Posting this too I gave a lot of unnecesary detailing lol

$\textbf{Diagram : }$[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.6, xmax = 10.6, ymin = -6.36, ymax = 6.4; /* image dimensions */ pen ttttqq = rgb(0.2,0.2,0); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-5.88,4.5)--(2.0402325145321587,-2.588486780423776), linewidth(2) + ttttqq); draw((-5.88,4.5)--(-5.68,-2.8), linewidth(2) + ttttqq); draw((-5.68,-2.8)--(2.0402325145321587,-2.588486780423776), linewidth(2) + ttttqq); draw((-5.78,0.85)--(-1.9198837427339206,0.955756609788112), linewidth(2) + ttttqq); draw((-1.9198837427339206,0.955756609788112)--(-1.8198837427339205,-2.694243390211888), linewidth(2) + ttttqq); draw((-5.88,4.5)--(-8.990992723813202,2.588397459621555), linewidth(2) + ttttqq); draw((-8.990992723813202,2.588397459621555)--(-5.78,0.85), linewidth(2) + ttttqq); draw((-8.990992723813202,2.588397459621555)--(0.20176229659374192,-5.984323825671564), linewidth(2) + ttttqq); draw((0.20176229659374192,-5.984323825671564)--(-1.9198837427339206,0.955756609788112), linewidth(2) + ttttqq); draw((-1.9198837427339206,0.955756609788112)--(-8.990992723813202,2.588397459621555), linewidth(2) + ttttqq); draw((-1.8198837427339205,-2.694243390211888)--(0.20176229659374192,-5.984323825671564), linewidth(2) + wrwrwr); /* dots and labels */ dot((-5.88,4.5),linewidth(4pt) + dotstyle); label("$A$", (-5.8,4.66), NE * labelscalefactor); dot((-5.68,-2.8),linewidth(4pt) + dotstyle); label("$B$", (-5.7,-2.64), NE * labelscalefactor); dot((2.0402325145321587,-2.588486780423776),linewidth(4pt) + dotstyle); label("$C$", (2.12,-2.42), NE * labelscalefactor); dot((-1.9198837427339206,0.955756609788112),linewidth(4pt) + dotstyle); label("$C_0$", (-1.84,1.12), NE * labelscalefactor); dot((-1.8198837427339205,-2.694243390211888),linewidth(4pt) + dotstyle); label("$A_0$", (-1.74,-3.54), NE * labelscalefactor); dot((-5.78,0.85),linewidth(4pt) + dotstyle); label("$B_0$", (-5.7,1), NE * labelscalefactor); dot((0.20176229659374192,-5.984323825671564),linewidth(4pt) + dotstyle); label("$C_2$", (0.28,-5.82), NE * labelscalefactor); dot((-8.990992723813202,2.588397459621555),linewidth(4pt) + dotstyle); label("$C_1$", (-7.90,2.74), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] $\textbf{Solution :}$ We know that $\angle B_0AC_1 = \angle AB_0C_1 = 60^{\circ}$ and $\angle A_0BC_2 = \angle BC_2A_0$ By Mid-Point Theroem : $B_0C_0 = \frac{1}{2} \cdot BC$ and $B_0C_0 \parallel BC \Longrightarrow \angle AB_0C_0 = 90^{\circ}$ and $B_0C_0 = A_0B = A_0C ----(i)$ By Mid-Point Theorem : $C_0A_0 = \frac{1}{2} \cdot AC$ and $C_0A_0 \parallel AC \Longrightarrow \angle BA_0C_0 = 90^{\circ}$ and $C_0A_0 = B_0A = B_0C_1 ---- (ii)$ From $(i)$ and $(ii)$ we get that $\triangle C_0A_0C_2 \cong \triangle C_0B_0C_1 \Longrightarrow \angle A_0C_2C_0 = \angle B_0C_0C_1 \stackrel{say}= \theta$ Thus, $\angle C_1C_0C_2 = 30^{\circ} - \theta + 90^{\circ} + \theta = 120^{\circ}$ Now we conclude that $\angle C_0C_1C_2 = \frac{180 - 12}{2} = 30^{\circ}$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.740597825698078, xmax = 24.117913912023365, ymin = -11.078891592498186, ymax = 13.262086613516368; /* image dimensions */ pen ffqqtt = rgb(1,0,0.2); pen qqzzcc = rgb(0,0.6,0.8); pen wwffqq = rgb(0.4,1,0); pen ttffcc = rgb(0.2,1,0.8); pen ffqqff = rgb(1,0,1); pen xfqqff = rgb(0.4980392156862745,0,1); pen wwccff = rgb(0.4,0.8,1); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen ttffqq = rgb(0.2,1,0); draw(arc((-11.154390903785789,2.7728163995621937),0.7180229559296385,-21.79364789786335,8.206352102136645)--(-11.154390903785789,2.7728163995621937)--cycle, linewidth(0.8)); draw(arc((-4.832271156362403,5.890172411393869),0.7180229559296385,326.2473141606852,356.2473141606852)--(-4.832271156362403,5.890172411393869)--cycle, linewidth(0.8)); draw(arc((-4.832271156362405,5.89017241139387),0.7180229559296385,-153.7526858393148,-123.7526858393148)--(-4.832271156362405,5.89017241139387)--cycle, linewidth(0.8)); draw(arc((-7.093427863778636,2.506455636466062),0.7180229559296385,176.2473141606852,326.2473141606852)--(-7.093427863778636,2.506455636466062)--cycle, linewidth(0.8)); draw(arc((0.22498763865054983,2.5106771976648057),0.7180229559296385,-123.7526858393148,26.24731416068522)--(0.22498763865054983,2.5106771976648057)--cycle, linewidth(0.8)); draw((-4.83002717029117,2.0000422728455383)--(-4.322308353576255,2.000335145811798)--(-4.322601226542514,2.508053962526713)--(-4.830320043257429,2.507761089560453)--cycle, linewidth(0.8)); /* draw figures */ draw((-4.832271156362405,5.89017241139387)--(-9.354584571194868,-0.8772611384617462), linewidth(0.8) + ffqqtt); draw((-9.354584571194868,-0.8772611384617462)--(5.282246433663505,-0.8688180160642585), linewidth(0.8) + ffqqtt); draw((-11.154390903785789,2.7728163995621937)--(-9.354584571194868,-0.8772611384617462), linewidth(0.8)); draw((-11.154390903785789,2.7728163995621937)--(-7.093427863778636,2.506455636466062), linewidth(0.8) + qqzzcc); draw((5.680345743214318,5.2006441807937716)--(5.282246433663505,-0.8688180160642585), linewidth(0.8) + wwffqq); draw((0.22498763865054983,2.5106771976648057)--(5.680345743214318,5.2006441807937716), linewidth(0.8) + ttffcc); draw((-2.0361690687656817,-0.8730395772630024)--(-11.154390903785789,2.7728163995621937), linewidth(0.8) + qqzzcc); draw((-11.154390903785789,2.7728163995621937)--(5.680345743214318,5.2006441807937716), linewidth(0.8) + blue); draw(circle((-7.093427863778636,2.506455636466062), 4.069689038295342), linewidth(0.8) + linetype("4 4") + green); draw(circle((0.22498763865054983,2.5106771976648057), 6.08250395966608), linewidth(0.8) + linetype("4 4") + ffqqff); draw((-11.154390903785789,2.7728163995621937)--(-4.832271156362405,5.89017241139387), linewidth(0.8) + blue); draw((-2.0361690687656817,-0.8730395772630024)--(5.680345743214318,5.2006441807937716), linewidth(0.8) + ttffcc); draw((-4.832271156362405,5.89017241139387)--(-4.828368930152454,-0.8746502322729661), linewidth(0.8) + blue); draw((-4.828368930152454,-0.8746502322729661)--(5.680345743214318,5.2006441807937716), linewidth(0.8) + xfqqff); draw((-11.154390903785789,2.7728163995621937)--(-4.828368930152454,-0.8746502322729661), linewidth(0.8) + blue); draw((-4.832271156362405,5.89017241139387)--(5.680345743214318,5.2006441807937716), linewidth(0.8) + blue); draw((0.22498763865054983,2.5106771976648057)--(-4.828368930152454,-0.8746502322729661), linewidth(0.8) + xfqqff); draw((-4.832271156362405,5.89017241139387)--(-2.0361690687656817,-0.8730395772630024), linewidth(0.8) + wwffqq); draw((-7.093427863778636,2.506455636466062)--(0.22498763865054983,2.5106771976648057), linewidth(0.8) + xfqqff); draw(circle((-2.0361690687656817,-0.8730395772630026), 7.318416720013612), linewidth(0.8) + xfqqff); draw((-7.093427863778636,2.506455636466062)--(-2.0361690687656817,-0.8730395772630024), linewidth(0.8) + qqzzcc); draw((0.22498763865054983,2.5106771976648057)--(-2.0361690687656817,-0.8730395772630024), linewidth(0.8) + ttffcc); draw((-7.093427863778636,2.506455636466062)--(-4.828368930152454,-0.8746502322729661), linewidth(0.8) + wwccff); draw((-4.832271156362405,5.89017241139387)--(5.282246433663505,-0.8688180160642585), linewidth(0.8) + ffqqtt); draw(circle((-3.4342201125640432,2.508566417065433), 