$3$ coins jumping over each other We are given $3$ coins in a plane such that they form a equilateral triangle,say with side lenght of $2$ units. Each coin can jump over another coin so that relative to the coin being jumped on, jumping coin is same distance away after the jump relative to the stationary coin. Problem asks if it is possible to produce a bigger equilateral triangle after series of jumps.
My answer is no, because, say it is possible to make a bigger triangle, then by reverse jumping we can also get a smaller triangle . Now lets construct an $xy$ system on the leftmost vertex of the triangle, such that one coin sits on coordinates $(0,0)$, other one at ($1$,$\sqrt{3})$, and last one at $(2,0)$. Now lets consider the $x$-coordinates of the coins: $\{0;1;2\}$. Now in each jump $x$-coordinates changes. Supposing $x$-coordinates change such that we are only multiplying adding and subtracting but not dividing we cannot get fraction $x$-coordinates. But a smaller triangle would have vertices locate on a fraction, that is a contradiction?
I would appreciate comments . Thanks.
 A: Your idea is right, but to make that work rigorusly, you need to define a lattice using the base equilateral triangle and continue your arguments on that lattice.
However, there is a little more tedious, but neater solution. Given any triangle, write the sides as vectors, and look at what happens to the area after one coin jump. You will find that the area is invariant of a coin jump (now that I've told you, you can even draw a few pictures an check it geometrically in case you find it hard to believe me :)).
This proves that a bigger triangle is not possible.
A: Your arguments do not seem correct. Firstly I do not understand why you claim that coordinates must be fractions. Even if that were true, coordinate systems are immaterial in this problem, and we can choose a coordinate system such that the coordinates no longer remain fractions in that system.
Even, so, your answer is correct. To see this, let the triangle be $ABC$. Consider the first jump. There are $6$ possible jumps- $A$ on $\overline{AB}$ or $\overline{AC}$, $B$ on $\overline{BA}$ or $\overline{BC}$, and $C$ on $\overline{CA}$ or $\overline{CB}$. Each of these jumps result in the formation of a congruent equilateral triangle. Hence, since the first jump brought no change, by induction, repeating this process will never bring any change to the configuration.
