The answer is the point $X$ such that $\boxed{XC \perp AB}$. [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(8); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.2243323376885145, xmax = 13.88365542137765, ymin = -5.922456616151421, ymax = 4.148826666177998; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); draw((3.587052950465511,2.4419513357090383)--(1.3742018932134419,-0.516290160847325)--(5.476150330531299,-1.7643148616665305)--cycle, linewidth(1.)); draw((4.06117451276288,2.165255786734233)--(3.587052950465511,2.4419513357090383)--(1.3742018932134419,-0.516290160847325)--(2.2111745127628804,-1.004744213265767)--cycle, linewidth(1.5) + xdxdff); /* draw figures */ draw((4.56,3.02)--(0.86,-3.32), linewidth(1.)); draw((0.86,-3.32)--(10.02,-3.52), linewidth(1.)); draw((10.02,-3.52)--(4.56,3.02), linewidth(1.)); draw(circle((5.476150330531299,-1.7643148616665305), 4.871242154081014), linewidth(1.)); draw((2.508651125842874,2.0987083836517533)--(10.02,-3.52), linewidth(1.)); draw(circle((3.587052950465511,2.4419513357090383), 1.1317094236094491), linewidth(1.)); draw(circle((1.3742018932134419,-0.516290160847325), 2.850472320361278), linewidth(1.)); draw((3.587052950465511,2.4419513357090383)--(1.3742018932134419,-0.516290160847325), linewidth(1.) + zzttqq); draw((1.3742018932134419,-0.516290160847325)--(5.476150330531299,-1.7643148616665305), linewidth(1.) + zzttqq); draw((5.476150330531299,-1.7643148616665305)--(3.587052950465511,2.4419513357090383), linewidth(1.) + zzttqq); draw((0.86,-3.32)--(2.508651125842874,2.0987083836517533), linewidth(1.)); draw((2.508651125842874,2.0987083836517533)--(4.56,3.02), linewidth(1.)); draw((4.06117451276288,2.165255786734233)--(3.587052950465511,2.4419513357090383), linewidth(1.) + xdxdff); draw((3.587052950465511,2.4419513357090383)--(1.3742018932134419,-0.516290160847325), linewidth(1.) + xdxdff); draw((1.3742018932134419,-0.516290160847325)--(2.2111745127628804,-1.004744213265767), linewidth(1.) + xdxdff); draw((2.2111745127628804,-1.004744213265767)--(4.06117451276288,2.165255786734233), linewidth(1.) + xdxdff); /* dots and labels */ dot((4.56,3.02),dotstyle); label("$A$", (4.618661193674296,3.1666199996334985), NE * labelscalefactor); dot((0.86,-3.32),dotstyle); label("$B$", (0.7044943284894912,-3.210393432409147), dir(260) * 3); dot((10.02,-3.52),dotstyle); label("$C$", (10.072107163145256,-3.3716512433343633), dir(-50) * 1.5); dot((2.508651125842874,2.0987083836517533),dotstyle); label("$X$", (2.24377343277565,2.331011343021014), dir(90) * 0.2); dot((3.5623490255257604,1.310511573468466),linewidth(4.pt) + dotstyle); label("$D$", (3.724413333089003,1.334144875483313), dir(-108) * 2.95); dot((5.476150330531299,-1.7643148616665305),linewidth(4.pt) + dotstyle); label("$O$", (5.527568855252789,-1.6417947261365882), NE * labelscalefactor); dot((3.587052950465511,2.4419513357090383),linewidth(4.pt) + dotstyle); label("$O_1$", (3.651114328122995,2.565568158912238), dir(120) * 2.35); dot((1.3742018932134419,-0.516290160847325),linewidth(4.pt) + dotstyle); label("$O_2$", (1.437484378149567,-0.395711641714462), dir(135) * 0.7); dot((4.06117451276288,2.165255786734233),linewidth(4.pt) + dotstyle); label("$M$", (4.120227959905444,2.2870319400414094), dir(-60) * 1.5); dot((2.2111745127628804,-1.004744213265767),linewidth(4.pt) + dotstyle); label("$N$", (2.273093034762053,-0.8941448754833125), dir(-60) * 1.5); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that the radical axis of two non-concentric circles is perpendicular to the line connecting their centers, so $$XA \perp OO_1, \quad XB \perp OO_2, \quad XD \perp O_1 O_2. $$This means $$\measuredangle (\overline{O_1 O_2}, \overline{O_1 O}) = \measuredangle (\overline{XD}, \overline{XA}) = \measuredangle CXA = \measuredangle CBA,$$$$\measuredangle (\overline{O_2 O}, \overline{O_2 O_1}) = \measuredangle (\overline{XB}, \overline{XD}) = \measuredangle BXC = \measuredangle BAC$$so $\Delta O O_1 O_2 \sim \Delta CBA$. Thus, all possible $\Delta O O_1 O_2$ are similar to each other, so it suffices to minimize $O_1 O_2$. Let $M,N$ be the midpoints of $AD,BD$. Note that $$O_1 M, O_2 N \perp MN \implies O_1 O_2 \geq MN$$with equality holding if $O_1 O_2 NM$ is a rectangle. In this case, $O_1 O_2 \parallel MN$, and since $O_1 O_2 \perp XD$, this is equivalent to $XC \perp AB$. $\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); 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 = -9.19, xmax = 12.01, ymin = -8.64, ymax = 5.6; /* image dimensions */ pen qqffff = rgb(0,1,1); /* draw figures */ draw(circle((1.1296663605334942,-1.841279453894974), 5.019806191877154), linewidth(0.8) + red); draw((-0.63,2.86)--(-2.37,-5.44), linewidth(0.8)); draw((-2.37,-5.44)--(4.67,-5.4), linewidth(0.8)); draw((4.67,-5.4)--(-0.63,2.86), linewidth(0.8)); draw((-3.593868592709285,-0.142258633097903)--(4.67,-5.4), linewidth(0.8)); draw(circle((-1.5469223379724755,0.8010835629503406), 2.2538596825503965), linewidth(0.8) + green); draw(circle((-3.978946076657836,-3.0214560388192346), 2.9048343102071406), linewidth(0.8) + qqffff); draw((-1.5469223379724755,0.8010835629503406)--(1.1296663605334942,-1.841279453894974), linewidth(0.8)); draw((1.1296663605334942,-1.841279453894974)--(-3.978946076657836,-3.0214560388192346), linewidth(0.8)); draw((-3.978946076657836,-3.0214560388192346)--(-1.5469223379724755,0.8010835629503406), linewidth(0.8)); draw((-3.593868592709285,-0.142258633097903)--(-0.63,2.86), linewidth(0.8)); draw((-3.593868592709285,-0.142258633097903)--(-2.37,-5.44), linewidth(0.8)); draw((-2.1119342963546424,1.3588706834510484)--(-1.5469223379724755,0.8010835629503406), linewidth(0.8)); /* dots and labels */ dot((-0.63,2.86),dotstyle); label("$A$", (-0.55,3.06), NE * labelscalefactor); dot((-2.37,-5.44),dotstyle); label("$B$", (-2.53,-5.84), NE * labelscalefactor); dot((4.67,-5.4),dotstyle); label("$C$", (4.75,-5.7), NE * labelscalefactor); dot((-3.593868592709285,-0.142258633097903),dotstyle); label("$X$", (-4.01,-0.08), NE * labelscalefactor); dot((-1.534116523752646,-1.452739739739633),linewidth(4pt) + dotstyle); label("D", (-1.41,-1.36), NE * labelscalefactor); dot((-1.5469223379724755,0.8010835629503406),linewidth(4pt) + dotstyle); label("$O_1$", (-1.69,0.28), NE * labelscalefactor); dot((-3.978946076657836,-3.0214560388192346),linewidth(4pt) + dotstyle); label("$O_2$", (-4.27,-3.44), NE * labelscalefactor); dot((1.1296663605334942,-1.841279453894974),linewidth(4pt) + dotstyle); label("$O$", (1.21,-1.68), NE * labelscalefactor); dot((-2.1119342963546424,1.3588706834510484),linewidth(4pt) + dotstyle); label("$M_{1}$", (-1.93,1.24), NE * labelscalefactor); dot((-2.9819342963546425,-2.7911293165489517),linewidth(4pt) + dotstyle); label("$M_{2}$", (-2.85,-3.08), NE * labelscalefactor); dot((-2.5639925582309653,-0.797499186418768),linewidth(4pt) + dotstyle); label("$M$", (-2.63,-1.22), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Denote the midpoints of $AX,BX,DX$ by $M_1,M_2,M$ respectively. Note that $\measuredangle OO_1O_2 = \measuredangle OO_1M = \measuredangle M_1O_1M = \measuredangle M_1XM= \measuredangle AXC = \measuredangle ABC$, and similarly $\measuredangle OO_2O_1 = \measuredangle BAC$, so $\triangle ABC \sim \triangle O_2O_1O$. Thus, we just aim to minimize $O_1O_2$. Note that $\angle XO_1O_2 = \angle XO_1M = \angle XAD = \angle XAB$, and similarly $\angle XO_2O_1 = \angle XBA$, so $\triangle XO_1O_2 \sim XAB$. Thus, we have: \begin{align*} \dfrac{O_1O_2}{XO_1} &= \dfrac{AB}{XA} \\ O_1O_2 &= \dfrac{XO_1 \times AB}{XA} \\ O_1O_2 &= \dfrac{AB}{2\sin(\angle XO_1M)} \\ O_1O_2 &= \dfrac{AB}{2\sin(\angle XDA)}, \end{align*}which is minimized when $\angle XDA=90^\circ,$ or when $CX \perp AB$.