3.6592083600068075), linewidth(0.8) + linetype("2 2") + wvvxds); draw(shift((-3.437876091805787,8.846500157618962))*xscale(9.820093443363424)*yscale(9.820093443363424)*arc((0,0),1,218.2063521021366,338.2063521021367), linewidth(0.8) + linetype("4 4") + ttffqq); /* dots and labels */ dot((-4.832271156362405,5.89017241139387),dotstyle); label("$C$", (-4.746608916348103,6.129725251282014), NE * labelscalefactor); dot((-9.354584571194868,-0.8772611384617462),dotstyle); label("$A$", (-9.270153538704825,-0.6436246329875217), NE * labelscalefactor); dot((5.282246433663505,-0.8688180160642585),dotstyle); label("$B$", (5.377514762259799,-0.619690534456534), NE * labelscalefactor); dot((0.22498763865054983,2.5106771976648057),linewidth(4pt) + dotstyle); label("$A_0$", (0.32741997222134206,2.707149161350765), NE * labelscalefactor); dot((-7.093427863778636,2.506455636466062),linewidth(4pt) + dotstyle); label("$B_0$", (-6.99641417826097,2.707149161350765), NE * labelscalefactor); dot((-2.0361690687656817,-0.8730395772630024),linewidth(4pt) + dotstyle); label("$C_0$", (-1.9463193882225132,-0.6914928300494972), NE * labelscalefactor); dot((-11.154390903785789,2.7728163995621937),linewidth(4pt) + dotstyle); label("$C_1$", (-11.615695194741644,2.204533092200022), NE * labelscalefactor); dot((5.680345743214318,5.2006441807937716),linewidth(4pt) + dotstyle); label("$C_2$", (5.784394437286594,5.387768196821394), NE * labelscalefactor); label("$30^\circ$", (-10.299319775537308,2.491742274571875), NE * labelscalefactor); dot((-4.828368930152454,-0.8746502322729661),linewidth(4pt) + dotstyle); label("$F$", (-4.722674817817115,-0.6914928300494972), NE * labelscalefactor); label("$30^\circ$", (-4.507267931038224,5.651043280662259), NE * labelscalefactor); label("$30^\circ$", (-5.225290886967862,5.507438689476333), NE * labelscalefactor); label("$150^\circ$", (-7.307557459163814,1.989126205421132), NE * labelscalefactor); label("$150^\circ$", (0.42315636634529386,2.084862599545083), NE * labelscalefactor); label("$90^\circ$", (-4.626938423693163,2.1087966980760706), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We claim that the answer is $30^\circ$. Firstly we'll define the measure of some angles that we'll use earlier. $\angle B_0C_0A_0=90^ \circ$ which follows since $\angle C_0B_0C=C_0A_0C=90$ By thales. \textcolor{blue}{Proof:} Clearly we have that $\angle CC_1A= 90^\circ$ since $C_1B_0=B_0A=B_0C$. Now let $\omega_1$ the circumcircle of $\Delta AC_1C$ and let $\omega_1\cap BC=F$ Now clearly since we have that $AC$ is a diameter of this circle $\implies \angle AFC=90^\circ \implies F$ is foot of altitude from $A$. Similiarly let $\omega_2$ be the circumcircle of $\Delta CC_2B$. We can similiarly obtain $F \in \omega_2$.Now observe that $$B_0C_0= \frac{1}{2} BC=A_0C=A_0C_2$$by thales theorem. Similiarly $$A_0C_0=\frac{1}{2}AC=B_0A=B_0C_1$$and since $$\angle C_1B_0C_0= \angle C_1B_0A+\angle AB_0C_0=60+90=150=\angle C_0A_0B=\angle BA_0C_2=\angle C_0A_0C_2 \implies \Delta C_1B_0C_0 \cong \Delta C_0A_0C_2$$by $SAS$ congruency. Now by this we have that $C_0C_1=C_0C_2$ and that $$\angle C_1C_0C_2= \angle C_1C_0B_0+\angle B_0C_0A_0+ \angle A_0C_0C_2=90^\circ+ (\angle C_1C_0B_0+\angle A_0C_0C_2)=90^\circ+30^\circ=120^\circ$$where $(\angle C_1C_0B_0+\angle A_0C_0C_2)=30^\circ$ follows from the congruency. Hence we have $C_0C_1C_2=\frac{1}{2}(180^\circ-120^\circ)=30^\circ$. Done . $\blacksquare$.

