riddle that involves elementary geometry $3$ frogs are positioned at the vertices of an equilateral triangle whos sides are of length $1$. We have $1$ frog on each vertex.
The frogs are able to "leap" one over another. When they do, they will land in the symmetric spot to where they jumped from as shown on this drawing.

they can jump in any order they want and also any number of times they want.
Is it possible to arrange the frogs such that they are on the vertices of an equialteral triangle whos sides are equal to $2$?
 A: It seems that it is imposiible because in the following coloring of a triangle (with side 1) grid we see that grid vertices of each triangle with side $1$ are 3-colored, 
grid vertices of each triangle with side $2$ are 1-colored, and a jump doesn't change the color of the jumping frog place.  

A: No, they can't. The process is clearly reversible (at each step the frog can jump back over the frog it just jumped over), so if it was possible, it would be possible to go from a triangle with sides of length 2 to one of length 1. Consider the following grid of equilateral triangles with sides of length 2 and the frogs being the green dots (sorry for my crude MSPaint work...):

Clearly the frogs always stay on the vertices, so the smallest (and only) possible equilateral triangle has sides of length 2. 
A: No it is not possible cause after each leap the area of the triangle spanned by the three frogs remains the same. Now the area of an equilateral triangle with sides $1$ obviously does not equal the area of a triangle with sides $2$. Thus this will not be possible.
Edit: The reason the area doesn't change is if you have triangle ABC then the area equals $b*h/2$ where $h$ is the height or distance from $C$ to the line through $A$ and $B$, $b$ is the base. In this case the base is $AB$ now when we move $B$ to the other side of $A$ then $|AB|=|BA|$ and the height also stays the same. Thus the area doesn't change.  
