The answer is $n$ odd only. First I prove that $n$ even fails. The idea is to show that the polygon can be very fat and have no height. Let $k=\frac n2$ and call the polygon $P_0P_1\cdots P_{n-1}$ where indices are taken modulo $n$. Let $P_0=(0,0)$. We can make sure $P_0,P_1,\cdots, P_k$ stays between the lines $y=0, y=\epsilon$ for any small $\epsilon$ and have $P_{-t}$ be the reflection of $P_t$ over the line $y=0$. An equi-lateral triangle requires a "height" of at least $\frac{\sqrt{3}}{2}$, which means this construction is a counterexample (i.e. $n$ even fails). It remains to show $n$ odd works. If $n=3$ then $P_0P_1P_2$ is an equilateral triangle of side length 1, so it works. Henceforth assume $n\ge 5$. Consider the longest diagonal of the polygon. Say it is $P_0, P_m$ where $m\ge \frac{n+1}{2}$. Then $P_0P_m\le \frac{n-1}{2}$ by triangular inequality. Setup coordinates $P_0=(0,0), P_m=(x_m,0)$ and $P_j=(x_j,y_j)$ where $y_1>0$. Since the polygon is convex, we can see $y_{i+1}-y_i$ is strictly decreasing and $x_i$ is strictly increasing. Let $t$ be the unique value that satisfies $y_{t-1}<y_t<y_{t+1}$. Let $X_t=(x_t,0)$ be the projection of $P_t$ onto the $x$-axis. We know at least one of the two holds: $$P_0P_1+\cdots+P_{t-1}P_t > P_0X_t+\frac 12$$ $$P_tP_{t+1}+\cdots+P_{m-1}P_m > X_tP_m + \frac 12$$ (If neither holds, summing yields $\frac{n+1}{2} < \frac{n-1}{2}+1$) WLOG the former holds. Then \begin{align*} y_t&= \sum\limits_{j=1}^t (y_j-y_{j-1})\\ &= \sum\limits_{j=1}^t\sqrt{1-(x_j-x_{j-1})^2} \end{align*} Let $d_j=x_j-x_{j-1}$ then $d_1+d_2+\cdots+d_t=P_0X_t$. Define $f(x)=\sqrt{1-x^2}$ where $x\in [0,1]$. Then $$f'(x)=\frac 12 (1-x^2)^{-\frac 12} (-2x)= -(x^{-2}-1)^{-\frac 12}$$ $$f''(x)=\frac 12 (x^{-2}-1)^{-\frac 32} (-2x^{-3}) = -(x^{-2}-1)^{-\frac 32} x^{-3} < 0$$ Since $f$ is concave, we can see that if $d_i+d_j\le 1$ then $f(d_i)+f(d_j) \ge f(d_i+d_j)+f(0)$ and if $d_i+d_j\ge 1$ then $f(d_i)+f(d_j)\ge f(d_i+d_j-1)+f(1)$. Therefore, we can see that if $d_1+\cdots+d_t$ is fixed, $f(d_1)+\cdots+f(d_t)$ is minimized when at most one of the $d_j$'s is not $0$ or $1$. We will use this frequently from now on. First, this immediately proves that $y_t\ge \frac{\sqrt{3}}{2}$. If $(x_t, \frac{\sqrt{3}}{2}), (x_t-\frac 12, 0), (x_t+\frac 12,0)$ are in the $(m+1)$-agon, then this equilateral triangle of side length 1 works since this $(m+1)$-agon is convex. Otherwise, since $(x_t,\frac{\sqrt{3}}{2})$ is in the polygon, we consider the following: $(x_t-\frac 12,0)$ is not in the $(m+1)$-agon is equivalent to $x_t<\frac 12$. $(x_t+\frac 12,0)$ not being in the $(m+1)$-agon implies $x_t>x_m-\frac 12$. Since $P_0P_m$ is the longest diagonal, $x_m>1$, so only one of the two points an lie in the exterior. WLOG $x_t<\frac 12$. Case 1: $t\ge 2$. Then $d_1+\cdots+d_t=x_t\le \frac 12$, so $$f(d_1)+\cdots+f(d_t)\ge \underbrace{f(0)+\cdots+f(0)}_{(t-1) \text{ } f(0) \text{'s}} +f(x_t)\ge 1+\frac{\sqrt{3}}{2}$$ I claim the triangle $(0,0), (1,0), (\frac 12, \frac{\sqrt{3}}{2})$ works. We just need to verify $(\frac 12, \frac{\sqrt{3}}{2})$ lies under the line segment $(x_t, 1+\frac{\sqrt{3}}{2})$ and $(1,0)$, which means it lies under $P_tP_m$. This is true since $x_t>0, y_t>\sqrt{3}$. Case 2: $t=1$ and $x_t<\frac 12$. Let $(x_1,y_1)=(\cos \theta, \sin \theta)$ where $\theta \ge \frac{\pi}{3}$. Consider the triangle $(0,0), (\cos\theta, \sin \theta), (\cos (\theta-\frac{\pi}{3}), \sin(\theta-\frac{\pi}{3}))$. It suffices to show the point $(\cos (\theta-\frac{\pi}{3}), \sin(\theta-\frac{\pi}{3}))$ lies within $P_1P_m$, or $\angle P_0P_1P_m\ge \frac{\pi}{3}$. Since $P_0P_m$ is the longest diagonal in $\triangle P_0P_1P_m$ (i.e. $P_0P_m > \max\{P_1P_m,P_0P_1\}$.) The conclusion follows by sine law.

