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

Step 1: Let the sides of the triangle be $a$, $a$, and $b$, where $a,b\in {\mathbb{Z}}^{+}$. The perimeter constraint is:

Step 2: By the triangle inequality, the sum of any two sides must be strictly greater than the third side:

1. $a+a>b⟹2a>23-2a⟹4a>23⟹a\ge 6$

2. $a+b>a⟹b>0⟹23-2a>0⟹2a<23⟹a\le 11$

Step 3: Thus, the possible integer values for $a$ are in the range $6\le a\le 11$. There are exactly $11-6+1=6$ possible values:

• $a=6⟹b=11$ (sides $6,6,11$)

• $a=7⟹b=9$ (sides $7,7,9$)

• $a=8⟹b=7$ (sides $8,8,7$)

• $a=9⟹b=5$ (sides $9,9,5$)

• $a=10⟹b=3$ (sides $10,10,3$)

• $a=11⟹b=1$ (sides $11,11,1$)

All these satisfy the triangle inequalities. So there are 6 such triangles.