I knew my solution is wrong : ( @below what is the construction?

I feel sorry about deleting my previous comment. But now I'm almost sure that the answer is $\frac{4\cos^2(10^\circ)}{\sqrt{3}}$. Perhaps I will upload an explicit solution later. The following proof is based on [1] and [2] to a large extent. In our proof, the lemma below plays a key role. We say convex closed set $C$ is a cover of a set of triangles $\mathcal{S}$ if all triangles in $\mathcal{S}$ is in the interior or on the boundary of $C$ in congruence sense. This question asks for the smallest equilateral triangular cover of a given set. Lemma. Let $\mathcal{S}$ be a set of triangles. Given a cover $C$, a triangle $\triangle \in \mathcal{S}$, and the smallest cover of $\mathcal{S}$ that is similar to $\triangle$, named $C_\triangle$. If $\frac{\mathrm{Area} \, C_\triangle}{\mathrm{Area} \, \triangle} = \left( \frac{\mathrm{Area} \, C}{\mathrm{Area} \, \triangle} \right)^2$, $C$ is the cover of $\mathcal{S}$ with smallest area. [2] Proof of Lemma. Since the entire proof is explained clearly in paper [2], here I only provide an outline. Choose a triangle $\triangle$ and a cover $C$. Let $C_\triangle'$ be the similar triangle each of whose sides tangent to $C$. We naturally deduce that $\mathrm{Area} \, \mathrm{C_\triangle'} \ge C$. Hence the ratio of similar triangles are larger than $\mathrm{Area} \, C : \mathrm{Area} \, \mathrm{\triangle}$. We can also note that the area of the hexagon formed by vertices of $\triangle$ and tangent points on $C$, is less than the area of $C$. By simple computation, we complete the proof. Now we construct the smallest cover of the set of all triangles whose circumscribed circle is the given unit circle $\mathcal{S}$. We first draw an equilateral triangle with side lengths $\sqrt{3}$ with an isosceles triangle whose base angles are $80^{\circ}$ and legs measure $2 \sin 80^{\circ}$ long. One of its legs coincides with a side of the equilateral triangle, and the left vertice lies on the adjacent side of the equilateral triangle. We claim that the convex hull of this figure, a triangle is precisely the smallest cover $C$ based on the lemma above (Note that in this part of the proof, the set of triangles $\mathcal{S}$ in lemma altered.) Then let $\triangle$ in the lemma be an equilateral triangle whose circumscribed circle is the given unit circle. Now we can compute the area of the smallest cover similar to it by the lemma and therefore deduce the side length of it. This is precisely what we want. By law of sines, we can calculate that the longest side of $C$ is $2 \left( \sin 80^{\circ} + \frac{\sin^2 20^{\circ}}{\sin 60^{\circ}} \right)$. So the area is $\mathrm{Area} \, C = \frac{1}{2} \times \sqrt{3} \times 2 \left( \sin 80^{\circ} + \frac{\sin^2 20^{\circ}}{\sin 60^{\circ}} \right) \times \sin 60^{\circ} = \sqrt{3} \cos^2 10^{\circ}$. So $$\frac{\sqrt{3}}{4} a^2 = \mathrm{Area} \, C_\triangle = \frac{\mathrm{Area}^2 \, C}{\mathrm{Area} \, \triangle} = \frac{4\cos^4 10^{\circ}}{\sqrt{3}} .$$Now we finally achieve that $a = \frac{4\cos^2(10^\circ)}{\sqrt{3}}$. Comment. A highly similar problem is mentioned in [1] where the author claimed it was an old problem proposed by Smith in 1922. That paper was referred, yet out of my search. [1] Wetzel, J. E. (1997). The smallest equilateral cover for triangles of perimeter two. Mathematics Magazine, 70(2), 125-130. [2] Park, J. W., & Cheong, O. (2021). Smallest universal covers for families of triangles. Computational Geometry, 92, 101686.

