Strike a chord

Suppose in general the two circles have radii $k$ and $2k$ . Center them both at the origin $O$ . Now without loss of generality, once the first random point is chosen, we can rotate the larger circle so that this point sits at $A(0,2k)$ . Now draw a chord $AB$ (down and to the right) that is tangent to the smaller circle at $P$ . Then $\Delta OPA$ is right-angled at $P$ with hypotenuse length $|OA| = 2k$ and adjacent leg length $|OP| = k$ . As $\cos(\angle AOP) = \dfrac{k}{2k} = \dfrac{1}{2}$ we see that $\angle AOP = 60^{\circ}$ , and so by symmetry $\angle AOB = 120^{\circ}$ .

Thus if the second point $B$ is chosen so that $\angle AOP \lt 120^{\circ}$ then the chord $AB$ will not intersect the smaller circle, and if $120^{\circ} \le \angle AOP \le 180^{\circ}$ the chord will intersect the smaller circle at at least one point. By symmetry the same result will occur if the chord is drawn down and to the left. In each case there is a $\dfrac{180 - 120}{180} = \dfrac{1}{3}$ probability that the chord will not intersect the smaller circle, so since there is a one-to-one mapping between the random choice of $B$ and $\angle AOB$ we can conclude that $P = \dfrac{1}{3}$ , and that $\lfloor 1000P \rfloor = \boxed{333}$ .

Not perfectly drawn

Use trigonometry to calculate the greatest angle $(A)$ at which a $\color{#D61F06}chord$ - whose 1st point starts at the base of a centre line - will intersect the smaller circle, relative to the centre. I.e. $\color{#20A900}2\pi$ is the hypotenuse, and $\color{#3D99F6}\pi$ is the opposite edge of the angle we are looking for (which represents the intersecting range for one half of the circles).

$\large \frac{\pi}{2\pi}$ $= 0.5$ $= \sin{A}$

$A = 30$

There are 90 degrees of possibilities, in either direction, for a potential position of the 2nd point on the circumference of the larger circle, ranging from the top of the centre line to a position that is infinitely close to the 1st point (at the base of the centre line).

As both halves of both circles have the same proportions, we don't need to unnecessarily extend the calculation for the probability $(P)$ for this problem beyond 90 degrees. Both sides are equivalent, and we already have the intersecting range that would apply to the other half (30 degrees).

$P = \large\frac{30}{90} = \frac{1}{3}$

The greatest integer equal to or less than $1000P$

$\large \lfloor\frac {1000}{3}\rfloor = \boxed{333}$

The region of interest is the one between the green points in the diagram. First, find out the slopes of the tangent lines which intersect the smaller circle. Equation of a tangent line is below:

$\large{y = m(x - 2\pi)}$

The equation of the lesser circle is:

$\large{x^2 + y^2 = \pi^2}$

Plug the first equation into the second:

$\large{x^2 + m^2(x^2 - 4 \pi x + 4 \pi^2) = \pi^2 \\ x^2(1+m^2) +x(-4 \pi m^2) + 4 \pi^2 m^2 - \pi^2 = 0}$

This is a quadratic equation in x, for which there are generally two solutions. However, we are particularly interested in the case in which there is only one solution. Use the usual $(b^2 - 4ac = 0)$ trick to get:

$\large{16 \pi^2 m^4 - 4(1+m^2)(4\pi^2 m^2 - \pi^2) = 0}$

Solving this equation yields $\large{m = \pm \frac{1}{\sqrt{3}}}$ . Now we need to see where this line intersects the larger circle. Plugging the $m$ value into the equation for the larger circle gives:

$\large{x^2 + \frac{1}{3}(x - 2\pi)^2 = 4 \pi^2}$

Solving this equation yields $\large{x = - \pi}$ . By inspection, when the $x$ coordinate has a value of $- \frac{R}{2}$ , this corresponds to $\theta = \pm 120^\circ$ . There is an angular separation of 120 degrees between these two angles, meaning that the region of interest occupies exactly one third of the entire angular space.

$\large{\lfloor 1000/3 \rfloor = \boxed{333}}$

Similar to @Steven Chase 's solution:

Let the $2$ circles be centered at $O$ . Once the first point $B$ is chosen, the second point has to lie in the minor arc $AC$ for the chord to intersect the small circle.The probability of the point lying on the arc is given by,

$P=\dfrac{\text{Length of minor arc AC}}{\text{Perimeter of the large circle}}=\dfrac{R(\angle{AOC})}{R\cdot2\pi}=\dfrac{\angle{AOC}}{2\pi}$

Consider the $\Delta OEB$ right angled at $E$

$\angle {OBE}=\sin^{-1}{\dfrac{r}{R}}$

by symmetry we have,

$\angle{ABC}=\angle{EBF}=2\angle{OBE}\\ \angle{ABC}=2\sin^{-1}{\dfrac{r}{R}}\\ \angle{AOC}=2\angle{ABC}\hspace{7mm}\color{#3D99F6}\small\text{Angle subtended by an arc at the center is double the angle subtended by it on any other point on the circle}\\ \angle{AOC}=4\sin^{-1}{\dfrac{r}{R}}$

thus the required probability is,

$P=\dfrac{4\sin^{-1}{\dfrac{r}{R}}}{2\pi}$

In the given problem $r=\pi$ and $R=2\pi$ , thus $\dfrac{r}{R}=\dfrac{\pi}{2\pi}=\dfrac{1}{2}$

$P=\dfrac{4\sin^{-1}{\dfrac{1}{2}}}{2\pi}\\ =\dfrac{1}{3}\\ \lfloor {1000P}\rfloor=\boxed{333}$

Let $A$ and $Z$ be the points picked. Rotate the figure so that $A$ is at the top. $Z$ remains uniformly distributed around the circle.

