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Source: 2020 USOMO Problem 1

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#1 h Jun 21, 2020, 10:59 PM • 10 Y

Y by Mateoaops, mira74, Frestho, poplintos, samrocksnature, a_n, I_am_human, megarnie, Lamboreghini, Rounak_iitr

Let $ABC$ be a fixed acute triangle inscribed in a circle $\omega$ with center $O$ . A variable point $X$ is chosen on minor arc $AB$ of $\omega$ , and segments $CX$ and $AB$ meet at $D$ . Denote by $O_1$ and $O_2$ the circumcenters of triangles $ADX$ and $BDX$ , respectively. Determine all points $X$ for which the area of triangle $OO_1O_2$ is minimized.

Proposed by Zuming Feng

This post has been edited 1 time. Last edited by djmathman, Jun 22, 2020, 4:47 AM
Reason: problem author

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budu

#2 h Jun 21, 2020, 10:59 PM • 16 Y

Y by trumpeter, mira74, OlympusHero, jeffisepic, franchester, Mobashereh, AMC_Kid, Frestho, Ultroid999OCPN, samrocksnature, centslordm, Reef334, Lamboreghini, Monoten, Rounak_iitr, Begli_I.

The answer is the point $X$ such that $\boxed{XC \perp AB}$ .
$[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$", (2.273093034762053,-0.8941448754833125), dir(-60) * 1.5); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
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 $XC \perp AB$ . $\blacksquare$

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rd123

#3 h Jun 21, 2020, 10:59 PM • 2 Y

Y by AMC_Kid, samrocksnature

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.19, xmax = 12.01, ymin = -8.64, ymax = 5.6; /* image dimensions */ pen qqffff = rgb(0,1,1); /* draw figures */ draw(circle((1.1296663605334942,-1.841279453894974), 5.019806191877154), linewidth(0.8) + red); draw((-0.63,2.86)--(-2.37,-5.44), linewidth(0.8)); draw((-2.37,-5.44)--(4.67,-5.4), linewidth(0.8)); draw((4.67,-5.4)--(-0.63,2.86), linewidth(0.8)); draw((-3.593868592709285,-0.142258633097903)--(4.67,-5.4), linewidth(0.8)); draw(circle((-1.5469223379724755,0.8010835629503406), 2.2538596825503965), linewidth(0.8) + green); draw(circle((-3.978946076657836,-3.0214560388192346), 2.9048343102071406), linewidth(0.8) + qqffff); draw((-1.5469223379724755,0.8010835629503406)--(1.1296663605334942,-1.841279453894974), linewidth(0.8)); draw((1.1296663605334942,-1.841279453894974)--(-3.978946076657836,-3.0214560388192346), linewidth(0.8)); draw((-3.978946076657836,-3.0214560388192346)--(-1.5469223379724755,0.8010835629503406), linewidth(0.8)); draw((-3.593868592709285,-0.142258633097903)--(-0.63,2.86), linewidth(0.8)); draw((-3.593868592709285,-0.142258633097903)--(-2.37,-5.44), linewidth(0.8)); draw((-2.1119342963546424,1.3588706834510484)--(-1.5469223379724755,0.8010835629503406), linewidth(0.8)); /* dots and labels */ dot((-0.63,2.86),dotstyle); label("$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$", (-2.63,-1.22), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

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 $CX \perp AB$ .

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799 posts

#4 h Jun 21, 2020, 10:59 PM • 1 Y

Y by samrocksnature

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

This post has been edited 1 time. Last edited by Overlord123, Jun 21, 2020, 11:00 PM

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#5 h Jun 21, 2020, 11:00 PM • 2 Y

Y by samrocksnature, Danielzh

We claim that $[OO_1O_2]$ is minimized when $CX \perp AB$ (i.e. when $D$ is the foot of the altitude).

$[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$", (2.473451110105351,-1.019582085855788), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

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.

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#6 h Jun 21, 2020, 11:00 PM • 6 Y

Y by samrocksnature, centslordm, Mango247, Mango247, Mango247, dbnl

$[asy] size(6cm); pair A=dir(270-58), B=dir(270+58), C=dir(90+27); pair O=(0,0), X=dir(270-19), D=extension(C,X,A,B), O1=circumcenter(A,D,X), O2=circumcenter(B,D,X); pair P1=foot(D,A,X), P2=foot(D,B,X); fill(O--O1--O2--cycle,rgb(230,255,242)); draw(A--B--C--cycle); draw(X--A); draw(X--B); draw(X--C); draw(O--O1,blue+dashed); draw(O--O2,blue+dashed); draw(O1--O2,blue+dotted); draw(D--P1,purple+dotted); draw(D--P2,purple+dotted); draw(P1--P2,purple+dotted); draw(circumcircle(A,B,C)); draw(circumcircle(A,D,X)); draw(circumcircle(B,D,X)); dot(A); dot(B); dot(C); dot(O); dot(X); dot(D); dot(O1); dot(O2); dot(P1); dot(P2); label("$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$",D,dir(40)); [/asy]$
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 $CX\perp AB$ .

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#7 h Jun 21, 2020, 11:01 PM • 11 Y

Y by mira74, k12byda5h, IAmTheHazard, samrocksnature, tigerzhang, ike.chen, khina, EpicBird08, Jack_w, ihatemath123, Funcshun840

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

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#8 h Jun 21, 2020, 11:01 PM • 1 Y

Y by samrocksnature

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.

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#9 h Jun 21, 2020, 11:01 PM • 3 Y

Y by Ha_ha_ha, samrocksnature, a_n

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

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#10 h Jun 21, 2020, 11:02 PM • 1 Y

Y by samrocksnature

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.

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mathisawesome2169

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mathisawesome2169

#11 h Jun 21, 2020, 11:06 PM • 1 Y

Y by samrocksnature

bad trig bash outline

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277546

#12 h Jun 21, 2020, 11:09 PM • 1 Y

Y by samrocksnature

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.

This post has been edited 1 time. Last edited by 277546, Jun 21, 2020, 11:10 PM

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jeff10

#13 h Jun 21, 2020, 11:10 PM • 3 Y

Y by Lcz, superagh, samrocksnature

Short

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MP8148

#14 h Jun 21, 2020, 11:10 PM • 1 Y

Y by samrocksnature

$[asy] size(8cm); defaultpen(fontsize(10pt)); pair C = dir(120), A = dir(210), B = dir(330), O = origin, X = dir(290), D = extension(C,X,A,B), O1 = circumcenter(A,D,X), O2 = circumcenter(B,D,X); draw(unitcircle^^X--C--B--A^^O1--O2); draw(circumcircle(A,D,X)^^circumcircle(B,D,X), gray); draw(A--X--B^^O1--X--O2, magenta); draw(A--C--B^^O2--O--O1, heavygreen); dot("$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$", O2, dir(15)); [/asy]$
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. $\blacksquare$

This post has been edited 3 times. Last edited by MP8148, Jun 22, 2020, 12:41 AM

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#15 h Jun 21, 2020, 11:11 PM • 2 Y

Y by samrocksnature, Danielzh

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.

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