A preview: Can you solve this?

To end in a pocket, the ball must travel an integer number of table widths and an integer number of table heights at the same time . Since $67$ and $100$ are coprime, the first time this happens is when the ball has travelled $67$ table widths and $100$ table heights.

At this point, the ball will have crossed the table horizontally an odd number of times, and so will be at the right-hand edge. It will also have crossed the table vertically an even number of times, and so will be at the bottom edge. Thus it will be in the bottom-right corner, and go into pocket $D$ .

In order to find the intersection points,instead of reflect the line, we can "reflect" the rectangles.

Thanks to Rohit Upadhyaya,I found the mistake in my origin solution, the "reflection" times are wrong. Sorry for the confusion caused by the diagram of mine.

Precisely speaking, the "reflection" times in horizontal direction is 67- 1 = 66, and the vertical direction the "reflection" time is 100-1 = 99. When the "reflection" times in horizontal direction is EVEN, so the Line AB still stand at the left side of the rectangle, the "reflection" times in vertical direction is ODD, so the Line BD will at the top side of the rectangle. So the answer is D.

I made a simpler version diagram of the this question, a 4x3 rectangle situation, hope it will help.

Hopefully this is right. Here it goes:

The 45 degree angle is very convenient, since it partitions the speed equally between the vertical and the horizontal. We therefore know that at any given time, the total distance traveled in the horizontal direction equals the total distance traveled in the vertical direction. Suppose that the ball will land inside a pocket. This means that the travel distance in either dimension must simultaneously be an integer multiple of 67 and an integer multiple of 100. The smallest number which satisfies this condition is 6700, which is the LCM of 67 and 100.

6700 = 67 x 100, which is an odd multiple of 100
6700 = 100 x 67, which is an even multiple of 67

Now we need to associate the sides with either even or odd multiples. The top side (AC) corresponds to odd multiples of 67, and the bottom side (BD) corresponds to even multiples of 67. The left side (BA) corresponds to even multiples of 100, and the right side (DC) corresponds to odd multiples of 100.

Since we have an odd multiple of 100, and an even multiple of 67, the pocket must be on the right side (DC) and the bottom side (BD).

We therefore have Pocket D as the answer.

Imagine the pool table as a Cartesian grid, with pocket B as the origin. Since the ball leaves (0,0) at a 45 degree angle, it is traveling along a line with slope 1. If we know (from physics) that the ball will bounce off any wall in such a way that the angle its path makes with that wall will be preserved, then we can say that the ball will also leave any wall it bounces off of at at 45 degree angle. This implies that the ball can either be traveling along a line with slope 1 or slope -1. In the first case, it is gaining or losing horizontal and vertical distance equally; in the second, horizontal distance and vertical distance are being converted into one another.

This gives us four types of motion the ball can be making at any time. After some increment of time along a given path, the ball's position will have changed by some multiple of either (1,1) or (-1,-1), if on a line of slope 1, and (1,-1) or (-1,1) if on a line of slope -1.

But this means that for integer coordinate pairs, the parity of the sum of the coordinates remains constant. In other words, adding a multiple of (1,1), (-1,-1), (1,-1), or (-1,1) to a point will give you a point with coordinates whose sum will be even if the starting point's sum was even, and odd if the starting point's sum was odd.

Since we start at (0,0), which has an even sum, the odd sum points (0,67) and (100,67) are unreachable. This rules out pockets A and C.

We also know that since the only path out of (0,0) that is on the table (under these rules), is in the direction (1,1), the only way into (0,0) would be the opposite direction (-1,-1). This means that the ball would have to reverse its path somewhere, which from angle preserving property of collisions we know would only be possible if it precisely hit another corner. Since that would end the path, we rule out pocket B.

Therefore, if the ball goes into a pocket it has to go into pocket D.

But before we can say it definitely goes into pocket D, we have to consider the possibility that "The ball will never land in a pocket."

But if it never lands in a pocket, it travels for an infinite distance on the table. We can rule out the possibility of the ball getting into an infinite loop by noticing that if at any point we reversed the balls motion (such as by placing a wall in its path perpendicular to its motion), it would trace out the exact same path on the table, except in to opposite direction. So it can't get into a loop because there's no way to reverse a loop that gets the ball to end up back at point (0,0). A loop like that would have to have two different histories of where the ball had been, which doesn't make sense here.

We also can rule out any way for the ball to travel an infinite distance without looping, because if we consider the ball as effectively moving alongs integer coordinate pairs to integer coordinate pairs along a path (1,1), (-1,-1), (1,-1), or (-1,1), then there are a finite number of states the ball-table system can actually be in (something like 68*101 points times 4 directions divided by 2 for the parity constraint = 13,736 distinct states as an upper bound). So the ball can't travel an infinite distance without the whole system repeating itself at some point. This seems to rule out the possibility that "The ball will never land in a pocket."

Therefore the ball lands in pocket D.

Imagine the table is on a grid, where pocket 'B' is at point (0,0), 'A' at (0,67), 'C' at (100,67), and 'D' at (100,0). Define even (odd) points as points where the sum of coordinates is even (odd). Among pockets 'B' and 'D' are even; 'A' and 'C' are odd. Since the ball moves at a 45 degree angle, every side point (i.e. not a pocket) is connected to exactly two other points (regardless of direction); pockets are connected to only one other point. Also, note that 'even' points only connect to 'even' points, and 'odd' to 'odd' (again b/c of 45 degree angle). Therefore, a path starting at 'B' (even) cannot end up in 'A' or 'C' (odd). Also, since no point reflects back, once a point has been hit, it can't be hit again, unless it's in a loop. The possibility for loops, however, can be ignored b/c a loop cannot contain a pocket. This means the path can't go reverse, and so a ball starting at 'B' can never come back to 'B'. Since there are a finite number of points, the path must end somewhere, and the only place it can end (by elimination) is 'D'.

