Billiards brain-teaser

Geometry Level 2

Consider the standard $5 \text{ cm}\times 5\text{ cm}$ mouse billiards table shown. The pockets are located at points $A, B, C,$ and $D$ .

A player has a very poor aim so he hits the ball from point D (his starting point) towards point E. The player hits it hard enough that it eventually lands in one of the pockets (corners). How many times does the ball bounce off the edges of the table before this occurs?

5 4 7 3

4 solutions

Isaac Reid
Dec 9, 2015

One way to solve this problem is the "unfolding" technique.

Every time there is a collision with the edge of the table, we introduce a new mirror image of the billiards table. This enables us to plot the ball's path without things getting too messy; a simple diagram yields the result above.

Before being potted at C, the ball clearly hits a boundary (i.e. goes through an edge) $5$ times, so the answer is $\boxed{5}$ .

For the sake of completeness and clarity, I have also represented this on a single diagram:

So can we conclude that for a square board of side M(in this case 5) in which the first collision takes place at a height of N (in this case 2), the minimum no. Of reflections required for slotting the ball is LCM(M,N)/N

Arijit ghosh Dastidar - 5 years, 6 months ago

Nice question with an awesome solution.

Anupam Nayak - 5 years, 6 months ago

How u have taken the mirror image . Can u pls explain ??

Chirayu Bhardwaj - 5 years, 6 months ago

If there is no air resistance or friction (as stated in the question), the ball's angle of incidence and angle of reflection will be equal.So it follows the laws of reflection.

Anupam Nayak - 5 years, 6 months ago

Pretty Simple!!! I loved this visual solution!

I used the principle of reflection too and a bit of trigonometry in it, not so simple as it.

Victor Paes Plinio - 5 years, 5 months ago

Great solution, nice approach. Kudos. To enhance the readability more, so others understand it clearly, particularly to understand the folded image you have drawn above,I may like to add to your solution where you say - "we introduce a new mirror image" . Please correct if any discrepancy. 1] We introduce reflection to maintain the direction such that we can count the edge crossings in the 'unfolded' images of billiards table 2] We obtain a mirror image/reflection only if the ball is going in opposite direction after bouncing of the edge of the table , to the direction of the ball in the unfolded image. i.e.in the given solution unfolded image, if the ball is travelling to left , take the reflection of the path after bouncing, along that particular edge of the table and draw the path in he folded image, till it hits the next edge. Continue doing so till ball ends up in C. 3] If after bouncing the ball travels to right, no reflection needed. just extend the path of the ball 4] If the ball hits the top edge, which is the case here in the 3rd bounce, take the reflection of the path after bouncing along the horizontal edge of table.

Once again nice solution. Loved figuring out your 'folding idea'

Ajit Deshpande - 5 years, 5 months ago

A Former Brilliant Member
Dec 25, 2015

Actually, we can work the common angle between every bounce that is arctan(2/5), and use some simple PT which will then lead to the answer 5.

Matthew Butt
Dec 9, 2015

Loved this

Achille 'Gilles'
Dec 30, 2015

5 walls. assuming enough momentum and no spin.

➊ at point E

➋ on left wall, 4 cm from point D

➌ on top wall, midway between A and B

➍ on right wall, 4 cm from point C

➎ on left wall, 2 cm from point D

✔ in pocket at corner C!

Billiards brain-teaser

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