MathSolver94 wrote: Let $N=\bar{abc}$ be a three-digit number. It is known that we can construct an isoceles triangle with $a,b$ and $c$ as the length of sides. Determine how many possible three-digit number $N$ there are. ($N=\bar{abc}$ means that $a,b$ and $c$ are digits of $N$, and not $N=a\times b\times c$.) solution for $aab$ is $61$ Hence $3*61-2*9$ because we don't want to count $aaa$ three times in permutations of $aab.$ So result $165$

To explain better: So we need to compute the number of $(a,b,c)$ such that at least two of them are equal (equilateral is considered to be isosceles), $1 \le a,b,c \le 9$, $a,b,c \in \mathbb{Z}$, and every number is less than the sum of the other two. Assume $a \ge b \ge c$. So $a < b + c$. The number of equilateral triangles is obviously 9. So now we will count the number of non-equilateral triangles. If $a=b$, there are 36 solutions, namely $(k,k,i)$ for all $1 \le i < k \le 9$. If $b=c$, there are 16 solutions: (3,2,2), (4,3,3), (5,3,3), (5,4,4), (6,4,4), (7,4,4), (6,5,5), (7,5,5), (8,5,5), (9,5,5), (7,6,6), (8,6,6), (9,6,6), (8,7,7), (9,7,7), (9,8,8). So there are 52 isosceles non-equilateral triangles. Since each non-equilateral triangles can be permuted in three ways, there are $3 \cdot 52 + 9 = \boxed{165}$ solutions.

chaotic_iak wrote: To explain better: So we need to compute the number of $(a,b,c)$ such that at least two of them are equal (equilateral is considered to be isosceles), $1 \le a,b,c \le 9$, $a,b,c \in \mathbb{Z}$, and every number is less than the sum of the other two. Assume $a \ge b \ge c$. So $a < b + c$. The number of equilateral triangles is obviously 9. So now we will count the number of non-equilateral triangles. If $a=b$, there are 36 solutions, namely $(k,k,i)$ for all $1 \le i < k \le 9$. If $b=c$, there are 16 solutions: (3,2,2), (4,3,3), (5,3,3), (5,4,4), (6,4,4), (7,4,4), (6,5,5), (7,5,5), (8,5,5), (9,5,5), (7,6,6), (8,6,6), (9,6,6), (8,7,7), (9,7,7), (9,8,8). So there are 52 isosceles non-equilateral triangles. Since each non-equilateral triangles can be permuted in three ways, there are $3 \cdot 52 + 9 = \boxed{165}$ solutions. But i calculated somewhere 216 (forgot alrdy since ive thrown the paper ). Starting with 2, if $a=b$ then (2,2,1). If $a=c$ then we have (2,1,2). And $b=c$ then (1,2,2). Again, as $2+2=4$ we have another 3 numbers, that are (2,2,3) (2,3,2) and (3,2,2). So for 2 we have 6 numbers. For 3, (3,3,1) (3,1,3) (1,3,3) and (3,3,2) (3,2,3) (2,3,3) and (3,3,4) ... and (3,3,5) ... . So we have 12 numbers for length 3. It will be tedious if we just calculate using this way, so I have found a way to find the answer. But again, ive thrown away the paper. So if u really want to know how i find it, just pm me . I will attempt to do it again.

My answer is still correct. Non-equilateral: $a=2$ gives 6. $a=3$ gives 12. $a=4$ gives 18. $a=5$ gives 24. $a=6$ gives 24. Note that 10 and 11 are not single digits. (I think you messed here and the rest of the cases.) $a=7$ gives 24. Again, 10,11,12,13 are not single digits. $a=8$ gives 24. $a=9$ gives 24. For a total of 156. Add the 9 equilateral triangles (which I think you also messed by not counting them) for a total of 165.

i dont think equilateral triangle should be counted

If so, the answer is 156.

Isoceles triangle of side length 1 cant exist, so we will start from 2. WLOG, confine our attention to $a=b$. Thus, $c\in\{1,2,\cdots, 2n-1:c\neq n ($cant be equilateral$) \}$, where $n$ is the side length of $a,b$ and $2\le n\le 9$. Thus, we have $2n-1-1=2n-2$ possibility for every side $n$. Therefore, $\displaystyle\sum_{n=2}^{9}2n-2$. Since there are also the case $a=c,b=c$, then there are 3 combination for every triplets. (ex: (2,2,1), (2,1,2) ,(1,2,2)). Thus, \begin{align*}3\displaystyle\sum_{n=2}^{9}2n-2 & =6\displaystyle\sum_{n=2}^{9}n-1 \\ & = 6(1+2+\cdots +8)=216 \end{align*}

The same error still exists. Put $n=6$ and see the possible values of $c$. It contains 10 and 11, which are not single digits and thus can't be the value of $c$. Similar for $n=7,8,9$.