A point $P$ lies inside a circle centered at $C$ such that $CP=6$. The radius of the circle is $10$. Find the number of chords passing through $P$ which has integral length.
Attempt:
One solution is a chord having a length of $16$ units. The midpoint of this chord is $P$. Let the the chord be $AB$. Let other chords be $XY$.
Using the property of intersecting chords, $$AP\times PB=XP\times PY=64$$
(AP=8=BP)
The only possible solutions are $(8,8)$ (already taken) and €$(4,16)$ which is the diameter.
So I got $2$ as the answer. But given answer is 8. Is it because I am not considering non-integral values of $XP$ and $PY$?
Best Answer
Let's approach the problem in this way. First find the minimum length of a chord through $P.$ Using Pythagorean Theorem (the minimum-lengthed chord is perpendicular to the radius at $P$), we find this length to be $2\sqrt{10^{2} - 6^{2}} = 2 \cdot 8 = 16.$ Since this is an integer, it counts.
Next find the maximum length of a chord through $P.$ This is simply the diameter, which we know to be $20.$ Since this too is an integer, it also counts.
Notice that we can construct chords of any length between $16$ and $20$ that passes through $P.$ And for each such chord we draw through $P,$ we can reflect it across the radius through $P$ to produce a chord with equal length. This means that there are $2$ chords each of length $17,$ $18,$ and $19.$
Our final answer is $1 + 1 + 2 \cdot 3 = \boxed{8}.$