# Yet another curious incident in the flea circus

The ringmaster of a flea circus puts four fleas $A$, $B$, $C$, $D$ on four different points in the plane that form the corners of a square.

Whenever the ringmaster shouts "Hop!", one of the four fleas jumps over the centroid of the three other fleas to the mirror point on the other side. (In other words, a flea sitting in point $x$ may jump over the centroid $y$ to the new point $z$, so that $y$ is the midpoint between $x$ and $z$.)

Question: Is it possible that after a finite number of jumps, two of the fleas are simultaneously sitting on the same point?

## 2 Answers

WLOG the fleas start at the corners of the unit square, with A at $(0,0)$, B at $(1,0)$, C at $(1,1)$, and D at $(0,1)$. I claim that after $n$ jumps, A will be at $(\frac{2a_1}{3^n},\frac{2a_2}{3^n})$, B will be at $(\frac{2b_1+1}{3^n},\frac{2b_2}{3^n})$, C will be at $(\frac{2c_1+1}{3^n},\frac{2c_2+1}{3^n})$, and D will be at $(\frac{2d_1}{3^n},\frac{2d_2+1}{3^n})$ for some integers $a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2$.

This is true in the starting position, with $n=0$. Now suppose that it is true after $n$ jumps, and A jumps next. The centroid of B, C, and D is $(\frac{2b_1+1+2c_1+1+2d_1}{3^{n+1}},\frac{2b_2+2c_2+1+2d_2+1}{3^{n+1}})$, and A ends up at the point $(\frac{4b_1+4c_1+4d_1+4-6a_1}{3^{n+1}},\frac{4b_2+4c_2+4d_2+4-6a_2}{3^{n+1}})=(\frac{2(2b_1+2c_1+2d_1+2-3a_1)}{3^{n+1}},\frac{2(2b_2+2c_2+2d_2+2-3a_2)}{3^{n+1}})$. Meanwhile, B stays at $(\frac{2b_1+1}{3^n},\frac{2b_2}{3^n})=(\frac{2(3b_1+1)+1}{3^{n+1}},\frac{2(3b_2)}{3^{n+1}})$, and similarly for C and D.

Similar calculations show that if B, C, or D jumps instead of A, the claim still holds. Therefore it holds for all $n$.

Therefore no two fleas can ever be at the same position, because at least one coordinate's numerator would have to be both even and odd.

• Very nice answer, clean solution. – Fimpellizieri Oct 30 '15 at 19:38

EDIT: I'm an idiot and answered the wrong question. Disregard.

No.

Call the initial positions of fleas A, B, C, and D (0,0), (1,0), (0,1), and (1,1) respectively. At any point in time, two fleas will be at a distance (n,m) away, where n and m are integers. Pivoting will move one flea a distance (2n,2m). Flea A can only occupy the points (2n,2m), flea B (1+2n,2m), flea C (2n,1+2m), and flea D (1+2n,1+2m).

• If flea D hops from your initial setup, it reaches the point $(-\tfrac13,-\tfrac13)$. – Fimpellizieri Oct 30 '15 at 18:50
• Yes I missread the question. I thought one flea was leapfrogging over another. – dashwhistle Oct 30 '15 at 18:53