its 6:59 where i am EDIT: nvm lol its 7 rn...

We claim that $[OO_1O_2]$ is minimized when $CX \perp AB$ (i.e. when $D$ is the foot of the altitude). [asy][asy] 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 = -3.704186819681157, xmax = 13.20119063123603, ymin = -4.589661795987416, ymax = 5.840571082632438; /* image dimensions */ pen qqwwtt = rgb(0,0.4,0.2); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen ccwwff = rgb(0.8,0.4,1); draw((4.020890197618199,4.998461300260606)--(0,0)--(7,0)--cycle, linewidth(2) + qqwwtt); draw((3.5,1.300994564817335)--(1.3602695664046482,-0.8107279732157674)--(4.860269566404648,-1.7212539237905704)--cycle, linewidth(2) + qqwwtt); /* draw figures */ draw((4.020890197618199,4.998461300260606)--(0,0), linewidth(2) + qqwwtt); draw((0,0)--(7,0), linewidth(2) + qqwwtt); draw((7,0)--(4.020890197618199,4.998461300260606), linewidth(2) + qqwwtt); draw(circle((3.5,1.300994564817335), 3.733977350987047), linewidth(2) + wrwrwr); draw((4.020890197618199,4.998461300260606)--(2.153005471018376,-2.1815609044577515), linewidth(2) + wrwrwr); draw(circle((1.3602695664046482,-0.8107279732157674), 1.5835444862210963), linewidth(2) + wrwrwr); draw(circle((4.860269566404648,-1.7212539237905704), 2.746117513621474), linewidth(2) + wrwrwr); draw((1.3602695664046482,-0.8107279732157674)--(4.860269566404648,-1.7212539237905704), linewidth(2) + qqwwtt); draw((4.860269566404648,-1.7212539237905704)--(3.5,1.300994564817335), linewidth(2) + qqwwtt); draw((0,0)--(2.153005471018376,-2.1815609044577515), linewidth(3.6) + dtsfsf); draw((2.153005471018376,-2.1815609044577515)--(7,0), linewidth(3.6) + dtsfsf); draw(circle((1.7566375187115124,-1.4961444388367593), 0.7917722431105482), linewidth(2) + ccwwff); draw(circle((3.506637518711512,-1.9514074141241615), 1.3730587568107366), linewidth(2) + ccwwff); draw((1.076502735509188,-1.0907804522288757)--(3.5,1.300994564817335), linewidth(2) + qqwwtt); /* dots and labels */ dot((0,0),dotstyle); label("$A$", (-0.2478596493331816,0.08294253050839115), NW * labelscalefactor); dot((4.020890197618199,4.998461300260606),dotstyle); label("$C$", (4.1184878392836275,5.105554671722985), NE * labelscalefactor); dot((7,0),dotstyle); label("$B$", (7.102304459761247,0.09169272587636083), NE * labelscalefactor); dot((3.5,1.300994564817335),linewidth(4pt) + dotstyle); label("$O$", (3.5322247496296675,1.3692212495999336), NE * labelscalefactor); dot((2.153005471018376,-2.1815609044577515),dotstyle); label("$X$", (2.1846946629623556,-2.095856116116058), NE * labelscalefactor); dot((2.7205391328092965,0),linewidth(4pt) + dotstyle); label("$D$", (2.753457361880377,0.07419233514042148), NE * labelscalefactor); dot((1.3602695664046482,-0.8107279732157674),linewidth(4pt) + dotstyle); label("$O_1$", (1.3359257122693084,-0.643323685033092), N * labelscalefactor); dot((4.860269566404648,-1.7212539237905704),linewidth(4pt) + dotstyle); label("$O_2$", (4.958506594608705,-1.6495961523496047), NE * labelscalefactor); dot((1.076502735509188,-1.0907804522288757),linewidth(4pt) + dotstyle); label("$M$", (1.029668874390374,-0.9408303275440609), NW * labelscalefactor); dot((4.576502735509188,-1.0907804522288757),linewidth(4pt) + dotstyle); label("$N$", (4.608498779889923,-1.019582085855788), NE * labelscalefactor); dot((2.4367723019138365,-1.0907804522288757),linewidth(4pt) + dotstyle); label("$I$", (2.473451110105351,-1.019582085855788), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Lemma: $\triangle O_1OO_2 \sim BCA.$ Proof: Let $M$ and $N$ denote the midpoints of $\overline{AX}$ and $\overline{BX},$ respectively, and let $I$ denote the intersection of $\overline{O_1O_2}$ and $\overline{CX}.$ WLOG suppose $\angle ADX$ is acute (and $\angle BDX$ acute). Then, quadrilateral $MO_1IX$ is cyclic, so $$ \angle ABC = \angle AXC = 180^{\circ} - \angle MO_1I = \angle OO_1O_2,$$where $M, O_1, O$ are collinear due to symmetry in $\triangle AOX.$ Similarly, quadrilateral $NO_2XI$ is cyclic, so $$\angle BAC = \angle BXC = \angle OO_2O_1.$$Moreover, quadrilateral $MONX$ is cyclic, so $\angle O_1OO_2 = 180^{\circ}-\angle MXN = \angle ACB.$ Since all three angles are congruent, we have $\triangle O_1OO_2 \sim \triangle BCA,$ as desired. Lemma: $[O_1OO_2]$ is minimized when $M=O_1$ and $N=O_2.$ Proof: Let $F_1$ and $F_2$ denote the feet of the perpendiculars from $O_1$ and $O_2$ onto $\overline{AB},$ respectively. Then, $$O_1O_2 \geq F_1F_2,$$with equality achieved when $\overline{F_1F_2} \parallel \overline{O_1O_2}.$ This occurs when $M=O_1$ and $N=O_2,$ as desired. Since $\triangle O_1OO_2 \sim \triangle BCA,$ it suffices to minimize $O_1O_2.$ This simply happens when $O_1O_2 = MN$ (i.e. when $\angle ADX = \angle BDX = 90^{\circ}$), so we are done.

