Three Concentric Circles

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See the above image. $A$ , $B$ and $C$ are three concentric circles with center $O$ and radii $a,b,c$ respectively such that $a>b>c$ .As given in the question, $\text{chord}~PQ=2$ of $\text{circle}~B$ is tangential to $\text{circle}~C$ at point $G$ .And $\text{chord}~MN=x$ of $\text{circle}~A$ is tangential to $\text{circle}~B$ at $\text{point}~D$ .

As we know,tangents form right angles with respect to the radius passing through the tangent point.Since $PQ$ is tangential to $\text{circle}~C$ at $\text{point}~G$ and $OG$ is the radius passing through the tangent point,we get $\angle OGQ = 90^{\circ}$ .And it's also known that in a circle,the perpendicular from the center to any chord bisects the chord.Since $O$ is also the center of $\text{circle}~B$ and $OG\perp \text{chord}~PQ$ ,we get $GQ=\dfrac{PQ}{2}=\dfrac{2}{2} = 1$ .So we get $\Delta OGQ$ is a right triangle.In the same way,we get $MD=\dfrac{1}{2} MN = \dfrac{x}{2}$ and $\Delta ODM$ is a right triangle.

From the Pythagorean theorem,we get the followings...

$OQ^2=OG^2+GQ^2~~~~~ \Longrightarrow ~~b^2=c^2+1 ~~~~~ \dots \dots\dots (*)$

$OM^2=OD^2+MD^2~~~~~\Longrightarrow ~~a^2=b^2+ \left(\dfrac{x}{2}\right) = c^2+1+\dfrac{x^2}{4} ~~~~~ \dots\dots\dots (**)$

OK,now we'll need another piece of information given in the question.The area between $\text{circle}~A$ and $\text{circle}~B$ is $2014$ times the area between $\text{circle}~B$ and $\text{circle}~C$ .So we get...

$(\text{Area of circle}~A)-(\text{Area of circle}~B) = 2014\cdot \left((\text{Area of circle}~B) - (\text{Area of circle}~C)\right)$

$\Longrightarrow ~~~ \pi a^2 - \pi b^2 = 2014\left(\pi b^2 - \pi c^2 \right)~~~~~ \Longrightarrow ~~~ \pi (a^2-b^2) = 2014 \cdot \pi (b^2-c^2)$

$\Longrightarrow ~~~ a^2-b^2 = 2014(b^2-c^2)$

Plugging in the values from $(*)$ and $(**)$ ,we have...

$c^2+1+\dfrac{x^2}{4} -c^2-1=2014(c^2+1-c^2)$

$\Longrightarrow ~~~ \dfrac{x^2}{4}=2014$

$\therefore ~~~ x=\sqrt{2014\times 4}=\sqrt{8056}=\sqrt{a}$

Therefore: $a=8\fbox{056}$

The area between two circles $\alpha$ and $\beta$ , in function of a chord of $\; \alpha$ tangent to $\beta$ is \begin{align} \dfrac{\pi \cdot c^2}{4} \end{align} where $\; c \;$ is the length of the chord.

From the problem, we extract that $\frac{\pi \cdot x^2}{4} = 2014 \cdot \frac{\pi \cdot 2^2}{4}$ , which yields $x^2 = 8056$ . The last three digits of it are $056$ , or simply $\boxed{56.}$

Let the radius of Circle A,Circle B and Circle C be a,b,c respectively.The difference between area of Circle B and Circle C is pi(b^2-c^2),Now note that the radius of B and C are hypotenuse and a side of a right triangle which includes the half of chord of B,So c^2+1^2=b^2 or b^2-c^2=1,so the difference between the areas of circle B and C is pi(b^2-c^2)=pi * 1=pi.The same way in Circle A and B,the area difference is pi(a^2-b^2).Now it also forms a right triangle which includes the half of chord x,so (x/2)^2+b^2=a^2 .So area difference between Circle A and Circle B is pi(a^2-b^2)=pi(x/2)^2,so according to the question pi(x/2)^2=2014pi or (x/2)^2=2014 or x^2/4=2014 or x^2=2014*4 or x^2=8056 or x=sqrt 8056,now last three digits is 056(ANSWER)

This problem can easily be solved by holditch's theorem which is Let a chord of constant length be slid around a smooth, closed, convex curve $C$ , and choose a point on the chord which divides it into segments of lengths p & q. This point will trace out a new closed curve $C'$ , as illustrated above. Provided certain conditions are met, the area between $C$ and $C'$ is given by $\pi$ pq

let the point of tangency of circle $C$ with chord in circle $B$ be $M$ . and point of tangency of circle $B$ with chord in circle $A$ be N .

It is clear that $M$ & $N$ bisect the chords . the area between circle $C$ and circle $B$ by the holditch's theorem is $\pi$ as M divides it into two each of 1 length .

the area between $A$ and $B$ from given is $2014\pi$ and by holditch's theorem it is $\frac{x}{2}\cdot\frac{x}{2}\cdot\pi=\frac{x^2\pi}{4}$ equating both areas we get $\frac{x^2\pi}{4}=2014\pi$ $\Rightarrow x^2=4\times2014$ $x=\sqrt{8056}$ hence , we get our answer as $\boxed{056}$