}}}{2} \\
\text{AC}\to -\frac{4}{\sqrt{11}} & \text{AP}\to 3 \sqrt{\frac{3}{11}} & \text{AB}\to 8 \sqrt{\frac{3}{11}} & \text{cosAPC}\to \frac{\sqrt{\frac{11}{3}}}{3} & \text{cosB}\to \frac{\sqrt{\frac{11}{3}}}{2} \\
\text{AC}\to \frac{4}{\sqrt{11}} & \text{AP}\to 3 \sqrt{\frac{3}{11}} & \text{AB}\to 8 \sqrt{\frac{3}{11}} & \text{cosAPC}\to \frac{\sqrt{\frac{11}{3}}}{3} & \text{cosB}\to \frac{\sqrt{\frac{11}{3}}}{2} \\
\end{array}\]

mathematica

mathematica 发表于 2021-1-19 08:54
求解结果:
\[\begin{array}{ccccc}
\text{AC}\to 0 & \text{AP}\to 1 & \text{AB}\to 4 & \text{cos ...

1. Clear["Global`*"];
   
2. ans=FullSimplify@Solve[AC==Tan[3*B]&&AC==4*Tan[B]&&0<=B<Pi/2,{AC,B}]
   

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求解结果
\[{{AC->0,B->0},{AC->4/Sqrt[11],B->2 ArcCot[2 Sqrt[3]+Sqrt[11]]}}\]

mathematica

假设题目修改成
Rt△ABC中∠C=90°，P在BC上满足PB=5，PC=1，∠APC=5∠B，求AC的长．
那么这个题目就有点难度了，用辅助线应该很难解决了！

1. Clear["Global`*"];(*mathematica11.2,win7(64bit)Clear all variables*)
   
2. ans=Solve[ArcTan[x/1]==5*ArcTan[x/6],{x}]
   

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求解结果
\[\left\{\{x\to 0\},\left\{x\to -6 \sqrt{\frac{25}{29}-\frac{2 \sqrt{149}}{29}}\right\},\left\{x\to 6 \sqrt{\frac{25}{29}-\frac{2 \sqrt{149}}{29}}\right\}\right\}\]
数值化结果

\[\{\{x\to 0.\},\{x\to -0.853552\},\{x\to 0.853552\}\}\]