Question 1059003
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This is a trick question.  The answer is 28.  What Iklyn solved 
instead was this problem:

Let A and B be endpoints of a diameter of a circle whose diameter 
has length {{{2sqrt(50)=10sqrt(2)}}}. How many points on the circumference of the 
circle are such that right triangle AXB has legs AX and BX whose 
lengths are both integers.

But that's not the given problem.

In solving the given problem, there is nothing to calculate but 
the diameter of the circle.  

The diameter of ANY circle is the longest possible chord of that 
circle.

Chords of ANY length greater than 0 and less than or equal to the
diameter of ANY circle can be drawn from ANY point on that circle.

Here the radius is {{{sqrt(50)}}} and the diameter is {{{2sqrt(50)}}} 
which is approximately 14.14 

The chords must be of integer length.  Therefore chords of lengths

 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14

can all be drawn from any point on that circle each in 2 ways,
clockwise or counter-clockwise.

so the answer is 28.

  :)

Edwin</pre></font></b>