peterdawson wrote: I knew my solution is wrong : ( @below what is the construction? I've updated my solution.

I think your answer is right for I also figure out the same number, but the answer is indeed weird...

Here is the main thought.(Not able to use LaTeX, I can only post this out.) If there are two angles not beyond 60 degrees, use the third side to create a equilateral triangle. If their is only one, while the angle is bigger than 20 degrees, use the height to the shortest side as the height of the new equilateral triangle. Otherwise, use the longest side of the original triangle, and another side passing through the left point.

Solved with Elliott Liu, Isaac Zhu, Kevin Wu, and Luke Robitaille. The answer is \(\frac4{\sqrt3}\sin^280^\circ\). Chapter I: Proof of optimality We will show that the smallest equilateral triangle containing a triangle \(ABC\) inscribed in the unit circle with \(\angle A=20^\circ\) and \(\angle B=\angle C=80^\circ\) has side length \(\frac4{\sqrt3}\sin^280^\circ\). Let \(ABC\) be inscribed in the smallest possible equilateral triangle \(PQR\). Claim: Each of \(A\), \(B\), \(C\) lies on a side of \(\triangle PQR\). Proof. If the sides of \(\triangle PQR\) contain exactly two of \(A\), \(B\), \(C\), it is possible to apply some rotation on \(\triangle ABC\) so that only one of \(A\), \(B\), \(C\) lies on the sides of \(\triangle PQR\). Now if the sides of \(\triangle PQR\) only contain at most one of \(A\), \(B\), \(C\), we may dilate \(\triangle PQR\) and make it smaller. \(\blacksquare\) Claim: One of the lines \(AB\), \(BC\), \(CA\) coincides with one of the lines \(PQ\), \(QR\), \(RP\). Proof. Otherwise we may assume without loss of generality \(A\), \(B\), \(C\) lie on the interiors of segments \(QR\), \(RP\), \(PQ\). Let \(M\) be the Miquel point of \(A\), \(B\), \(C\) with respect to \(\triangle PQR\). Then \(\angle MAR=\angle MBP=\angle MCQ\), so the three angles are either all acute or right/obtuse, thus some rotation on \(\triangle ABC\) results in the three vertices no longer all lying on the sides of \(\triangle PQR\). From here, return to Claim 1. \(\blacksquare\) Finally, we may check that the heights in the two potentially ``optimal'' \(\triangle PQR\) are at least \(2\sin^280^\circ\). [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pen pri=blue; pen pri2=lightblue; pen sec=lightred; pen tri=heavygreen; pen fil=invisible; pen fil2=invisible; pen sfil=invisible; pen tfil=invisible; pair A,B,C,M1,Q1,R1,M2,Q2,R2; A=dir(90); B=dir(250); C=dir(-70); M1=(B+C)/2; Q1=M1+(-abs(A-M1)/sqrt(3),0); R1=2M1-Q1; M2=reflect(A,C)*M1; Q2=M2+rotate(90)*((A-M2)/sqrt(3)); R2=2M2-Q2; filldraw(unitcircle,fil2,pri2); fill(A--Q1--R1--cycle,sfil); fill(A--Q2--R2--cycle,tfil); draw(B--Q1--A--R1--C,sec); draw(B--Q2--R2--A,tri); draw(A--M1,sec+linewidth(.4)+dashed); draw(A--M2,tri+linewidth(.4)+dashed); filldraw(A--B--C--cycle,fil,pri); dot("\(A\)",A,N); dot("\(B\)",B,SW); dot("\(C\)",C,dir(300)); [/asy][/asy] Chapter