This problem was proposed by Burii.

Here is a simpler proof. Solved with p_square, guptaamitu The answer is all odd $n$. To show even $n$ fail, make the polygon very wide and have super small height, so the area it covers is not large enough to contain an equilateral triangle of side length $1$. Let $\mathcal{P}$ denote an equilateral polygon with $n$ sides of side length $1$ that does not contain an equilateral triangle of side length $1$. It suffices to show that $n$ must be even. Suppose $A_1A_k$ is the longest diagonal of $\mathcal{P}$. On one side of this diagonal, construct points $X,Y$ such that $XA_1 = 1 = YA_k$ and $\angle XA_1A_k = \angle YA_kA_1 = 60^{\circ}$. I claim that the polygon on this side of the diagonal is contained in this trapezoid $\mathcal{T}$. First observe that $\angle A_2A_1A_k \leqslant 60^{\circ}$ since otherwise let $Z$ be the point on $A_1A_k$ such that $ZA_1 = ZA_2$ and $E$ be such that $EA_1A_2$ is equilateral. Then we have that $\triangle EA_1A_2$ is contained in $\triangle ZA_1A_2$, which is contained in $\triangle A_kA_1A_2$ (since $Z$ lies on segment $A_1A_k$ as $A_1A_k$ is the largest diagonal) which is contained in $\mathcal{P}$, a contradiction. So if this side of $\mathcal{P}$ is not contained in $\mathcal{T}$, then there is a side $A_iA_{i+1}$ which intersects segment $XY$, say at a point $M$. But then construct points $P,Q$ on $A_1A_k$ such that $MPQ$ is equilateral with side length $1$ and is contained in $\mathcal{P}$, again a contradiction. Therefore, we have that indeed the polygon on this side is contained in $\mathcal{T}$. Let $s$ denote the number of sides on this side of $A_1A_k$ and suppose $A_1A_k = \ell$. By the triangle inequality, we must have that $s \geqslant \ell$ since each side has length $1$. Now, for every side $A_iA_{i+1}$ with $1 \leqslant i \leqslant k-1$, let $B_{2i}, B_{2i+1}$ denote the points on $\mathcal{T}$ such that $\angle B_{2i}A_{i}A_{i+1} = \angle B_{2i+1}A_{i+1}A_i = 90^{\circ}$. Note that $B_{2i}B_{2i+1} \geqslant A_iA_{i+1}$. Since $\mathcal{P}$ is convex, we have that $s \leqslant \sum_{i=1}^{k-1} B_{2i} \leqslant A_1X + XY + XA_k = \ell + 1$ and equality cannot hold since then the polygon coincides with $\mathcal{T}$ and so contains an equilateral triangle of side length $1$. Therefore, we have $\ell \leqslant s < \ell + 1$ implying that $s = \lceil \ell \rceil$, so $s$ is uniquely determined by $\ell$. But by an analogous argument on the other side, the other side of $A_1A_k$ must have $s$ sides too, implying that there are a total of $2s = n$ sides, implying that $n$ must be even, as desired. $\blacksquare$

Very Nice Problem! Sketch: The answer is all odd $n \geq 3$. Evens clearly fail, just take a equilateral polygon which is very fat and very short. Assume $n = 2k+1 \geq 5$ since the case of $3$ is trivial. Denote the polygon $\mathcal{P}$ as $P_0P_1 \cdots P_{2t}$, let $\mathcal{D}=P_0P_k$ denote the largest diagonal, this splits the polygon into two parts we in fact show that the part with greater number of sides contain an equilateral triangle, denote this part by $P_0P_1 \cdots P_k$, now if the corner angle is greater than $60^{\circ}$, then we're done, else inscribe the polygon in isosceles trapezoid with angles $60,60,120,120$ and equal sides $1$, then by applying the following lemma: Lemma If a polygon $\mathcal{P}$ is inside $\mathcal{Q}$ then $Per(P) \leq Per(Q)$

Proof

this yields contradiction, and we're done.