[asy][asy] size(6cm); pair A=dir(270-58), B=dir(270+58), C=dir(90+27); pair O=(0,0), X=dir(270-19), D=extension(C,X,A,B), O1=circumcenter(A,D,X), O2=circumcenter(B,D,X); pair P1=foot(D,A,X), P2=foot(D,B,X); fill(O--O1--O2--cycle,rgb(230,255,242)); draw(A--B--C--cycle); draw(X--A); draw(X--B); draw(X--C); draw(O--O1,blue+dashed); draw(O--O2,blue+dashed); draw(O1--O2,blue+dotted); draw(D--P1,purple+dotted); draw(D--P2,purple+dotted); draw(P1--P2,purple+dotted); draw(circumcircle(A,B,C)); draw(circumcircle(A,D,X)); draw(circumcircle(B,D,X)); dot(A); dot(B); dot(C); dot(O); dot(X); dot(D); dot(O1); dot(O2); dot(P1); dot(P2); label("$A$",A,dir(185)); label("$B$",B,B); label("$C$",C,C); label("$O$",O,dir(120)); label("$X$",X+(0,-0.05),dir(270)); label("$D$",D,dir(40)); [/asy][/asy] Note that $O_1O\perp AX$ and $O_2O\perp BX$. Using directed angles modulo $\pi$, \[ \measuredangle{O_1OO_2}=\measuredangle{(O_1O,O_2O)}=\measuredangle{(AX,BX)}=\measuredangle{ACB}. \]Let $P_1,P_2$ be the feet from $D$ to $AX,BX$; by a similar argument, $\measuredangle{P_1DP_2}=\measuredangle{ACB}$. From $\measuredangle{O_1OO_2}=\measuredangle{ACB}$, it suffices to minimize $OO_1\cdot OO_2$. Let $\rho$ denote unsigned power. Claim. \[ OO_1=\frac{\rho(D,\omega)}{2DP_1}\quad\text{and}\quad OO_2=\frac{\rho(D,\omega)}{2DP_2} \]Proof. We proceed using complex numbers with $\omega$ as the unit circle. The circumcenter of $\triangle{ADX}$ is at \[ \frac{ \begin{vmatrix} a & 1 & 1 \\ d & |d|^2 & 1 \\ x & 1 & 1 \end{vmatrix} }{ \begin{vmatrix} a & a^{-1} & 1 \\ d & d^{-1} & 1 \\ x & x^{-1} & 1 \end{vmatrix} } = \frac{ \begin{vmatrix} a & 0 & 1 \\ d & |d|^2-1 & 1 \\ x & 0 & 1 \end{vmatrix} }{ \begin{vmatrix} a & a^{-1} & 1 \\ d & d^{-1} & 1 \\ x & x^{-1} & 1 \end{vmatrix} } = \frac{(|d|^2-1)(a-x)}{\frac{4}{i}[ADX]}. \]Taking the magnitude, we have that \[ OO_1=\frac{\rho(D,\omega)\cdot AX}{4[ADX]}=\frac{\rho(D,\omega)}{2DP_1} \]as desired. The other result follows by symmetry. $\square$ Note that \[ \frac{1}{2}DP_1\cdot DP_2\cdot\sin\angle{C}=[DP_1P_2]=\frac{\rho(D,\omega)\cdot[XAB]}{4R^2} \]since $\triangle{DP_1P_2}$ is the pedal triangle of $D$ w.r.t. $\triangle{XAB}$. Then \[ OO_1\cdot OO_2\propto\frac{\rho(D,\omega)^2}{DP_1\cdot DP_2}\propto\frac{\rho(D,\omega)}{[XAB]}\propto\frac{DA\cdot DB}{XA\cdot XB}=\frac{\sin\angle{DXA}}{\sin\angle{XDA}}\cdot\frac{\sin\angle{DXB}}{\sin\angle{XDB}}=\frac{\sin\angle{B}\sin\angle{A}}{\sin^2{\angle{ADX}}} \]so it suffices to maximize $\sin\angle{ADX}$. This occurs when $CX\perp AB$.

