# Do a barrel roll! (i.e. a Euclidean plane rotation puzzle)

One of my favorite Putnam problems due to a slick solution.

$R$ is at $(3, 4)$ on the cartesian plane. To try to confuse $R$, the devious $S$ decides to rotate $R$ about the point $(1, 0)$ by $36^\circ$. $S$ then rotates $R$ by $36^\circ$ about the point $(2, 0)$, then $36^\circ$ about the point $(3, 0)$, then $(4, 0)$, etc., until finally rotating her $36^\circ$ about the point $(10, 0)$. Where does $R$ end exactly and why?

(Edit) Additional hint: Narmer and xnor have the correct solution below, but there is still a clever proof it works that no one has gotten. If you're curious, it involves only very basic geometry, and doesn't require much more than

putting a regular polygon in the right starting location.

• I assume all the rotations are in the same direction? May 8, 2015 at 11:57
• @psmears Yes, good point, that's needed. May 8, 2015 at 14:45

Place a regular decagon in the plane with one side being from (0,0) to (1,0). Attach the point (3,4) to it. Now roll the decagon along the x-axis. This has the same effect as the 36 degree turns. The decagon ends up in the same orientation, moved to the right by 10 units. The point is in the same location relative to the decagon, so it is at (13,4).

• Nice! See Problem B4 from 2004 to see the original source. May 12, 2015 at 3:14

There are $10$ points used by $S$ and for each of them $S$ rotates $R$ by $36°$ and since $36°\times10 = 360°$ the sum of the rotations gives a full circle.

If the point of rotation had not changed in the rotations, the $R$ point would end in the same position it started. But the point of rotation did change, but only in the $X$ axis and the sequence of the $X$ axis values used by $S$ forms an arithmetic progression. This mean that the $Y$ axis of $R$ after the rotations will be the same, $4$.

What happens to the changing axis then? Well, adding $1$ to each point in the $X$ axis shifts the $R$ point on the $X$ axis by the same amount.

The $R$ point will be in $(13, 4)$.

• what tool did you use to make the image? May 8, 2015 at 9:06
• GeoGebra May 8, 2015 at 9:06
• I guess your solution is correct but I do not understand the reasoning. You say the Y coordinate doesn't change because the Y coordinate of the rotation points don't change. But let's say you take rotation points (10,0),(100,0), ..., (10000000000,0) then the resultating Y will be not the same I guess. Or does it only work when moving the rotation point with a fixed distance every time? or what is the criteria for this to work? May 8, 2015 at 9:15
• @Narmer Actually the X coordinates must form an arithmetic progression for this to work - for Y to be the same.
– dmg
May 8, 2015 at 9:22
• I guess my next question is: why does it work and only work when there is an arithmetic progression? but maybe that's something for math.stackexchange.com May 8, 2015 at 9:35

Let $R_k$ be the position of $R$ after $k$ rotations, represented as a complex number, starting at $R_0 =3+4i$. Let $\alpha=e^{2\pi i/10}$ be the complex number representing rotation by $36^\circ$. Then, rotating around the $k^{th}$ point, whose coordinate is $k+0i$, gives the relationship

$$R_k-k =\alpha (R_{k-1} - k)$$

If we switch to coordinates relative to the current rotation point $S_k=R_k-k$, the recursion becomes

$$S_k = \alpha (S_{k-1}-1)$$

which is intuitively, shifting coordinates by $1$ due to switching rotation points, then rotating.

Applying it ten times takes us back to the starting value because

$$S_{10} = -(\alpha + \alpha^2 + \dots + \alpha^9 + \alpha^{10}) + \alpha^{10} S_0 = S_0,$$

using the fact that $a^{10}=1$ and so the geometric progression is $0$. Substituting back for the $R$'s, we have

$$R_{10} - 10 = R_0$$

so $R$ ends up ten spaces right of its starting point, at $(13,4)$.

• I didn't know this could be done with the geometric sum formula and complex numbers -- very cool. But despite this being elegant and correct, I might hold off accepting for a little bit since it isn't the intended solution. May 8, 2015 at 20:51