II: Proof of sufficiency Now we show any triangle \(ABC\) may be inscribed in an equilateral triangle \(PQR\) of side length \(\frac4{\sqrt3}\sin^280^\circ\). Without loss of generality \(\angle A\le\angle B\le\angle C\). First case If \(\angle B\le60^\circ\), then we may construct an equilateral triangle with base \(AB\) containing point \(C\). This triangle has side length at most \(AB\le2<\frac4{\sqrt3}\sin^280^\circ\). Second case If \(\angle A\ge20^\circ\) but \(\angle B,\angle C>60^\circ\), I claim the equilateral triangle \(AQR\) with \(B\) and \(C\) on segment \(QR\) has height at most \(2\sin^280^\circ\). (This triangle exists since \(\angle B,\angle C>60^\circ\).) [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pen pri=blue; pen pri2=lightblue; pen sec=lightred; pen tri=orange; pen fil=invisible; pen fil2=invisible; pen sfil=invisible; pen tfil=invisible; pair A,B,C,M1,Q1,R1,M2,Q2,R2,Ap,Qp,Rp; A=dir(90); B=dir(250); C=dir(-70); M1=(B+C)/2; Q1=M1+(-abs(A-M1)/sqrt(3),0); R1=2M1-Q1; M2=reflect(A,C)*M1; Q2=M2+rotate(90)*((A-M2)/sqrt(3)); R2=2M2-Q2; /* Ap=A-(.4,0); Qp=Q1+Ap-A; Rp=R1+Ap-A; */ Ap=dir(120); Qp=extension(B,C,Ap,Ap+Q1-A); Rp=extension(B,C,Ap,Ap+R1-A); filldraw(unitcircle,fil2,pri2); fill(B--Ap--C--cycle,fil2); draw(B--Ap--C,pri2); draw(Rp--Ap--Qp--Q1,tri); fill(Ap--Qp--Rp--cycle,tfil); fill(A--Q1--R1--cycle,sfil); draw(B--Q1--A--R1--C,sec); draw(A--M1,sec+linewidth(.4)+dashed); filldraw(A--B--C--cycle,fil,pri); dot("\(A'\)",A,N); dot("\(B\)",B,S); dot("\(C\)",C,S); dot("\(A\)",Ap,N); dot("\(Q'\)",Q1,S); dot("\(R'\)",R1,SE); dot("\(Q\)",Qp,SW); dot("\(R\)",Rp,S); [/asy][/asy] Let \(A'\) be the midpoint of arc \(BAC\) on the circumcircle. Then the distance from \(A\) to \(\overline{BC}\) is at most the distance from \(A'\) to \(\overline{BC}\), which is maximized when \(\angle A'\) is minimized, a case we've settled in Chapter I. Third case If \(\angle A\le20^\circ\) and \(\angle B,\angle C>60^\circ\), I claim \(\triangle AQR\) shown below works. [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pen pri=blue; pen pri2=lightblue; pen sec=lightred; pen tri=orange; pen fil=invisible; pen fil2=invisible; pen sfil=invisible; pen tfil=invisible; pair R,A,Q,C,B,M; R=dir(90); A=dir(210); Q=dir(330); C=extension(Q,R,A,A+dir(20)); B=intersectionpoint(A--Q,circle(A,abs(C-A))); M=(Q+R)/2; draw(A--M,red+linewidth(.4)+dashed); fill(A--R--Q--cycle,sfil); draw(A--R--Q--B,sec); filldraw(A--B--C--cycle,fil,pri); dot("\(A\)",A,SW); dot("\(R\)",R,N); dot("\(Q\)",Q,SE); dot("\(B\)",B,S); dot("\(C\)",C,dir(30)); dot("\(M\)",M,dir(30)); [/asy][/asy] It will suffice to show \(AM\le2\sin^280^\circ\). Indeed since \(\angle B\le\angle C\), we have \[\angle A+\angle B\le 90^\circ+\tfrac12\angle A\le100^\circ,\]It follows that \[AM=AC\sin(60^\circ+\angle A)=2\sin B\sin(60^\circ+A)\le2\sin^280^\circ,\]as desired.