Let $AB$ be a chord that is tangent to the inner circle. Draw the diameter $CD$ through the point of tangency $E$ . Obviously $CE = 3\pi$ and $ED = \pi$ . By the intersecting chord theorem, $AE = \sqrt{3} \pi$ , therefore $AEO$ is a 90-60-30 triangle. So is $AFO$ by the same argument. Therefore $\angle BAC = 60^\circ$ , hence $\angle BOC = 120^\circ$ .

$AZ$ intersects the inner circle whenever $Z$ is on the minor arc $BC$ . Since $\angle BOC = 120^\circ$ , this is $\frac13$ of the circle, therefore $\boxed{P = \frac13}$ .

I used 2-dimensional geometric probability by angles. The central angle TOT1 is 120 degrees. P=120/360=1/3 1000xP=1000/3 The answer is 333.

This was weird, because the solution came to me intuitively and I barely know how to justify it. It is based on the Euler-inequality, that states that for a triangle's circumradius R, and inradius r, $R \geq 2r$ and it is equal if and only if it's an equilateral triangle. This last part was something that seemed really obvious, and it pops out when you actually take a random point on the outer circle and draw the equilateral triangle that touches the inner circle. Since this triangle divides the outer circle in three equal parts, all the lines that don't cross the triangle, and thus the second chosen point is covered by the triangle from the first one's viewpoint, are the ones that also intersect the circle. These are the points in one third of the outer circle, thus 1000*1/3 floored, 333.

The radii of the circles $R$ and $r$ do not matter much; only the ratio $r/R = 1/2$ is relevant here.

If both points are chosen uniformly on the circle, the arc distance $\theta$ between them will be uniformly distributed on $[0^\circ, 180^\circ]$ .

The chord between the points reaches it smallest distance to the center of the circle at its midpoint. The distance between that midpoint and the center of the circle is well-known to be $d = R \cos \frac\theta 2.$ (For proof, connect the two points and the midpoint to the center of the circle, and analyze one of the two right triangles.)

We require $d \leq r$ , which means $R \cos \frac\theta 2 \leq \frac R 2, \\ \cos \frac\theta 2 \leq \frac 12, \\ \frac\theta 2 \geq 60^\circ, \\ \theta \geq 120^\circ.$ Thus we find that $\theta \in [120^\circ,180^\circ]$ , which is one-third of the interval.

Conclusion: $P = 1/3$ and $\lfloor 1000P \rfloor = \boxed{333}$ .

Points on the larger circle are uniquely definded by $\theta_1,\theta_2\in[0,2\pi]$ , if the two circles are centered in $O$ in the $xy$ plane. Given the first point $P_1$ on the bigger circle, it's proven, with a bit of trigonometry, that the angle between the two lines passing from $P_1$ to the tangent points on the smaller circle is $\frac{\pi}{3}$ . Now, the two tangent lines also intersect the bigger circle, creating an arc 'containing' all the points $P_2$ that, if connected to $P_1$ , create chords that intersect the smaller circle. Due to the fact that the probability density function is $\frac{1}{4\pi^2}$ ( $\int_{0}^{2\pi} \int_{0}^{2\pi} \frac{1}{4\pi^2} d\theta_1 d\theta_2=1)$ and that the arc is $\frac{2\pi}{3}$ (I omit the proof), the probability can be calculated as

$\displaystyle \mathbb{P}= \int_{0}^{2\pi} \int_{0}^{\frac{2\pi}{3}} \frac{1}{4\pi^2} d\theta_2 d\theta_1=\frac{1}{3} \approx 0.333$

Solution area angle must be between 0 and 60. Therefore the arc can be 2pi/3 of the circle; which is 1/3 ==> 0.3333... probability.

Call the first point A, 2nd point B, 3rd point C and the center D. The 2 random points formed in the outer circle of radius 2r form an angle in the interval [0, 360] with the center of the two circles. In order for the 2 points to cross the smaller circle of radius r, the second point must be between the 2 tangent lines between the inner circle and the first point. We can draw line AB and line AC. We know the line creates a right angle with the point of intersection, center and the first point. We see the hypotenuse is the bigger radius 2r and one of the legs is the smaller radius r. This is the ratio of a 30-60-90 triangle, therefore we know that the 2 angle formed is $60 \cdot 2$ and $60 \cdot 4$ , or $120$ and $240$ . This is an interval of size 120, therefore the probability is 120/360 or 1/3. multiplying by 1000 and taking the floor, we get $\boxed{333}$ .