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Let point A be a point on the circle whose equation is (x - 3)^2 + (y + 4)^2 = 50.
How many different points X lie on the circle so that the chord AX has integer length?
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The given circle has the radius of and the diameter of = .
Let us draw the diameter of the circle through the point A and through the center of the circle .
Let B be the point diametrically opposite to the point A (i.e. B is the other endpoint of the diameter).
Then, if X is the point on the circle different of the points A and B, the triangle AXB is a right-angled triangle.
Now, if we translate the geometry situation to the Algebra language, then this translation sounds like this:
Find all integer positive solutions "x" and "y" to the equation
= 200,
where "x" and "y" represent the lengths of the legs AX and BX.
It is not difficult to check directly that these solutions are
(x,y) = (2,14) and
(x,y) = (10,10).
Of them, one solution (x,y) = (2,14) corresponds to 4 different points X on the circle.
The other solution (x,y) = (10,10) corresponds to 2 different points X on the circle.
In all, there are 6 = 4 + 2 such points X on the circle.
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Comment from student: Did you see the other tutor's answer? It's really 28.
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My responce: Shortly speaking, Edwin is right. I am not.
Slightly more extended explanation is this:
When reading the problem, I mistakenly put into my mind another formulation:
Find the points X such that BOTH CHORDS AX and BX have integer lengths.
In other words, I solved different problem.
In reality, if to interpret the problem as is, as it is written, then it is TRULY TRIVIAL.
The correct answer is 28.
Thanks to Edwin for noticing my mistake.
Thanks to you for your feedback.
After couple of days I, probably, will delete my solution in order for do not confuse future readers/visitors.
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 . 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 and the diameter is
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