So, uh... why is there a C-labeled problem?

I claim that $X$ is the unique point that satisfies $CX\perp AB$. The main claim is that $\triangle OXO_1 \sim\triangle CAD$ and $\triangle OXO_2\sim \triangle CBD$. By radical axes, we have that $OO_1$ and $OO_2$ perpendicularly bisect $AX$ and $BX$, respectively. Using the fact that $\triangle OAX$ and $\triangle OBX$ are isosceles, \[\angle XOO_1=\frac{1}{2}\angle XOA=\frac{1}{2}(2\angle XCA)=\angle XCA=\angle DCA\]\[\angle XOO_2=\frac{1}{2} \angle XOB=\frac{1}{2}(2\angle XCB)=\angle XCB=\angle DCB\] Now, let $\angle BDX=\theta$. Suppose that $\theta$ is acute for now, the obtuse case follows similarly. If $\angle BDX$ is acute, then $\triangle ADX$ is obtuse and $\triangle BDX$ is acute, meaning $O_1$ lies outside of the triangle and $O_2$ lies inside. Using the relation between central angles and inscribed angles, \[360^{\circ}-\angle AO_1X=2(\angle ADX)=2(180^{\circ}-\theta)=360^{\circ}-2\theta \implies \angle AO_1X=2\theta\]and $\angle BO_2X=2\angle BDX=2\theta$. Since $OO_1$ and $OO_2$ perpendicularly bisect $AX$ and $BX$, \[\angle OO_1X=\frac{1}{2}\angle AO_1X=\theta=\angle BDX=\angle CDA\]\[\angle OO_2X=180^{\circ}-\frac{1}{2}\angle XO_2B=180^{\circ}-\theta=\angle XDA=\angle CDB\]which combined with $\angle XOO_1=\angle DCA$ and $\angle XOO_2=\angle DCB$ gives $\triangle OXO_1\sim \triangle CAD$ and $\triangle OXO_2\sim \triangle CBD$, as claimed. We have that $[OO_1O_2]=\frac{1}{2}OO_1\cdot OO_2\sin(\angle O_1OO_2)$. Note that $\angle O_1OO_2=\angle O_1OX+\angle XOO_2=\angle DCA+\angle BCD=\angle ACB$, or $\angle O_1OO_2$ is fixed. Since $\frac{1}{2}\sin(\angle O_1OO_2)$ is fixed, it suffices to minimize $OO_1\cdot OO_2$. Using similarity ratios, $\frac{OO_1}{OX}=\frac{CD}{CA}$ and $\frac{OO_2}{OX}=\frac{CD}{CB}$. Thus, \[OO_1\cdot OO_2=\frac{OX^2\cdot CD^2}{CA\cdot CB}\]Clearly $CA\cdot CB$ is fixed since $\triangle ABC$ is fixed, but also note that $OX^2$ is fixed since $OX$ is simply the radius of $(ABC)$. This means minimizing $[OO_1O_2]$ is equivalent to minimizing $CD^2$, or $CD$ since $CD>0$. This occurs when $CD$ is the altitude from $C$ to $AB$, or $CX\perp AB$, as claimed.

dont tell me that you guys wrote that solution and drew the asymptote that fast

Solution using trig: I claim that the answer is \(X\) such that \(CX\perp AB.\) Use directed angles when necessary. First observe that \(\measuredangle O_1OO_2=\measuredangle AOB\) is constant, so by the area formula \([OO_1O_2]=\frac12 OO_1\cdot OO2 \cdot \sin \measuredangle OO_1O_2,\) it suffices to minimize \(OO_1\cdot OO_2.\) We now show \(\measuredangle OXO_2\) is constant. Indeed, \(\measuredangle OXO_2=-\measuredangle XO_2O-\measuredangle O_2OX=-\measuredangle XDA-\measuredangle DAX=\measuredangle AXD=\measuredangle ABC.\) By symmetry, this shows \(\measuredangle OXO_1\) is constant as well. By law of sines, \(OO_1\cdot OO_2=\frac{OX\cdot \sin \measuredangle OXO_1}{\sin \measuredangle OO_1X}\cdot \frac{OX\cdot \sin \measuredangle OXO_2}{\sin \measuredangle OO_2X},\) and the numerator is constant. Furthermore, \(\measuredangle OO_1X=\measuredangle ADC\) and \(\measuredangle OO_2X=\measuredangle BDC,\) so we wish to minimize \(\frac{1}{\sin \measuredangle ADC\cdot \sin \measuredangle BDC}.\) This occurs when \(\sin \measuredangle ADC=\sin \measuredangle BDC=1\) or \(CD\perp AB,\) as desired.

Pure angle chase outline: Show $\triangle{OO_1O_2} \sim \triangle{CBA}$ by extending $OO_1$ to meet $AX$ at $U$ and $OO_2$ to meet $BX$ at $V$ and note $OUXV$ is cyclic so then you get a free angle. Some more angle chase to get $\angle{CBA}=\angle{O_1OO_2}$. So to minimize we just need to minimize $CD$ (the cevian) so we show $\triangle{OO_1A} \sim \angle{CDA}$ through more boring angle chase. So the minimum occurs when $CD$ is minimized or when its the projection. So intersection of $CH$ and $(ABC)$ is it.

[asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair C = dir(120), A = dir(210), B = dir(330), O = origin, X = dir(290), D = extension(C,X,A,B), O1 = circumcenter(A,D,X), O2 = circumcenter(B,D,X); draw(unitcircle^^X--C--B--A^^O1--O2); draw(circumcircle(A,D,X)^^circumcircle(B,D,X), gray); draw(A--X--B^^O1--X--O2, magenta); draw(A--C--B^^O2--O--O1, heavygreen); dot("$A$", A, dir(180)); dot("$B$", B, dir(0)); dot("$C$", C, dir(120)); dot("$D$", D, dir(60)); dot("$X$", X, dir(285)); dot("$O$", O, dir(90)); dot("$O_1$", O1, dir(210)); dot("$O_2$", O2, dir(15)); [/asy][/asy] I claim that $X \in \omega$ is the unique point such that $\overline{CX} \perp \overline{AB}$. Claim: $\triangle OO_1O_2 \sim \triangle CBA$. Proof. Since $\overline{OO_1}$ is the perpendicular bisector of $\overline{AX}$ and $\overline{O_1O_2}$ is the perpendicular bisector of $\overline{DX}$, we have $$\angle OO_1O_2 = \angle(\overline{OO_1},\overline{O_1O_2}) = \angle(\overline{AX},\overline{CX}) = \angle CXA = \angle CBA.$$Similarly $\angle OO_2O_1 = \angle CAB$ and the conclusion follows. $\square$ Thus, we just need to minimize $O_1O_2$. It is well known that $\triangle XO_1O_2 \sim \triangle XAB$. Then by the Extended Law of Sines, we have $$O_1O_2 = AB \cdot \dfrac{XO_1}{XA} = \dfrac{AB}{2 \sin(\overline{CX}, \overline{AB})}.$$Since $AB$ is constant, this is minimized when $k = \sin(\overline{CX}, \overline{AB})$ is maximized, which occurs when $k=1$ i.e. $\overline{CX} \perp \overline{AB}$ as desired. $\blacksquare$

