# Reflections in a Square

An ideal billiards table (no friction, ideal reflections off of the walls, no pockets) is shaped like a square. From the bottom-left corner, shoot a point-sized cue ball at some angle.

What is the shortest path that hits all 4 edges of the table at least once each and ends in the top-right corner?

Note 1: hitting a corner ends the path without counting as hitting the walls on either side of it.

Note 2: I'm aware of this puzzle, which is higher-dimensional and more mathematically involved but has a similar answer to this one. I felt that this warranted its own puzzle, though, because I want more people to have the chance to find its elegant, visual solution without being intimidated by the apparent need to use math (and it's not asking for exactly the same thing).

• Um, sorry, but ideal billiards table is shaped like a rectangle. – Zereges Apr 22 at 20:46
• It has no pockets either - it's such an ideal billiards table that you can't play billiards with it. I'm a physicist. Making useful, non-ideal things is the engineer's job. :P – Gilad M Apr 22 at 22:10

Suppose we label the corner on the table like this: Now we want to move from $$A$$ to $$D$$.

Now, imagine the table like this:

Here,

We can simulate the path as a single straight line.
As the line passes the border, it will simulate the result of the ball's reflection hence the table will be mirrored as above.

Now, to hits all $$4$$ edges, that means

The path should pass $$\overline{AB}$$, $$\overline{CD}$$, $$\overline{AC}$$, and $$\overline{BD}$$.
In other words, the path must go beyond $$2 \times 2$$ tables.

To get the shortest path,

We should choose the nearest $$D$$ (beyond $$2 \times 2$$ tables).
One of the options is the $$D$$ located $$3$$ above and $$3$$ right. But we can't choose this because the line will hit only both corners $$A$$ and $$D$$.

Thus,

We can choose the next nearest $$D$$ which is located $$3$$ above and $$5$$ right (or $$5$$ above and $$3$$ right). This is the shortest path. The overall length will be $$\sqrt{5^2 + 3^2} = \sqrt{34}$$ times the table length.

Which is like this:

• Can you add another picture at the end with the line drawn over the first image? Just to save us from rotating our heads repeatedly :-D – Ahmed Abdelhameed Apr 22 at 12:50
• @AhmedAbdelhameed added :) – athin Apr 22 at 12:57
• Nice! That's exactly right! – Gilad M Apr 22 at 14:08