Every integer n>=3 is correct.

Small_Potato wrote: Every integer n>=3 is correct. Can you explain?

@above Try it yourself before asking. It's easy.

I don't understand the question...

It's asking for a construction of a concave pentagon using \(3,4,5,...\) congruent triangles

Ah ok.

ThE-dArK-lOrD wrote: @above Try it yourself before asking. It's easy. Yes, I see that proving infinite n exist is easy, but how do you prove for all n?

Every $n\geq 3$ is right. Since at least 3 triangles are needed to construct a pentagon, we only needed to consider $n\geq 3$. $n=3$: Consider 3 isosceles triangles, combine the legs. $n=4$: Consider 2 isosceles triangles, combine the legs and cut the altitude of the vertex angle of each triangle. $n=8$: Consider 4 isosceles triangles, combine the legs and cut the altitude of the vertex angle of each triangle. For cases below, pentagons are made up of 45°–45°–90° triangles and their length of legs is 1. $n=4k+1(k \in \mathbb{N^*})$: The coordinates of 5 vertex is $(0,0), (1,0), (k+1,k), (k,k+1), (0,1)$. $n=4k+2(k \in \mathbb{N^*})$: The coordinates of 5 vertex is $(0,0), (k,0), (k+1,1), (k,2), (0,2)$. $n=4k+3(k \in \mathbb{N^*})$: The coordinates of 5 vertex is $(0,0), (k+1,0), (k+1,1), (k,2), (0,2)$. $n=8k+4(k \in \mathbb{N^*})$: The coordinates of 5 vertex is $(0,0), (2,0), (k+2,k), (k,k+2), (0,2)$. $n=8k+8(k \in \mathbb{N^*})$: The coordinates of 5 vertex is $(0,0), (k,0), (k+2,2), (k,4), (0,4)$. And we are done.

P.S.

guoh064 wrote: Every $n\geq 3$ is right. Since at least 3 triangles are needed to construct a pentagon, we only needed to consider $n\geq 3$. $n=3$: Consider 3 isosceles triangles, combine the legs. $n=4$: Consider 2 isosceles triangles, combine the legs and cut the altitude of the vertex angle of each triangle. $n=8$: Consider 4 isosceles triangles, combine the legs and cut the altitude of the vertex angle of each triangle. For cases below, pentagons are made up of 45°–45°–90° triangles and their length of legs is 1. $n=4k+1(k \in \mathbb{N^*})$: The coordinates of 5 vertex is $(0,0), (1,0), (k+1,k), (k,k+1), (0,1)$. $n=4k+2(k \in \mathbb{N^*})$: The coordinates of 5 vertex is $(0,0), (k,0), (k+1,1), (k,2), (0,2)$. $n=4k+3(k \in \mathbb{N^*})$: The coordinates of 5 vertex is $(0,0), (k+1,0), (k+1,1), (k,2), (0,2)$. $n=8k+4(k \in \mathbb{N^*})$: The coordinates of 5 vertex is $(0,0), (2,0), (k+2,k), (k,k+2), (0,2)$. $n=8k+8(k \in \mathbb{N^*})$: The coordinates of 5 vertex is $(0,0), (k,0), (k+2,2), (k,4), (0,4)$. And we are done.

P.S.

For n=4k we can consider the fact that each triangle can be divided into 4 smaller replicas of itself by joining the midpoints of all 3 sides. Since 4, 8, 4k+1,4k+2 and 4k+3 have been covered already, this covers all multiples of 4 as well.