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Ring of triangles I (Posted on 2016-03-02) Difficulty: 3 of 5
ABCDEF is a hexagon and O is an interior point.

Triangle ABO is equilateral with sides of length 4.
Triangle CDO is equilateral with sides of length 3.
Triangle EFO is equilateral with sides of length 7.

Given: BC, DE, and FA are integer lengths. Find them.

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Solution computer solution -- spoiler | Comment 1 of 2
Sides bc, de and fa are respectively: 3, 6 and 7.

The angles in degree measure,

48.1896851042214 58.4118644947988 73.3984504009798

add up to the required 180.


The program uses what could be valid triangles between the given triangles and uses the law of cosines to find the total of the angles at O, which should total 180.


DefDbl A-Z
Dim crlf$, pi


Private Sub Form_Load()
 Form1.Visible = True
 
 crlf = Chr(13) + Chr(10)
 Text1.Text = ""
 
 pi = 4 * Atn(1)
 
 For bc = 2 To 6
 For de = 5 To 9
 For fa = 4 To 10
    a1 = angle(bc, 4, 3)
    a2 = angle(de, 3, 7)
    a3 = angle(fa, 4, 7)
    diff = Abs(180 - a1 - a2 - a3)
    If diff < 0.001 Then
      Text1.Text = Text1.Text & bc & Str(de) & Str(fa) & crlf
      Text1.Text = Text1.Text & a1 & Str(a2) & Str(a3) & "     " & Str(a1 + a2 + a3) & crlf & crlf
    End If
 Next
 Next
 Next


 Text1.Text = Text1.Text & crlf & " done"
  
  
End Sub

Function angle(a, b, c)
  cosa = (b * b + c * c - a * a) / (2 * b * c)
  If cosa = 0 Then angle = 90: Exit Function
  sina = Sqr(1 - cosa * cosa)
  a1 = Atn(sina / cosa) * 180 / pi
  If cosa < 0 Then a1 = a1 + 180
  angle = a1
End Function


  Posted by Charlie on 2016-03-02 15:41:57
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