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Two Digit Diameter Trial (Posted on 2016-07-11) Difficulty: 3 of 5
Consider a circle, where the length of diameter PQ is a 2-digit integer. Reversing the digits of PQ one obtains the length of a perpendicular chord RS.

PQ intersects RS at the point T. The center of the circle is denoted by O and it is known that the length of TO is a positive rational number.

Determine the length PQ.

No Solution Yet Submitted by K Sengupta    
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Solution computer solution (spoiler) | Comment 1 of 2
DefDbl A-Z
Dim crlf$


Private Sub Form_Load()
 Form1.Visible = True
 
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)
 
 For a = 1 To 9
  For b = 0 To a - 1
    ab = 10 * a + b
    ba = 10 * b + a
    sq = ab * ab - ba * ba
    sr = Int(Sqr(sq) + 0.5)
    If sr * sr = sq Then
      Text1.Text = Text1.Text & ab & Str(ba) & "     " & sq & Str(sr) & crlf
    End If
    DoEvents
  Next
 Next
 
 
 Text1.Text = Text1.Text & crlf & " done"
  
End Sub

finds

65 56     1089 33

giving the answer as PQ has length 65. The chord RS has length 56 and the length of OT is 33/2.

This is the result of segment OR (or OS) being a radius of the circle (half the length of PQ) and the hypotenuse of a right triangle with legs being to TO and half the length of RS. The halvings require the halving of the 33 result.

The important consideration is that in working with the Pythagorean result sqrt(ab^2/4 - ba^2/4) (the ab and ba being concatenations, not multiplications), the difference of the numerators is an integer which must be a square as no multiplication by a square will turn a non-square into a square or the square of a rational.

  Posted by Charlie on 2016-07-11 15:31:47
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