We claim that the answer is all odds. Clearly, if $n$ is even then we can make a polygon that is nearly degenerate. Suppose that $n=2k+1$ is odd and there is an equilateral $n$-gon in which a unit equaliteral triangle doesn't fit. Because our $n$-gon is convex, if all the vertices of our equilateral triangle are in the $n$-gon we are done. Let the longest diagonal of the polygon be $AB$. One side of $AB$ has at least $k+1$ edges, and the other side at most $k$, meaning that the length of $AB\le k$. Look on the side with $k+1$ edges. Construct trapezoid $A'B'BA$, on this side, such that $A'B'\parallel AB$, $AA'=BB'=1$, and $\angle A'AB=\angle B'BA=60^\circ$. If segment $A'B'$ intersects the $n$-gon at any point, $X$, then since the distance from $X$ to $AB$ is $\tfrac{\sqrt{3}}{2}$ we can construct points $Y,Z$ on $AB$ such that $XYZ$ is an equilateral triangle, and it is clear that $XYZ$ is inside our $n$-gon, contradiction. Therefore, $A'B'$ does not intersect our polygon. Let $C$ be the vertex of the $n$-gon adjacent to $A$ on the side with $k+1$ edges. Since $AB$ is the longest diagonal, $\angle ACB$ is smaller than $\angle CAB$. In particular, if $\angle CAB\ge 60^\circ$ then $\angle ACB\ge 60^\circ$, which means that if we construct $X$ such that $ACX$ is an equilateral triangle then $\triangle ACX$ is inside $\triangle ACB$, contradiction. Therefore, the part of the $n$-gon with $k+1$ edges is strictly inside trapezoid $A'B'BA$. The part of the $n$-gon with $k+1$ edges, including $AB$, has perimeter exactly $k+1+AB$ while trapezoid $A'B'BA$ has perimeter $2AB+1$ but $AB\le k$ so this is impossible. We are done.

Attachments:

We claim the answer is odd $n$. If $n$ is even, we can "flatten" the polygon so that it doesn't contain any equilateral triangles. Claim: Let $n\in \mathbb{N}$ and $m\in\mathbb{R}^+$ such that $n-1\ge m\ge 1$. Let $\mathcal{P}=A_0\dots A_n$ be a convex polygon such that: (i) $A_iA_{i+1}=1$ for all $0\le i<n$ (ii) $A_0A_n=m$ (iii) $\angle A_nA_0A_1, \angle A_0A_nA_{n-1}\le 90^\circ$ Then, $\mathcal{P}$ contains an equilateral triangle of side length $1$. Proof: The case where $n=2$ is obvious, so assume $n\ge 3$. Assume toward a contradiction the claim is false. Let $A_0=(0, 0)$ and $A_n=(m, 0)$ in the coordinate plane. Let $X=(\frac12, \frac{\sqrt{3}}2)$ and $Y=(m-\frac12, \frac{\sqrt{3}}2)$. All points on segment $XY$ form an equilateral triangle with two points on segment $A_0A_n$. Therefore, segment $XY$ lies strictly outside of $P$. Because $\angle A_nA_0A_1, \angle A_0A_nA_{n-1}<90^\circ$, $\mathcal{P}$ is bounded by $x=0$ and $x=m$. Case 1: $\angle A_nA_0A_1, \angle A_0A_nA_{n-1}\le 60^\circ $ Then, $\mathcal{P}$ is contained in $A_0XYA_n$ and $\mathcal{P}\neq A_0XYA_n$. However, the perimeter of $A_0XYA_n$ is at most the perimeter of $\mathcal{P}$, a contradiction. Case 2: $\angle A_nA_0A_1>60^\circ$ Assume toward a contradiction $\angle A_0A_1A_2\ge 60^\circ$. Let $Z$ be the point with positive $x$ coordinate such that $\triangle A_0A_1Z$ is equilateral. Let $W=(0, 1)$. Then, $\angle A_0ZW=90-\frac12\angle ZA_0W \ge 75$. Similarly, $\angle A_2ZA_1\ge 75$. Therefore, $\angle A_2ZW$ is a reflex angle, so $Z$ is contained in $A_0A_1A_2W$ which is contained in $\mathcal{P}$, a contradiction. Therefore, $\angle A_0A_1A_2<60^\circ$. Let $R$ be the intersection of $A_1Z$ with the $x$-axis. Then, $\mathcal{P}$ is contained in $A_0A_1R$. So, $A_3$ is in $A_0A_1R$. However, the only point in $A_0A_1R$ with distance $1$ from $A_2$ is $A_1$, a contradiction. Claim: Let $n\ge 5$ be odd. Let $\mathcal{P}=A_0\dots A_n$ be an equilateral convex polygon with side length $1$. Then, $\mathcal{P}$ contains an equilateral triangle of sidelength $1$. Let $A$ and $B$ be the two points in $\mathcal{P}$ with the highest distance. Then, $AB\ge 1$ and by sliding, $A=A_i$ and $B=A_j$ for some $i, j$. Because $AB$ is maximal, $\mathcal{P}$ is bounded by the perpendiculars to $AB$ at $A$ and $B$. WLOG assume $j-i>\frac12n$. Then, $j-i\ge n+i-j+1 = A_jA_{j+1}+A_{j+1}A_{j+2}+\dots+A_{i-1}A_i +1\ge A_iA_j+1$. By the claim, an equilateral triangle of sidelength $1$ is contained in $A_iA_{i+1}\dots A_j$.