How many darts do you need to throw?

We're looking for the maximum number of darts, so they need to be as close as possible from each other (r). It is possible that 6 darts hit the edge and one hits the center keeping distance r from each other, because that's the configuration of a hexagon (6 equilateral triangles). So the 8th will be closer.

The best way to see that the eighth dart will land within a distance smaller than the radius from some other dart is this: imagine that each dart is the center of a circle of radius $\frac{r}{2}$ . For the darts to be in the circle, the circles must be inside a circle of radius $r+\frac{r}{2}$ concentric to the first, and if we want them to be separated by a distance bigger than $r$ , the little circles must be exterior to each other. As it isn't possible to fit more than seven circles of radius $\frac{r}{2}$ inside one of radius $r+\frac{r}{2}$ , the minimum number of darts to "pack" the dartboard is seven as shown above, with the 8th dart guaranteeing that the distance between a pair of darts is less than $r$ .

Note To be more convinced, you can check this Wikipedia article on parking circles in a circle .

in the worst case scenario the darts hit the border of the target - there should be 6 darts when the distance between the darts is equal to R. Only the dart in the middle of the target will be R way from each of other darts. If any of the 6 dart is not at the border of the target then the distance between this dart and the 7th one in the middle is less than R.. So the answer is 7 darts

The answer is less than ten. Think of the problem as placing circles radius $r/2$ within a circle of radius $3r/2$ without overlap. Simply comparing the areas gives a 9:1 ratio; but it is clear that there will be holes in the covering, no matter how it is done.

The answer is at least eight. Placing six darts equidistant at the edge of the dartboard results in minimal distance of $r$ between neighbors, and there is precisely one point left for a second dart, at distance $r$ from all of them, in the center.

This gives $8 \leq x \leq 9$ . Proving that $x \not = 9$ is more challenging.

The easiest method of constructing a hexagon is to draw a circle and use the compass to mark 6 equispaced points on the circumference. If an arc is drawn in the construction this immediately gives the 7 locations that the initial darts have to hit and the area that is excluded for subsequent darts. This gives the minimum number of darts required as 8.

The distance between darts will be maximized when they form a regular polygon inscribed on the circle of the dartboard. The next dart can always be placed at a distance of r from the first set of darts, if each of the first darts are placed on the circle itself at the vertices of the polygon and the final dart is in the center. So the task becomes finding the polygon whose side length is less than r, when inscribed on the circle with radius r. Because of its symmetry, a hexagon will have radius r and side length r, and the n=7th dart will be in the center, a distance of r from each vertex. However, a regular heptagon will have side length of approximately 0.86r, making the task impossible with n=8 darts.

You can throw a dot at the center and another one at the perimeter of the board, that make it two. Now you can start throwing them all around the circle at a distance R from the center and also from previous thrown dart on the perimeter of the board. So the we simply have to find the number of R's in the perimeter of a circle I.e, 2πR / R = 2π . which is approximately to 6. 32 plus the dart at the center. That makes the total to 7 darts. Now the eighth dart thrown will be closer than R to atleast any two darts and hence the answer.