Using a ruler, the ball will hit around the 3rd dot from the bottom right corner, then it will move 90 degrees to a line parallel to the 2nd line, then it will hit around the 1st dot down from the top left corner, then to the 1st dot to the right of that corner, which will aim it straight at Pocket D. Another way to figure it out is that every time it hits a wall, it will form triangles that get smaller to the point where it will go into Pocket D.

Pretty new to this website, (and math puzzles) so forgive me if my answer is all over the place. My method got me the correct solution, but I'm not sure if my reasoning was legitimate, or I was just lucky. Here we go: After a minimum of three bounces off of the walls of the table, a polygon can be formed by connecting each point where the ball hits the wall. In other words, after three bounces, connect the points where the ball hits the wall between pockets AC, CD, and BD to form a triangle. After four bounces, a four-sided polygon can be formed, this time using an additional point between AB, etc.

For every bounce, the sum of the internal angles of that polygon is 180(B-2), where B>/ 3, and equals the number of bounces (not including the 45 degree initial angle). When the ball lands in a pocket, another 45 degree angle is formed between the ball and the pocket, and we have a closed figure. As each individual angle formed at a bounce is 90 degrees (assuming equal angles before and after the bounce), then the sum of the interior angles of the polygon will be 45+90B+45, where the two 45 degree angles are from the initial and final pockets.

Finally, to make a long story short, set 180(B-2)=90B+90, B=5 bounces before landing in a pocket. Starting at pocket B, the ball goes to AC (bounce one), CD (bounce two), BD (bounce three), AB (bounce four), AC (bounce five), and, lands in pocket D. Let me know what you guys think!

Let (x, y) denote the ball's current position on the coordinate plane. Note that the x direction and y direction are both increasing at either the exact same rate or their inverse—that is, y=±x because it is a 45 degree angle. Also notice that for a ball to land in any hole, ball must have either 0 or 100 units as the x-coordinate, equivalent to 0 (mod 100). Similarly, the y-coordinate of the ball must be either 0 or 67, both of which are 0 (mod 67). Thus, x must be 0 (mod 100) and y must be 0 (mod 67). lcm(67,100) = 6700, so the ball goes back and forth horizontally 6700/100 = 67 times, or an odd number, and vertically 6700/67 = 100 times, or an even number. Thus, the vertical movement cancels out, but the horizontal movement is equivalent to crossing the board exactly once, meaning the answer is D.

I just used cad and drew the lines. Took about 3 minutes.

Just for variety, I wrote a simple simulator in Excel. First few rows of the worksheet are as follows: link text and then you just copy the bottom row down far enough. I'm aware this is a programmatic solution rather than a mathematical one - just though someone might be interested.

I realized that to go in a pocket the ball must come off a long cushion 67 from a "far" end. I traced the ball around a couple of times and realized that it was bouncing of the top long side at (99-2n) from pocket C. It will therefore go in pocket D.

A corollary of the elegant LCM and traversal parity methods is that the ball can never go in pocket B! This is a bit of a surprising result to me but the logic is correct. If the table measures X length and Y width and both measures are even then in some other integral unit of measurement the table will have at least one of X or Y to be odd and so the answer must be A, C or D, never B.

At every reflection the ball would get a 90 angular projection and thus u can easily trace the ball by summing up the angles made by it as the total angle of quadrilateral.

Let’s put our table on a grid. It is 100 in the X direction and 67 in the Y the angle of 45º This which mean the increase in X to Y is 1:1 we then get out first bounce at 0,0 to 67,67 this then bounces down to 100,34 this bounces down and left to 66,0 which again bounces up to 0,66 which quickly goes infront of A to 1,67 which then bounces to 68,0 to 100,32 to 65,67 to 0,2 to 2,0 to 69,67 to 100,36 to 64,0 to 0,64 to 3,67 to 70,0 to 100,30 to 63,67 to 0,4 and on until it will reach 100,0 eg D

at 45 angle it means the ball will cover maximum distance so it will go to D

Due to the initial +x trajectory of the ball, it cannot enter pockets B or C. If it makes a full vertical traverse an even number of times and enters a pocket, it will be D. If it makes a full vertical traverse an odd number of times and enters a pocket, it will be A. To enter a pocket it must travel a common multiple of 67 and 100, the first of which is 6700, which involves an even number of vertical traverses, so pocket D.

first i thought that the ball will end up in the point of A, but think a little more...it cannot exactly go straight away into this point, so it will end up in the D.

Use the plane as a covering space for the rectangle in the following manner: $(x,y) \rightarrow (x',y')$ where $x' = (x\mod 200)$ if $(x\mod 200) < 100$ , otherwise $x' = 200 - (x\mod 200)$ Similarly let $y' = (y\mod 134)$ if $(y\mod 134) < 67$ , otherwise $y' = 134 - (y\mod 134)$ .

Then the line $y = x$ is an exact cover for the path of the billiard ball. The billiard ball is going right when $(x\mod 200) < 100$ and left when $100 < (x\mod 200) < 200$ . Similarly it is going up (according to the picture) when $(y\mod 134) < 67$ and down when $67 < (y\mod 134) < 134$ . It reaches a pocket when $x'=y'=0$ . The first time this happens is (since $67$ and $100$ are relatively prime) at the point $(6700,6700)$ .

To determine the pocket, consider what direction the ball is traveling at $(6699,6699)$ . Now $6699 = (99 \mod 200)$ , so the ball is going to the right here. And $6699 = (133\mod 134)$ , so the ball is going down here. Thus the pocket the ball ends up is in the lower right corner, pocket $D$ .