mofumofu wrote: Find the smallest positive real constant $a$, such that for any three points $A,B,C$ on the unit circle, there exists an equilateral triangle $PQR$ with side length $a$ such that all of $A,B,C$ lie on the interior or boundary of $\triangle PQR$. Let $T$ be the fermat Torricelli point of $\triangle ABC$ which lies on the unit circle. and also consider that $a \ge b \ge c$. let the circumcircles of $\triangle BTC$, $\triangle ATC$ and $\triangle ATB$ be $ \omega_a$, $\omega_b$ and $\omega_c$. now we can construct all the equilateral triangles which the points $\triangle ABC$ lie on the boundary of the triangle $\triangle PQR$. just consider a point $P$ on $\omega_a$ and draw the line $PC$ to intersect $\omega_b$ at a second point named $Q$ and finally let $R=QA \cap PB$. $\textbf{Note:}$ It is trivial that if we want the equilateral triangle to be optimal we should have all the vertices of $\triangle ABC$ to be on the boundery of $\triangle PQR$. now not that the length of $PQ$ gets minimized when $P$ is located on $C$.(because we considered that $a \ge b \ge c$) so we want the maximum of the minimums of $PQ$ through all configurations of $\triangle ABC$ on the unit circle. so we want to maximize $\frac{2b \times Sin(120^{\circ}-\angle C)}{\sqrt{3}}$ which is $\frac{4 Sin(80^\circ)^2}{\sqrt{3}}$. $\textbf{Construction:}$ the construction is just to draw a random $\triangle PQR$ as defined and then put $P$ on $C$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7.082294608893418cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ real xmin = -3.247866531824416, xmax = 3.8344280770690027, ymin = -1.6611958430221936, ymax = 1.9313344936534758; /* image dimensions */ pen ffdxqq = rgb(1.,0.8431372549019608,0.); pair C = (0.971746448787996,-0.23602720027132176), A = (-0.7949738756566301,0.606643665608961), B = (-0.9846776079534837,-0.17438465642081474), T = (-0.6643622858207507,0.16136675414014057), R = (-1.2736732483979034,-0.16527905051839537), Q = (-0.0896937048481774,1.7439373745260915); draw(A--B--C--cycle, linewidth(0.4) + ffdxqq); /* draw figures */ draw(circle((0.,0.), 1.), linewidth(0.4)); draw(A--B, linewidth(0.4) + ffdxqq); draw(B--C, linewidth(0.4) + ffdxqq); draw(C--A, linewidth(0.4) + ffdxqq); draw(circle((-1.1152891977893635,0.27089225504800574), 0.4640376227060344), linewidth(0.4)); draw(circle((0.33164441219280083,0.6953164601192632), 1.1301024869466707), linewidth(0.4)); draw(T--C, linewidth(0.4)); draw(circle((-0.02426024922555557,-0.769976906250444), 1.1301024869466703), linewidth(0.4)); draw(T--B, linewidth(0.4)); draw(B--R, linewidth(0.4)); draw(T--R, linewidth(0.4)); draw(R--Q, linewidth(0.4)); draw(Q--C, linewidth(0.4)); /* dots and labels */ dot(C,linewidth(4.pt)); label("$C \thickspace {or} \thickspace P$", (0.9869517077134261,-0.20106134384888819), NE * labelscalefactor); dot(A,linewidth(4.pt)); label("$A$", (-0.8714012912344239,0.689577881453304), NE * labelscalefactor); dot(B,linewidth(4.pt)); label("$B$", (-1.0298323072737567,-0.3466466018309773), NE * labelscalefactor); dot(T,linewidth(4.pt)); label("$T$", (-0.6358957268516318,0.2100029139828928), NE * labelscalefactor); dot(R,linewidth(4.pt)); label("$R$", (-1.3381305006475936,-0.27813589219234713), NE * labelscalefactor); dot(Q,linewidth(4.pt)); label("$Q$", (-0.07068237233293088,1.7771853969665579), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]