Note that altitude from $O$ to $O_1O_2$ is $\frac{CD}2$, and $\triangle OO_1O_2$ is similar to $\triangle ABC$. Thus just minimize $CD$, which is when it's the altitude.

Here's a complex bash. Is it overkill? Almost certainly. Do I regret it? Not in the slightest. Identify $\omega$ with the unit circle. Then $$|a|=|b|=|c|=|x|=1$$$$o = 0$$$$d = \frac{ab(c+x) - cx(a+b)}{ab-cx}$$Now we compute the coordinate of $O_1$ as follows: \begin{align*} o_1 &= \frac{a\overline{a}(d-x) + d\overline{d}(x-a) + x\overline{x}(a-d)}{\overline{a}(d-x) + \overline{d}(x-a) + \overline{x}(a-d)} \\ &= \frac{d-x+d\overline{d}(x-a)+a-d}{\frac{d}a-\frac{x}a+\overline{d}x-a\overline{d}+\frac{a}x-\frac{d}x} \\ &= \frac{(x-a)(d\overline{d}-1)}{(x-a)\left(\overline{d}+\frac{d}{ax}-\frac1a-\frac1x\right)} \\ &= \frac{ax(d\overline{d}-1)}{d-a-x+ax\overline{d}} \\ &= \frac{ax\left[\frac{(abc+abx-acx-bcx)(a+b-c-x)}{(ab-cx)^2}-1\right]}{\frac{ab(c+x) - cx(a+b)}{ab-cx} - a - x + \frac{ax(a+b-c-x)}{ab-cx}} \\ &= \frac{ax[(abc+abx-acx-bcx)(a+b-c-x) - (ab-cx)^2]}{(ab-cx)(abc+abx-acx-bcx-a^2b+acx-abx+cx^2+a^2x+abx-acx-ax^2)} \\ &= \frac{ax\begin{pmatrix}a^2bc+a^2bx-a^2cx-abcx+ab^2c+ab^2x-abcx-b^2cx\\-abc^2-abcx+ac^2x+bc^2x-abcx-abx^2+acx^2+bcx^2-a^2b^2+2abcx-c^2x^2\end{pmatrix}}{(ab-cx)(abc-bcx-a^2b+cx^2+a^2x+abx-acx-ax^2)} \\ &= \frac{ax\begin{pmatrix}a^2bc+a^2bx-a^2cx+ab^2c+ab^2x-b^2cx\\-abc^2-2abcx+ac^2x+bc^2x-abx^2+acx^2+bcx^2-a^2b^2-c^2x^2\end{pmatrix}}{(ab-cx)(abc-bcx-a^2b+cx^2+a^2x+abx-acx-ax^2)} \\ &= \frac{ax(a-c)(b-c)(a-x)(b-x)}{(ab-cx)(a-c)(a-x)(b-x)} \\ &= \frac{ax(b-c)}{ab-cx} \end{align*}Similarly, $$o_2 = \frac{bx(a-c)}{ab-cx}$$Then the (signed) area of $OO_1O_2$ is given by \begin{align*} \frac{i}4\begin{vmatrix}o & o_1 & o_2 \\ \overline{o} & \overline{o}_1 & \overline{o}_2 \\ 1 & 1 & 1\end{vmatrix} &= \frac{i}4\begin{vmatrix}0 & \frac{ax(b-c)}{ab-cx} & \frac{bx(a-c)}{ab-cx} \\ 0 & \frac{b-c}{ab-cx} & \frac{a-c}{ab-cx} \\ 1 & 1 & 1\end{vmatrix} \\ &= \frac{i}4\begin{vmatrix}\frac{ax(b-c)}{ab-cx} & \frac{bx(a-c)}{ab-cx} \\ \frac{b-c}{ab-cx} & \frac{a-c}{ab-cx}\end{vmatrix} \\ &= \frac{i}4\left[\frac{ax(b-c)(a-c)}{(ab-cx)^2} - \frac{bx(a-c)(b-c)}{(ab-cx)^2}\right] \\ &= \frac{ix(a-b)(a-c)(b-c)}{4(ab-cx)^2} \end{align*}Thus, taking absolute values, the magnitude of the area of $OO_1O_2$ is $$\frac{|a-b||a-c||b-c|}{4|ab-cx|^2}$$The numerator is fixed. Thus to minimize the area, we must maximize the value of $|ab-cx|$. Thus we want to maximize the distance from $x$ to the point $\frac{ab}c$. This is achieved (for $X$ on the circumcircle) when $x$ is the point diametrically opposite it, that is, $x = -\frac{ab}c$, the second intersection of the altitude from $C$ with $\omega$. Because the triangle is acute, this point is indeed on minor arc $AB$, so it is the unique $X$ that minimizes the area. $\blacksquare$

The similarity lemma some people used also showed up as 2013 USAMO p6.

Proposed by Zuming Feng

This reminds me of 2018 AIME I/13.

