How many ways can the vertices of an equilateral triangle be colored using three different colors? Its not $3^3$ because some of the colorings are equivalent. How would I apply Burnside's theorem to this?
 A: In this case it suffices to select the number of vertices of each color (since all vertices are connected amongst themselves). Use stars and bars
Look at the three vertices as stars and we need to bars to separate the stars, thus the answer is $\binom{5}{2}=10$

Edit: I assume two triangles are the same if you can take the two triangles and move them in $3d$ space so that they look the same. In other words this is when the group acting on the triangle is $D_3$. The answer when the triangle can only be moved in $2d$ space is different, since one triangle becomes two different triangles (namely $RGB$ splits into $RGB$ and $RBG$. So in this case the answer is $11$. This corresponds to when the group that is acting is $Z_3$.

Also notice problems of this type are usually solved via Burnside’s lemma, you can look up the same thing for the faces of cubes and there the lemma actually is useful. In this case however it is not useful because it is very simple.
A: I'll assume reflections of colorings are considered the same. This means that $S_3$ acts on colorings of your triangle by permuting the vertices, and you want to count the number of orbits of this action. Letting $X$ be the set of vertices, then Burnside's Lemma gives
$$
\text{no. of orbits}=\frac1{|S_3|}\sum_{\sigma\in S_3}\text{no. of fixed points of }\sigma
$$
So, now you look at each permutation $\sigma$ of the vertices, and ask how many colorings are fixed by $\sigma$. For example, when $\sigma$ is the identity, all $3^3$ colorings are fixed. When $\sigma$ is a rotation, all vertices must be the same color, so there are 3 fixed points there. Finally, when $\sigma$ is a transposition, there are 3 choices for the switched vertices, and 3 for the vertex that was untouched. So, you add up $3^3+3+3+3^2+3^2+3^2$, and divide by $|S_3|=6$.
A: By enumeration of the cases the answer is 10. Proof:
RGB
RRR
GGG
BBB
GGB
GGR
BBG
BBR
RRG
RRB
A: Let $X$ be the set of all $3^3=27$ possible colorings of the points of the equilateral triangle. Let the set $G=\{(1), (12), (13), (23), (123), (132) \}$ be the permutations for the 3 verticies on the triangle.
We consider the fixed point sets for all $g\in G$, i.e., we will consider the number of colorings (out of 27) that are fixed by each of the elements of $G$.
 The identity fixes all colorings of the vertices of the equilateral triangle. So, $|X_{(1)}|=27$.
The elements $(123)$ and $(132)$ will only fix the colorings that have all vertices painted the same color. Out of all 27 colorings, there are 3 with this property. So, $|X_{(123)}| =|X_{(132)}|=3$.
The element $(12)$ will fix all the colorings for which verticies 1 and 2 are colored the same. We have three options to color vertices 1 and 2, and three color options for vertex 3. So, there are $3\cdot 3=9$ colorings that are fixed by $(12)$. Using very similar arguments for elements (13) and (23), we get $|X_{(12)}|=|X_{(13)}|=|X_{(23)}|=9$.
By Burnside's theorem, 
$$k=\frac{\sum_{g\in G}|X_g|}{|G|}=\frac{27+3+3+9+9+9}{6}=10$$
So, there are $k=10$ distinct orbits of $X$ under $G$, which means there are 10 distinct colorings of the vertices of the equilateral triangle with 3 colors. 
A: Using all three colors it's one or two ways, depending on whether you are allowed to flip the triangle over. Using exactly two colors we have three possibilities to choose the color used twice and two possibilities for the remaining color, makes six ways in all; finally there are three way to use just one color. In all, there are $10$, or $11$, ways to color the triangle with $\leq3$ colors.
