Make isosceles triangle with matchsticks 
Alan is using matchsticks to make isosceles triangles. Isosceles triangles have at least two sides of the same length. How many different single isosceles triangles can he make with B matchsticks, using all the matchsticks each time? 

Not sure I understand the answer here where $B = 45$ (or 33?):
Why do we have $X+2X=B$? What if B is not divisible by 3?
What I tried:
If it's 3,5,6, Alan can make only 1 isosceles triangle. If it's 7, he can make 2.
Is this something like $\lfloor \frac{x}{3} \rfloor$? Or  $\lfloor \frac{x}{3}+1 \rfloor - 1$?

Well, if we call the number of matches in the base side of the triangle $a$, and the number of matches in the other two sides (which totals to $2b$ for both sides) then we need to find the number of solutions to $a+2b=B$. This is simply the number of numbers you can fill in for $b$, since then $a=B-2b$ follows.
Thus, in general, the number of triangles you can make is $\lfloor\tfrac 12(B-1)\rfloor$, because that's how many $b$'s there are such that $B-2b\geq1$.
As correctly noted, we need to take the triangle inequality $2b>a$ into account. When we try to find the number of $b$'s that do not satisfy that property, we look at $2b\leq a$, or $2b\leq B-2b$, in other words, all $b$ less than or equal to $\frac{B}{4}$. Subtract this from $\lfloor\tfrac 12(B-1)\rfloor$ to find the final answer:
$$\lfloor\tfrac 12(B-1)\rfloor-\lfloor\frac{B}{4}\rfloor$$

Note that now for $B=45$, we get the correct answer $$\lfloor\tfrac 12(45-1)\rfloor-\lfloor\frac{45}{4}\rfloor=\lfloor22\rfloor - 11=11$$