We prove $[OO_1O_2] \ge \frac14 [ABC]$, with equality if and only if $\overline{CX} \perp \overline{AB}$. To see this, we need two claims. Claim: We have $\triangle O O_1 O_2 \sim \triangle CBA$, with opposite orientation. Proof. Notice that $\overline{OO_1} \perp \overline{AX}$ and $\overline{O_1 O_2} \perp \overline{CX}$, so $\measuredangle OO_1O_2 = \measuredangle AXC = \measuredangle ABC$. Similarly $\measuredangle OO_2O_1 = \measuredangle BAC$. $\blacksquare$ Therefore, the problem is equivalent to minimizing $O_1 O_2$. [asy][asy] pair C = dir(130); pair A = dir(200); pair B = dir(340); pair X = dir(280); pair D = extension(C, X, A, B); pair O_1 = circumcenter(A, D, X); pair O_2 = circumcenter(B, D, X); pair O = origin; filldraw(O--O_1--O_2--cycle, invisible, deepcyan); filldraw(A--X--B--cycle, invisible, red); filldraw(A--B--C--cycle, invisible, blue); draw(X--C, red); filldraw(O_1--X--O_2--cycle, invisible, orange); draw(unitcircle, grey); dot("$C$", C, dir(C)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$X$", X, dir(X)); dot("$D$", D, dir(245)); dot("$O_1$", O_1, dir(O_1)); dot("$O_2$", O_2, dir(20)); dot("$O$", O, dir(45)); /* TSQ Source: C = dir 130 A = dir 200 B = dir 340 X = dir 280 D = extension C X A B R245 O_1 = circumcenter A D X O_2 = circumcenter B D X R20 O = origin R45 O--O_1--O_2--cycle 0.1 lightcyan / deepcyan A--X--B--cycle 0.1 lightred / red A--B--C--cycle 0.1 lightcyan / blue X--C red O_1--X--O_2--cycle 0.1 yellow / orange unitcircle grey */ [/asy][/asy] Claim: [Salmon theorem] We have $\triangle X O_1 O_2 \sim \triangle XAB$. Proof. It follows from the fact that $\triangle AO_1 X \sim \triangle BO_2 X$ (since $\measuredangle ADX = \measuredangle XDB \implies \measuredangle XO_1A = \measuredangle XO_2B$) and that spiral similarities come in pairs. $\blacksquare$ Let $\theta = \angle ADX$. The ratio of similarity in the previous claim is equal to $\frac{XO_1}{XA} = \frac{1}{2\sin \theta}$. In other words, \[ O_1 O_2 = \frac{AB}{2 \sin \theta}. \]This is minimized when $\theta = 90^{\circ}$, in which case $O_1 O_2 = AB/2$ and $[OO_1O_2] = \frac14 [ABC]$. This completes the solution. Remark: [Bobby Shen] Alternatively, one can simply note that since the projection of $O_1 O_2$ onto line $AB$ has length $\frac12AB$ \[ O_1 O_2 = \frac{\frac{1}{2} AB}{\cos\angle(O_1O_2, AB)}. \]Thus we just need $\overline{O_1 O_2} \parallel \overline{AB}$ to maximize the area (where the cosine is $1$) which occurs when $\overline{CDX} \perp \overline{AB}$. This eliminates the use of Salmon theorem entirely.

unique approach?

Credits to budu for the figure

latex is just a sketch from overleaf so here is pasted Notice the angle formed by $O_1O_2\cap O_1O$ is equivalent to the angle formed by $XD\cap XA$ (since the respective lines are perp. from rad. ax, meaning the angle is just rotated by 90 degrees from the old lines to new lines forming same angle). Then, $OO_1O_2=DXA=CBA$ etc. so the triangles $OO_1O_2$ and $CBA$ are similar, which rephrases the problem into minimizing $O_1O_2$. Indeed, notice that length is at least the length connecting their projections onto AB, which is the connection of midpoint of BD and DA, which is just 1/2AB. Indeed, this only happens if XD perp. $O_1O_2$ parallel to MN means XC perp. AB.

The answer is the $X$ such that $D$ is the foot of the altitude from $C$ to $AB.$ We first claim that $\triangle OO_1 O_2 \sim \triangle CBA.$ To prove this, we note that by perpendicular bisector conditions we have $OO_1 \perp AX,$ and because $O_1 O_2$ is perpendicular to the radical axis if $(AXD)$ and $(BXD),$ we get $O_1 O_2 \perp CX.$ Hence by these perpendicularities, we get $$\angle OO_1 O_2 = \angle AXC = \angle ABC.$$Similarly, $\angle OO_2 O_1 = \angle CAB,$ proving the claim. Thus it suffices to minimize $O_1 O_2.$ I think this last part is not a common approach, but I'm not sure. Take a homothety centered at $D$ of scale $2,$ and set $C'$ to be the $C$-antipode with respect to $(ABC).$ Also let $\ell_A$ be the line through $A$ perpendicular to $AB,$ and let $\ell_B$ be the line through $B$ perpendicular to $AB.$ Then this homothety sends $O_1$ to the intersection of lines $\ell_A$ and $C'X,$ and the homothety sends $O_2$ to the intersection of lins $\ell_B$ and $C'X.$ Hence the segment $O_1 O_2$ gets mapped to another line segment such that 1. it passes through $C'$, and 2. it has endpoints on $\ell_A$ and $\ell_B.$ To minimize the length of this new line segment, for obvious reasons, we want it to be perpendicular to each of $\ell_A$ and $\ell_B,$ and hence we want $XC' \parallel AB.$ Again for obvious reasons, this happens precisely when $CD \perp AB,$ as desired.

