# How to solve this riddle?

How to solve this riddle?

All rooms have the same shape and size. Every room has one entrance, and two exits: one on the left, and one on the right (on opposite directions). Both exits lead to a new room.

You are in the first room. You turn right, and arrive to a new room. In the second room, you turn again right. In the third room, you turn left, and arrive at where you started.

What shape the rooms have?

Rooms are usually rectangular but I think it is not the case here.

• What is the source of this puzzle? Where did you find it? – user135229 Jun 19 '15 at 20:13
• Are you sure you typed this correctly? If every room has two doors, then turning left after turning right will lead you to the previous room. Maybe you should turn right in the third room? – A.P. Jun 19 '15 at 20:16
• This sounds a bit like "You go south 1km, then east 1km, then north 1km - what colour is the bear you see?" (It's an old puzzle - in our times walking there might become difficult and the bear might be a Capuccino bear) – Hagen von Eitzen Jun 19 '15 at 20:22
• I feel like there is some sort of miscommunication. I feel like an orientation would be needed in the first room. – anakhro Jun 19 '15 at 20:24
• We could try to deduce the concavity of the rooms from the turn angles + some sort of radial symmetry – FisherDisinformation Jun 19 '15 at 20:33

As stated, all you can tell is that the second room shares the left wall with the first room and the right wall with the third room. This is because turning right from room $A$ into room $B$ and then turning left from room $B$ will lead you again in room $A$.

On the other hand, I suppose that the actual path is:

You are in the first room. You turn right, and arrive to a new room. In the second room, you turn again right. In the third room, you turn right, and arrive at where you started.

Let's call the rooms $A,B$, and $C$ for brevity. Then in this case, you can conclude that $A$ shares the left wall with $C$ and the right wall with $B$, which in turn shares its right wall with $C$. In other words, you can imagine the three rooms arranged in a circle.

Note, though, that in either case you cannot really tell anything about the shape of the walls. All you can say is how the rooms are connected to each other.

First let us examine the orientation of the three rooms. We start in room $A$ with our back towards the entrance. As we step through the right door into room $B$ our orientation changes by an angle $R$. From there going through the right door into room $C$ means another rotation over an angle $R$. Finally we go through the left door, over an angle $L$, into room $A$. Note that our orientation is the same as the begin situation, since our back is once again oriented to the entrance of room $A$. So we have: $R + R + L = n*360$ degrees, with $n$ an integer. Now it is given that $L$ is opposite $R$, hence $L = R + 180$. It follows that $R$ is either $60$ degrees or $180$ degrees. The last possibility leads to a contradiction: as we exit room $B$ we return to room $A$, however it is given that we enter room $C$. Therefore we can exclude an angle of $180$ degrees. Conclusion: room $B$ must be rotated $60$ degrees with respect to $A$, and similarly room $C$ is rotated $60$ degrees with respect to $B$.

The next step is to find the location of the doors. Let us denote the position of the entrance of room $A$ by $(0,0)$ and the right door by $(1,0)$. We then find that the entrance of room $B$ is $(1,0)$ and the right door is $(3/2, 1/2 \sqrt3)$. For room $C$ we find that the entrance is $(3/2, 1/2 \sqrt3)$, the left door is $(0,0)$, and the right door is $(1, \sqrt3)$. Finally we obtain that the left door in room $A$ is located at $(0, \sqrt3)$ and the left door in room $B$ is at $(-1/2, 1/2 \sqrt3)$. To summarize: the six doors are located at the corners of a regular hexagon, which has its centre at $(1/2, 1/2 \sqrt(3)$.

We should now be able to allocate the space in and around the hexagon to the three rooms. We can indeed connect the entrance of each room to the right door. However this is not possible when it comes to the left doors! The three left doors are positioned such that they interfere with each other. Hence no solution (in $2$ dimensions) is possible.