Quite Similar to USATST 2000. Sharygin 2020 CR P1 wrote: Let $ABC$ be a triangle with $\angle C=90^\circ$, and $A_0$, $B_0$, $C_0$ be the mid-points of sides $BC$, $CA$, $AB$ respectively. Two regular triangles $AB_0C_1$ and $BA_0C_2$ are constructed outside $ABC$. Find the angle $C_0C_1C_2$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.37, xmax = 11.05, ymin = -6.25, ymax = 7.05; /* image dimensions */ pen ffdxqq = rgb(1,0.8431372549019608,0); /* draw figures */ draw((-0.43,3.67)--(-6.01,0.03), linewidth(1)); draw((-0.43,3.67)--(1.93,0.05), linewidth(1)); draw((1.93,0.05)--(-6.01,0.03), linewidth(1)); draw((-0.43,3.67)--(2.87,1.99), linewidth(1) + blue); draw((2.87,1.99)--(1.93,0.05), linewidth(1) + blue); draw((-0.43,3.67)--(-6.33,3.29), linewidth(1) + blue); draw((-6.33,3.29)--(-6.01,0.03), linewidth(1) + blue); draw((-6.33,3.29)--(-3.22,1.85), linewidth(1) + ffdxqq); draw((-6.33,3.29)--(-2.04,0.04), linewidth(1) + green); draw((-2.04,0.04)--(-3.22,1.85), linewidth(1) + red); draw((-2.04,0.04)--(2.87,1.99), linewidth(1) + green); draw((2.87,1.99)--(0.75,1.86), linewidth(1) + red); draw((0.75,1.86)--(-2.04,0.04), linewidth(1) + ffdxqq); /* dots and labels */ dot((-6.01,0.03),dotstyle); label("$A$", (-6.29,-0.29), NE * labelscalefactor); dot((1.93,0.05),dotstyle); label("$B$", (2.07,-0.13), NE * labelscalefactor); dot((-0.43,3.67),dotstyle); label("$C$", (-0.35,3.87), NE * labelscalefactor); dot((-3.22,1.85),dotstyle); label("$B_0$", (-3.33,2.15), NE * labelscalefactor); dot((0.75,1.86),dotstyle); label("$A_0$", (0.73,2.15), NE * labelscalefactor); dot((2.87,1.99),dotstyle); label("$C_2$", (2.95,2.19), NE * labelscalefactor); dot((-6.33,3.29),dotstyle); label("$C_1$", (-6.25,3.49), NE * labelscalefactor); dot((-2.04,0.04),dotstyle); label("$C_0$", (-2.13,-0.31), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] First we will prove that triangles $C_1B_0C_0$ and $C_0A_0C_2$ are congruent. For this just notice that $$\begin{cases} B_0C_0=BA_0=A_0C_2 \\ A_0C_0=AB_0=B_0C_1 \\ \angle C_1B_0C_0=\angle C_1B_0A+\angle AB_0C_0=60^\circ+\angle ACB=\angle BA_0C_2+\angle C_0A_0B=\angle C_2A_0C_0\end{cases} \implies \triangle C_1B_0C_0\cong\triangle C_0A_0C_2$$So from here we get that $C_0C_1=C_0C_2$. Now let $\angle CAB=\theta\implies\angle ABC=90^\circ-\theta$ and $\angle B_0C_0C_1=\angle A_0C_2C_0=\alpha$. So, we get that $\angle C_1C_0A=90^\circ-\theta-\alpha$ and $\angle C_2C_0B=180^\circ-(60^\circ-\alpha)-(150^\circ-\theta)=\theta+\alpha-30^\circ$. So, from here it's easy to get that $\angle C_1C_0C_2=180^\circ-\angle C_1C_0A-\angle C_2C_0B=120^\circ$. Hence, $\angle C_0C_1C_2=30^\circ$. $\blacksquare$

so here is my solution

Very complex bashable!!

My 550th post!

I hope this works. Since $AB_0=B_0C=C_1B_0$ it can be seen that $C_1$ lies on the circle centered at $B_0$ with radius $B_0A$. Same for $C_2$. Since $CC_1=AC*\frac{3^{1/2}}{2} $, $CC_2=CB* \frac{3^{1/2}}{2} $ and $\angle C_1B_0C_0=\angle C_1CC_2=150^\circ$. We get that triangles $C_1B_0C_0$ and $C_1CC_2$ are similar, hence $\angle C_0C_1C_2=30^\circ$.

Gotcha ans 30