We claim that $[OO_1O_2] \ge \frac{1}{4} [ABC]$ if and only if $\boxed{\overline{CX} \perp \overline{AB}}$. Let $M$ be the midpoint of $CX$, $N$ be the midpoint of $DX$, $E$ be the projection of $O_1$ onto $AB$, $F$ be the projection of $O_2$ onto $AB$, and let $h_O$ and $h_C$ be the $O$-altitude in $\triangle OO_1O_2$ and $C$-altitude in $\triangle ABC$, respectively. It is clear that \[h_O = MN = \frac{1}{2}CD \ge \frac{1}{2}h_C\] with equality when $\overline{CX} \perp \overline{AB}$. Also, \[O_1O_2 \ge EF = \frac{1}{2}AB\] with equality when $\overline{CX} \perp \overline{AB}$. Hence, we must have \[\frac{[OO_1O_2]}{[ABC]} = \frac{\frac{1}{2} O_1O_2 \cdot h_O}{\frac{1}{2} AB \cdot h_C} \ge \frac{1}{4}. \] Thus, the minimum is $\frac{1}{4}[ABC]$ when $\overline{CX} \perp \overline{AB}$. [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 = -7.817017346188641, xmax = 24.19460421850487, ymin = -5.654482358756065, ymax = 13.7674940084797; /* image dimensions */ /* draw figures */ draw(circle((7,4.125), 8.125), linewidth(1)); draw((5,12)--(3.8834401390220306,-3.378511153649402), linewidth(1)); draw((0,0)--(14,0), linewidth(1)); draw((14,0)--(5,12), linewidth(1)); draw((5,12)--(0,0), linewidth(1)); draw((7,4.125)--(2.0643688066462773,-1.548276604984709), linewidth(1)); draw((2.0643688066462773,-1.548276604984709)--(9.064368806646277,-2.0565129972307163), linewidth(1)); draw((9.064368806646277,-2.0565129972307163)--(7,4.125), linewidth(1)); draw((2.0643688066462773,-1.548276604984709)--(2.0643688066462773,0), linewidth(1)); draw((9.064368806646277,-2.0565129972307163)--(9.064368806646277,0), linewidth(1)); /* dots and labels */ dot((5,12),dotstyle); label("$C$", (5.107077628810755,12.238580739718213), NE * labelscalefactor); dot((14,0),dotstyle); label("$B$", (14.189443082784458,-0.0461672878702996), NE * labelscalefactor); dot((0,0),dotstyle); label("$A$", (0.09033096568714462,0.1461672878702996), NE * labelscalefactor); dot((3.8834401390220306,-3.378511153649402),linewidth(4pt) + dotstyle); label("$X$", (3.984281947064042,-3.1938875668430464), NE * labelscalefactor); dot((7,4.125),linewidth(4pt) + dotstyle); label("$O$", (7.089887024235801,4.307343158018), NE * labelscalefactor); dot((4.128737613292555,0),linewidth(4pt) + dotstyle); label("$D$", (4.223174645308023,0.19838874822150313), NE * labelscalefactor); dot((2.0643688066462773,-1.548276604984709),linewidth(4pt) + dotstyle); label("$O_{1}$", (2.118697440409783,-2.0544137903643823), NE * labelscalefactor); dot((9.064368806646277,-2.0565129972307163),linewidth(4pt) + dotstyle); label("$O_{2}$", (9.168253498958439,-1.856088456676745), NE * labelscalefactor); dot((2.0643688066462773,0),linewidth(4pt) + dotstyle); label("$E$", (2.168697440409783,0.19838874822150313), NE * labelscalefactor); dot((9.064368806646277,0),linewidth(4pt) + dotstyle); label("$F$", (9.168253498958439,0.19838874822150313), NE * labelscalefactor); dot((4.441720069511016,4.310744423175299),linewidth(4pt) + dotstyle); label("$M$", (4.5337351530252,4.498457316613186), NE * labelscalefactor); dot((4.006088876157293,-1.689255576824701),linewidth(4pt) + dotstyle); label("$N$", (4.103728296186033,-1.4977494093107715), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Let $D_0$ denote the foot of the altitude of $C$ onto $AB$, and $X_0$ denote the intersection of ray $AD_0$ and $\omega$. I claim that $[OO_1O_2]$ is minimized when $X=X_0$. Firstly, we note that \[ O_1O_2 = \frac{AB}{2 \sin(\angle ADB)} \ge \frac{AB}{2} = \frac{AB}{2 \sin (\angle AD_0B)}. \]On the other hand, let $h_O$ denote the length of the altitude from $O$ in $\triangle OO_1O_2$ and $h_C$ denote the length of the altitude from $C$ in $\triangle ABC$. Let $\ell_O$ and $\ell_C$ be the segments that correspond to those altitudes. Denote $M$ by the midpoint of $CX$ and $M'$ by the midpoint of $DX$. It is easy to see that $\ell_O \parallel MM'$ and that the projection of $\ell_O$ onto $\overline{CX}$ is precisely $MM'$, and thus \[ h_O = MM' = \frac{CX-DX}{2} = \frac{CD}{2} \ge \frac{h_C}{2} = \frac{CD_0}{2}. \]Hence, $[OO_1O_2]$ is minimized precisely when $D=D_0$, or $X=X_0$, as desired.

Denote the intersection of $DX$ and $O_1O_2$ by $P$. Next, denote the intersection of $OO_1$ and $AX$ by $Q$, and denote the intersection of $O_1O_2$ and $AX$ by $R$. Then we have $\triangle O_1QR \sim \triangle O_1PX$ by AA Similarity, so $\angle OO_1O_2 = \angle CXA = \angle CBA$. Next, extend $OO_2$ to intersect $BX$ at $S$. We know that $PO_2SX$ is a cyclic quadrilateral, so $\angle PO_2S = 180 - \angle DXB = 180 - \angle OO_2O_1$, meaning that $\angle CXB = \angle OO_2O_1 = \angle CAB$. Using these two relations, we know that $\triangle OO_2O_1 \sim \triangle CAB$. Since the side lengths and area of $\triangle ABC$ are fixed, we should try to minimize one side of $\triangle OO_2O_1$ - let's choose side $O_1O_2$. If we drop altitudes from $O_1$ to the midpoint of $AD$ and $O_2$ to the midpoint of $BD$, we get an trapezoid; $O_1O_2$ would be minimized if it was the height of the trapezoid, or in other words if the trapezoid was a rectangle. This would mean that $O_1O_2$ would have to be parallel to $AB$, and since $O_1O_2$ is perpendicular to $XC$, it would mean that $AB$ would be perpendicular to $XC$. Thus, the answer to the problem is that the area is minimized when $AB$ is perpendicular to $XC$.

The desired $X$ is reached when $CX\bot AB$ By radical axis theorem, we have $OO_1\bot AX, OO_2\bot BX, O_1O_2\bot CX$. Let the intersections of line $OO_1, OO_2$ and $AX, BX$ at $P,Q$ respectively, it is clear to see $OPXQ$ is cyclic, $\angle{O}+\angle{PXQ}=180^{\circ}\implies \angle{C}=\angle{O}$. Moreover, note $\angle{OO_2O_1}=180^{\circ}-\angle{O_1O_2X}+\angle{XO_2Q}=180^{\circ}=\angle{ABX}+\angle{XDB}+\angle{O_1O_2O}\implies \angle{OO_2O_1}=\angle{DXB}=\angle{CAB}, \triangle{OO_1O_2}\sim \triangle{CBA}$. Since $\triangle{ABC}$ is fixed, we have to minimize the length of $O_1O_2$. Drop altitude from $O_1, O_2$ to $AB$, we could see $O_1O_2\geq \frac{AB}{2}$, where the equality reaches when $O_1O_2$ is parallel to $AB$. Since $O_2O_1$ is perpendicular to $XC$, so $AB\bot CX$ in the minimal case, which finished the proof.

