How many isosceles integer-sided triangles are there with perimeter $$23$$?

Let the equal sides be $$a,a$$ and the third side (base) be $$b$$, all in integers.

Perimeter condition: $$2a + b = 23$$ $$-(1)$$

Triangle inequality for an isosceles triangle:
$$a + a \gt b \;\;\Longrightarrow\;\; 2a \gt b$$ $$-(2)$$
The other two inequalities, $$a + b \gt a$$ and $$a + b \gt a$$, simply give $$b \gt 0$$, which is already implied by positive side lengths.

From $$-(1)$$, express $$b$$ in terms of $$a$$:
$$b = 23 - 2a$$ $$-(3)$$

Positive base: $$b \gt 0 \;\Longrightarrow\; 23 - 2a \gt 0 \;\Longrightarrow\; a \lt 11.5 \;\Longrightarrow\; a \le 11$$

Using $$-(2)$$ with $$-(3)$$:
$$2a \gt 23 - 2a$$
$$4a \gt 23$$
$$a \gt 5.75 \;\Longrightarrow\; a \ge 6$$

Therefore $$a$$ can take any integer value from $$6$$ to $$11$$ inclusive:
$$a = 6,\,7,\,8,\,9,\,10,\,11$$

Each admissible $$a$$ gives a unique integer $$b$$ via $$b = 23 - 2a$$, and all satisfy $$2a \gt b$$:
6 cases in total.

Hence, the number of isosceles integer-sided triangles with perimeter $$23$$ is 06.