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$2009$ concentric circles are drawn with radii from $1$ unit to $2009$ units. From a point on the outermost circle, tangents are drawn to the inner circles. Discover the number of tangents which will have integer measure in the problem.

My approach to the problem is that since the tangents are perpendicular to the radius at the point of contact. This means that I could find all the Pythagorean triplets with $2009$ as one measure. However, I wasn't getting anywhere with that approach.

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I will use the approach of Pythagorean triples $a^2+b^2=c^2=2009^2$. In any primitive triple ($a,b,c$ are coprime), all prime factors of $c$ are of the form $4k+1$, but $2009=7×7×41$. Thus the values of $a$ and $b$ must themselves be multiples of 49, and we get $$(49c)^2+(49d)^2=(49\cdot41)^2$$ $$c^2+d^2=41^2$$ where $c$ and $d$ are coprime. Now there are only two solutions for $(c,d)$, $(9,40)$ and $(40,9)$. Therefore the number of tangents with integer lengths is also two, and these are of lengths $49\cdot9=441$ and $49\cdot40=1960$.

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  • $\begingroup$ Right, I see where I went wrong now, thank you so much. $\endgroup$ – Prakhar Nagpal Dec 8 '17 at 9:33
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From the Tangent Secant Theorem you can say that if $t_r$ is the length of the tangent to the $r^{th}$ circle one with radius $r$, then $$t_r^2=2009\times 2r$$ So you only need to find integers, $1\le r\le2009$ such that $2009\times 2r$ is a perfect square.

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