Claim: $\triangle O_2O_1O \sim \triangle ABC.$ Proof. The idea is that the radical axis of two circles is perpendicular to the segment connecting their centers. Consequently, $O_1O_2 \perp XC, O_2O \perp BX, OO_1 \perp AX.$ The angle between $O_2O_1$ and $OO_1,$ then, due to a $90^\circ$ rotation is the same as the angle between $AX$ and $CX,$ which is precisely $\angle ABC.$ The same logic follows similarly for the other angles. Thus, it suffices to minimize $O_1O_2$ as $X$ varies across minor arc $AB,$ as the similarity between the two triangles would imply it has minimal area. But observe that $O_1O_2$ is at least half the length of $AB$ (one can see this through dropping the altitdues from $O_1$ and $O_2$ onto $AB$), with equality being achieved if and only if $O_1O_2 \parallel AB.$ Thus, the condition for $[OO_1O_2]$ to be minimized is achieved precisely when $XC \perp AB.$

alr so like $O_1O_2$ is perp bisector of $DX$ so the length of altitude from $O$ to $O_1O_2$ is $\frac{CX-DX}{2}=\frac{CD}{2}$ and $O_1O_2=\frac{AB}{2\sin\angle ADX}$ so the area of $OO_1O_2$ is $\frac{AB\cdot CD}{8\sin\angle ADX}$ which is clearly minimized when $\angle ADX=90^\circ$ and $CD$ is the altitude from $C$ to $AB$.

Horrible angle chasing solution: We claim $\triangle OO_1O_2 \sim \triangle CBA$. Since $OO_2$ is the perpendicular bisector of $XB$ and $OO_1$ is the perpendicular bisector of $XA$, we have \[ \angle O_2OO_1 = \angle O_2OX + \angle XOO_1 = \frac12 (\angle BOX + \angle XOA) = \frac12 \angle BOA = \angle C. \] The other angles are a bit more tricky. All arcs are with respect to the circle with center $O_1$. Since $O_1O_2$ is the perpendicular bisector of $XD$, we see that $\angle XO_1O_2 = \frac12 \overarc{XD}$. Similarly, $\angle XO_1O = \frac12 \overarc{XA}$. Thus, \[ \angle O_2O_1O = \angle XO_1O - \angle XO_1O_2 = \frac12 (\overarc{XA} - \overarc{XD}) = \frac12 \overarc{AD}. \]Furthermore, \[ \frac12 \overarc{AD} = \angle DXA = \angle CXA = \angle B. \]Thus, we have AA similarity. This means minimizing the area is equivalent to minimizing the length of $O_1O_2$. Place triangle $ABC$ into a coordinate plane such that side $AB$ is horizontal. It is easy to see that the horizontal distance between $O_1$ and $O_2$ is constant (it equals half the length of side $AB$). Thus, we want to minimize the vertical distance. This distance is minimized when $\overline{O_1O_2} \parallel \overline{AB}$, meaning that line $CX$ is perpendicular to line $AB$. In other words, $X$ must be the reflection of the orthocenter over line $AB$.

Funny geo, first note that by Salmon Theorem we have $\triangle XO_1O_2 \sim \triangle XAB$, $\triangle XOO_2 \sim \triangle BOO_2 \sim \triangle BCD$, $\triangle XOO_1 \sim \triangle AOO_1 \sim \triangle ACD$ and that by perpendicular bisectors we have $OO_1 \perp AX$ and $OO_2 \perp BX$ which gives $\angle O_2OO_1=\angle ACB=180-\angle BXA=180-\angle O_1XO_2$ so we have that $XO_1OO_2$ is cyclic. But we also have $\angle OO_1O_2=\angle OXO_2=\angle CBA$, so we have $\triangle OO_1O_2 \sim \triangle CBA$, so all we need to do is to minimize the ratio of similarily which is $\frac{O_1O_2}{AB}$, but if we let $M,N$ midpoints of $BD, AD$ respectively then by dropping perpendiculars we note that $O_1O_2 \ge MN$, equallity being achieved when $XD \perp AB$ so minimal happens when $CX \perp AB$ thus we are done .

nice! The answer is $X$ such that $CX\perp AB$. First, note that because $O$ is the circumcenter of $\triangle AXB$, $\overleftrightarrow{OO_1}$ is the perpendicular bisector of $AX$ and $\overleftrightarrow{OO_2}$ is the perpendicular bisector of $BX$. We also have that $\overleftrightarrow{O_1O_2}$ is the perpendicular bisector of $DX$. Let $\overleftrightarrow{OO_1}$ intersect $AX$ at $M$ and $\overleftrightarrow{OO_2}$ intersect $BX$ at $N$. Also, let $\overleftrightarrow{O_1O_2}$ intersect $DX$ at $L$. Clearly, $O_1MXL$ and $O_2NLX$ are cyclic because $\angle MO_1L + \angle LXM=180^\circ$ and $\angle XLM=\angle O_2NX=90^\circ$. Thus, $\angle OO_1O_2=\angle MXL=\angle ABC$ and $\angle OO_2O_1=\angle LXN=\angle BAC$, so $\triangle CBA\sim \triangle OO_1O_2$. Since $[ABC]$ is fixed, it suffices to minimize $O_1O_2$. Let $Y$ and $Z$ be the projections of $O_1$ and $O_2$ onto $AB$, respectively. Also, let $W$ be the projection of $O_1$ onto $\overleftrightarrow{ZO_2}$. We have that $O_1O_2^2=O_2W^2+O_1W^2$ by the Pythagorean Theorem. However, $O_1W=YZ=\frac 12 AB$, so $O_1W$ is fixed. Therefore, the minimum value of $O_1O_2$ occurs when $O_2W=0$, or when $O_2=W$. This means that $O_1O_2Z=90^\circ\implies O_2O_2\parallel AB$. To finish, note that $O_1O_2\perp CX$, so $CX\perp AB$, as desired. $\blacksquare$

Claim: $OO_1O_2 \sim ABC$ Proof: $\angle B = \angle AXC$ and also $XA$ is perpendicular to $O_1O$ from radical axis stuff so if $O_1 O_2 \cap XA = Y$, then $\angle OO_1O_2 = \angle AXC = 90 - \angle XYH$ from similarity so we are done. The other angle follows similarly. So we just need to maximize $O_1O_2$ which occurs when $O_1O_2FE$ is a rectangle where $E$ and $F$ are the projections of $O_1$ and $O_2$ on $AB$ resp. This is because $EF$ is always constant and $O_1O_2 \geq EF$ by pythag. Therefore the desired $X$ is when $OX \perp AB$

great complex bash exercise but hellish to write up...

Get $OO_1O_2 \sim CBA$, then trig bash to oblivion.