China Mathematical Competition (Hainan)

If $a$, $b$ and $c$ are the lengths of the sides of a triangle, all of them are not zero, it follows that $a,b,c\in \{1,2,\dots ,9\}$.

i. If the triangle we construct is equilateral, let $n_1$ be the number of such 3-digit numbers. Since the three digits in such 3-digit number are equal, we have $$n_1=C_9^1=9.$$

ii. If the triangle we construct is isosceles but not equilateral, let $n_2$ be the number of such 3-digit numbers. Since there are only 2 different digits in such a 3-digit number, denote them by $a$ and $b$. Note that the equal sides and the base of an isosceles triangle can be replaced by each other, thus the number of such pairs $(a,b)$ is $2C_3^2$. But if the bigger number, say $a$, is the length of the base, then $a$ must satisfy the condition $b<a<2b$. All pairs that do not satisfy this condition we list in the following table. There are 20 pairs.

a	9	8	7	6	5	4	3	2	1
b	4, 3, 2, 1	4, 3, 2, 1	3, 2, 1	3, 2, 1	2, 1	2, 1	1	1

On the other hand, there are $C_3^2$ possible 3-digit numbers with digits taken from a given pair $(a,b)$. Thus $$n_2=C_3^2(2C_3^2-20)=6(C_3^2-10)=156.$$ Consequently, $n=n_1+n_2=165$. Answer: C.