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发表于 2025-2-27 09:24
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本帖最后由 天山草 于 2025-2-27 09:27 编辑
主楼中的两个关于 R 的方程,当给定 AH、BH、CH 后,对于钝角三角形可能解出两个正根,还需要进行判别舍弃其中一个正根,因而解方程的方法并不太实用。
下面给出一个不用解方程,用 mathematica 软件计算三角形各边长、外接圆半径、内切圆半径的程序,程序也相当于给出了计算公式。
当已知量 AH、BH、CH 都化为有理数输入时,可达到任意计算精度。
- Clear["Global`*"];(*不解方程,给定 AH、BH、CH,求边长、外接圆半径、内切圆半径的程序*)
- AH = 0.7; BH = 1.1; CH = 1.5;(*小数,无法算到任意精度*)
- AH = 7/10; BH = 11/10; CH = 15/10;(*有理数,可算到任意精度*)
- u = AH^2 + BH^2 + CH^2; v = AH^2 BH^2 CH^2;
- w1 = Power[-(u^3/27) + 2 v -
- Sqrt[(2 v - u^3/27)^2 - u^6/729], (3)^-1]; w2 = Power[-(u^3/27) +
- 2 v + Sqrt[(2 v - u^3/27)^2 - u^6/729], (3)^-1];
- t1 = (2 u)/3 + w1 + w2;
- t2 = 1/6 (4 u + 3 I (I + Sqrt[3]) w1 - 3 I (-I + Sqrt[3]) w2);
- If[N[Re[t1] > AH^2 && Re[t1] > BH^2 && Re[t1] > CH^2],
- Print["a1 = ", Sqrt[N[Re[t1], 100] - AH^2]] &&
- Print["b1 = ", Sqrt[N[Re[t1], 100] - BH^2]] &&
- Print["c1 = ", Sqrt[N[Re[t1], 100] - CH^2]]];
- If[N[Re[t1] > AH^2 && Re[t1] > BH^2 && Re[t1] > CH^2],
- a1 = Sqrt[N[Re[t1], 100] - AH^2] ;
- b1 = Sqrt[N[Re[t1], 100] - BH^2]; c1 = Sqrt[N[Re[t1], 100] - CH^2];
- p1 = (a1 + b1 + c1)/2;
- Print["R1 = ", a1 b1 c1/4/Sqrt[p1 (p1 - a1) (p1 - b1) (p1 - c1)] ]];
- If[N[Re[t1] > AH^2 && Re[t1] > BH^2 && Re[t1] > CH^2],
- a1 = Sqrt[N[Re[t1], 100] - AH^2] ;
- b1 = Sqrt[N[Re[t1], 100] - BH^2]; c1 = Sqrt[N[Re[t1], 100] - CH^2];
- p1 = (a1 + b1 + c1)/2;
- Print["r1 = ", Sqrt[((p1 - a1) (p1 - b1) (p1 - c1))/p1] ]];
- Print["。。。。。。。。。。。。。。 "];
- If[N[Re[t2] > AH^2 && Re[t2] > BH^2 && Re[t2] > CH^2],
- Print["a2 = ", Sqrt[N[Re[t2], 100] - AH^2]] &&
- Print["b2 = ", Sqrt[N[Re[t2], 100] - BH^2]] &&
- Print["c2 = ", Sqrt[N[Re[t2], 100] - CH^2]]];
- If[N[Re[t2] > AH^2 && Re[t2] > BH^2 && Re[t2] > CH^2],
- a2 = Sqrt[N[Re[t2], 100] - AH^2] ;
- b2 = Sqrt[N[Re[t2], 100] - BH^2]; c2 = Sqrt[N[Re[t2], 100] - CH^2];
- p2 = (a2 + b2 + c2)/2;
- Print["R2 = ", a2 b2 c2/4/Sqrt[p2 (p2 - a2) (p2 - b2) (p2 - c2)] ]];
- If[N[Re[t2] > AH^2 && Re[t2] > BH^2 && Re[t2] > CH^2],
- a2 = Sqrt[N[Re[t2], 100] - AH^2] ;
- b2 = Sqrt[N[Re[t2], 100] - BH^2]; c2 = Sqrt[N[Re[t2], 100] - CH^2];
- p2 = (a2 + b2 + c2)/2;
- Print["r2 = ", Sqrt[((p2 - a2) (p2 - b2) (p2 - c2))/p2] ]];
复制代码
在上面的例子中,给定 AH = 0.7; BH = 1.1; CH = 1.5; 按 AH = 7/10; BH = 11/10; CH = 15/10; 化为有理数输入,计算精度为 100 位的结果是:
a1 = 2.120059563663967377742424850341356237336932802640686722401888290426853014029935336920632968373456083....
b1 = 1.942846507957602848069348736674465426972174318385771770681211925121813371259398979904206151114691419....
c1 = 1.653678491570550646873306568153420921307077526295614026349241730524120403027783877224738987442955181....
R1 = 1.116316773308886072616698531505218038556770139742717759697302904379461299508024170809850051037441176....
r1 = 0.533683226691113927383301468494781961443229860257282240302697095620538700491975829190149948962558824....。
a2 = 1.4122564007156504478701407285719650452733289280277422556017498520880501745207773980803956807398757634....
b2 = 1.128923443534734932225674111298961535882002582503426926564853106110144715578697348745554627918694950....
c2 = 0.484219104706045082381343974023271090343795084628634765315337629255123047436019662936144775933350162....
R2 = 0.78810978634996086610533636330979167357865438548781731953760981922833473031838881534245939573864999....
r2 = 0.161890213650039133894663636690208326421345614512182680462390180771665269681611184657540604261350013....。
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