圆内两条弦AC=4和BD=3，AB与CD交于圆内P点，∠APC=60°，求圆的直径。

天山草@

下图中，已知两条弦 AC=4 和 BD=3，AB 与 CD 交于圆内 P 点，∠APC=60°，求圆的直径 R=?





此问题摘自【数学研发】网站。

ccmmjj

这题简单。它等价于“已知三角形两边分别等于3，4。这两边夹角为120度，求三角形外接圆直径。”
这是简单的解三角形题目。

ccmmjj

没人算？直径等于(2√111)/3

王守恩

批量生产。
\(2R=\frac{n}{\sin(\theta)}=\frac{m}{\sin(60°-\theta)}\)
\(R=\sqrt{\frac{(n+m)^2-n\ m}{3}}\)

ysr

各位老师好！天山草@老师老师您好！谁能登录数学研发论坛？请您帮忙问一下高手如下程序哪里错了？是我在网上抄录的vb程序。我被数学研发论坛禁言好几年了，不能在该论坛发言讨论了，请求帮忙，谢谢您！

Private Sub Command1_Click()
Dim xr() As Double, a As String
a = Trim(Text1)
ReDim xr(0 To Len(a) - 1)
For i1 = 0 To Len(a) - 1
xr(i1) = Mid(a, i1 + 1, 1)
  Next
Dim l As Long, le As Long, le1 As Long, n As Long, r As Long, p As Long, q As Long, m As Byte
Dim wr As Double, w1 As Double, wlr As Double, wl1 As Double, tr As Double, t1 As Double
Dim pi As Double, t As Double
Dim xi()
n = Len(a) '求数组大小，其值必须是2的幂
m = 0
l = 2
pi = 3.14159265358979
Do
l = l + l
m = m + 1
Loop Until l > n
n = l / 2
ReDim xi(n - 1)

l = 1
Do
  le = 2 ^ l
  le1 = le / 2
  wr = 1
  wi = 0
  t = pi / le1
  w1r = Cos(t)
  w1i = -Sin(t)
  r = 0
Do
  p = r
  Do
   q = p + le1
   
   tr = xr(q) * wr
   ti = xr(q) * wi
   
   xr(q) = xr(p) - tr
   xi(q) = xi(p) - ti
   xr(p) = xr(p) + tr
   xi(p) = xi(p) + ti
   p = p + le
Loop Until p > n - 1

wr = w1r
wi = w1i
r = r + 1
Loop Until r > le1 - 1
l = l + 1
Loop Until l > m

For i = 0 To n - 1 '仅输出模
   
   Text2 = Text2 & "  " & xr(i) & "+" & xi(i) & "i"
   Next
End Sub

Private Sub Command2_Click()
Text1 = ""
Text2 = ""
End Sub


输入：Text1 =80607000，结果：
21+0i  12.9497474683058+-10.9497474683058i  6.94974746830584+-4.94974746830583i  12.9497474683058+1.05025253169417i
  7+0i  3.05025253169417+-1.05025253169417i  -2.94974746830584+4.94974746830583i  3.05025253169415+10.9497474683058i
与正确值比较：12.9+10.9i, 2+7i, 3.1-1.1i, 7, 3.1+1.1i, 2-7i, 12.9-10.9i, 21也是不对。

ysr

这个程序就是快速傅立叶变换，将多项式系数表示法快速变换为点值表示法！
谢谢！我换不了，都能被封号。谢谢老师！