Claim: Given a line $\ell$ and a point $P\in\ell$, a line perpendicular to $\ell$ can be constructed through $P$ Select arbitrary points $X$, $Y$, and $Z$ on $\omega$, and locate points $X'$, $Y'$, and $Z'$ on $\omega$ satisfying $\overline{XX'}\parallel\overline{YY'}\parallel\overline{ZZ'}\parallel\ell$ (as shown in the figure below). The line through $P$ parallel to the one through $\overline{YZ'}\cap\overline{Y'Z}$ and $\overline{XY'}\cap\overline{X'Y}$ satisfies the above. Proving that this works is trivial and therefore left as an exercise for the reader. [asy][asy] size(6cm); defaultpen(fontsize(9pt)); pair A=(-90,-80), B=(60,-80), X=(-40,-30), Y=(-47.7,15), Z=(-30,40), XX=(40,-30), YY=(47.7,15), ZZ=(30,40), P=(-60,-80); draw(A--B, linewidth(0.5)); draw(circle((0,0), 50), linewidth(0.5)); draw(X--XX--Y--YY--Z--ZZ--Y, linewidth(0.5)); draw(X--YY, linewidth(0.5)); draw((-60,-100)--(-60,60), linewidth(0.4)+grey); draw((0,-100)--(0,60), linewidth(0.4)+grey); label("$\ell$", A, SW); dot("$X$", X, SW); dot("$Y$", Y, (-1,0)); dot("$Z$", Z, NW); dot("$X'$", XX, SE); dot("$Y'$", YY, (1,0)); dot("$Z'$", ZZ, NE); dot((0,30.4)); dot((0,-9.5)); dot("$P$", P, SW); [/asy][/asy] Claim: The midpoint of any given line segment $\overline{XY}$ can be constructed. Construct an arbitrary line $\overline{X'Y'}$ parallel to $\overline{XY}$, and let $P\equiv\overline{XX'}\cap\overline{YY'}$ and $Q\equiv\overline{XY'}\cap\overline{X'Y}$. By the theorem of Ceva, we see that $\overline{PQ}$ must, in fact, pass through the midpoint of $\overline{XY}$. [asy][asy] size(6cm); defaultpen(fontsize(9pt)); pair X=(-50,0), Y=(50,0), XX=(-35,20), YY=(25,20), P=(-12.5,50), Q=(-3.1,12.5); draw(X--Y, linewidth(0.5)); draw(XX--YY, linewidth(0.5)); draw(X--P--Y, linewidth(0.5)); draw(X--YY, linewidth(0.5)); draw(Y--XX, linewidth(0.5)); draw((-20,80)--(5,-20), linewidth(0.4)+grey); dot("$X$", X, SW); dot("$Y$", Y, SE); dot("$X'$", XX, NW); dot("$Y'$", YY, NE); dot("$M$", (0,0), SW); dot("$P$", P, (-1,0.5)); dot("$Q$", Q, (-0.5,-1)); [/asy][/asy] Observe that the two claims above, when used in conjunction with each other, allows one to construct the perpendicular bisector of any line segment; by extension, the cirucumcenter of any triangle may also be located. We now return to the actual problem. To begin, locate the center of $\omega$ and the circumcenter of $ABC$ (the triangle bound by $\ell_1$, $\ell_2$, and $\ell_3$)—denote these by $O'$ and $O$ respectively. Draw a line through $O'$ parallel to $\overline{OA}$, and mark $A'$ as one of its intersections with $\omega$. Draw $B'$ and $C'$ on $\omega$ satisfying $\overline{A'B'}\parallel\overline{AB}$ and $\overline{A'C'}\parallel\overline{AC}$: triangle $A'B'C'$ is the one we wished to construct. I'm too lazy to prove that this construction works, so we're done. [asy][asy] size(7cm); defaultpen(fontsize(9pt)); pair A=(-60,-40), B=(-80,-100), C=(-20,-100), O=(-50,-76.7), OO=(0,0), AA=(-10.5,38.6), BB=(-31.6,-24.6), CC=(31.6,-24.6); draw(A--B--C--cycle, linewidth(0.5)); draw(circle((0,0),40), linewidth(0.5)); draw(O--A, linewidth(0.5)); draw(OO--AA, linewidth(0.4)+dashed); draw(AA--BB, linewidth(0.4)+grey); draw(AA--CC, linewidth(0.4)+grey); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$A'$", AA, NW); dot("$B'$", BB, SW); dot("$C'$", CC, SE); dot("$O'$", OO, SW); dot("$O$", O, SW); [/asy][/asy]

Solution. We start with the following claim. Claim 1. Given a segment $AB$, we can construct the midpoint $M$ of $AB$ using the tools we have. (Proof): Consider any point $C$ not on line $AB$. Draw lines parallel to $CA$ through $B$ and parallel to $CB$ through $A$. Let these lines meet at $D$, so $CADB$ is a parallelogram. Consider the intersection of segments $AB$ and $CD$ which is the midpoint $M$ of $AB$ and we proved the claim. $~\square$ Claim 2. Given a circle $\Gamma$ and a segment $AB$, we can construct the midpoint of minor and major arcs of $AB$ in $\Gamma$ using the tools we have. (Proof): Consider a point $C$ on $\Gamma$ distinct from $A$ and $B$ (and $CA\neq CB$). Draw a line parallel to $AB$ through $C$ and say it intersects $\omega$ again at $D$. From claim 1, construct the midpoints $M$ and $N$ of $AB$ and $CD$. The line $MN$ meets $\Gamma$ at the midpoint of arcs $AB$ of $\Gamma$ (as $MN$ passes through the center of $\Gamma$). $~\square$ Claim 3. Given a circle $\Gamma$, we can construct the center of the circle. (Proof): Pick a chord $AB$ of $\Gamma$, from Claim 2, we can draw the perpendicular bisector of $AB$. Similarly we can pick another chord $CD$ of $\Gamma$ which is not parallel to $AB$ and draw its perpendicular bisector. Intersecting these lines, we get the center of the circle. $~\square$ Main Problem. Let $O$ be the center of $\omega$. Construct $O$ from Claim 3. Draw a line parallel to $l_1$ through $O$ and say it meets $\omega$ at $A$ and $B$. Draw a line parallel to $l_2$ through $A$ which meets $\omega$ again at $C$ and parallel to $l_3$ through $B$ which meets $\omega$ again at $D$. If $l_2\perp l_1$, then $C = A$ and if $l_3\perp l_1$, then $D = B$. If $C = D$, then we can just consider the triangle $\triangle CAB$ and we would be done. Now from claim 2, we construct any one of the midpoint of arc $CD$ in $\omega$, name it $P$. We draw parallel lines through $P$ to $CA$ and $DB$ which meets $\omega$ again at $Q,R$ respectively. We prove that $\triangle PQR$ is the required triangle. Firstly it $\triangle PQR$ is inscribed in $\omega$. Now if $P = Q$ then $AC$ is parallel to tangent at $P$ to $\omega$. But $P$ is the midpoint of arc $CD$, so $AC||CD$ or $A = D$ which would give lines $BA$ and $BD$ are same. But then $l_1$ is parallel to $l_3$ which is not true. Therefore $P\neq Q$ and similarly $P\neq R$. Suppose $Q = R$. Then $PQ$ and $PR$ are same or $AC$ is parallel to $DB$ which eventually gives that $l_2$ is parallel to $l_3$. Therefore $\triangle PQR$ is not degenerate and is inscribed in $\omega$. From our construction, $l_2||AC||PQ$, $l_3||BD||PR$ and $l_1||AB$, so we just remain to show that $AB||QR$. As $PQ||AC$ we get that $AQ=CP=PD = BQ$, so $QR||AB$ and we are done. $~\blacksquare$ [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 = -8.88, xmax = 15.52, ymin = -6.86, ymax = 6.12; /* image dimensions */ pair O = (0.,0.), B = (5.,0.), A = (-5.,0.), C = (1.9196256316882558,4.616821139503399), D = (-4.384062043356595,-2.4041630560342613), P = (3.72047261450566,-3.3403717644467985), R = (-1.6559968766197932,-4.717803974798602), Q = (1.6559968766197937,-4.717803974798603); /* draw figures */ draw(circle(O, 5.), linewidth(2.)); draw(A--B, red); draw(A--C, blue); draw(B--D, green); draw(C--D, linewidth(1.)); draw(R--Q, red); draw(R--P, green); draw(P--Q, blue); draw(P--D,dotted); draw(P--C,dotted); draw(R--B,dotted); draw(Q--A,dotted); /* dots and labels */ dot(O,dotstyle); label("$O$", (-0.06,-0.42), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (5.1,-0.24), NE * labelscalefactor); dot(A,linewidth(4.pt) + dotstyle); label("$A$", (-5.5,-0.32), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (1.9,4.7), NE * labelscalefactor); dot(D,dotstyle); label("$D$", (-4.8,-2.76), NE * labelscalefactor); dot(P,linewidth(4.pt) + dotstyle); label("$P$", (3.9,-3.4), NE * labelscalefactor); dot(R,linewidth(4.pt) + dotstyle); label("$R$", (-1.76,-5.14), NE * labelscalefactor); dot(Q,linewidth(4.pt) + dotstyle); label("$Q$", (1.7,-5.06), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

First, we need to invent another tool. Claim: Given a chord $AB$ of $\omega$, we can draw the perpendicular bisector of $AB$. Proof: Take another chord $CD\parallel AB$. Then the perpendicular bisector is the line joining $AC\cap BD$ and $AD\cap BC$. Let $ABC$ be the (currently invisible) triangle that we want. Draw the perpendicular bisector of any chord parallel to $BC\parallel\ell_1$. This induces two midpoints of arc $BC$, call them $M_A, N_A$. Similarly, we can construct $M_B, N_B, M_C, N_C$. Now, we need incenter $I$ of $\triangle ABC$. To construct that, recall the fact that $I$ is orthocenter of $\triangle M_AM_BM_C$. So all we have to do is just construct line through $M_A$ parallel to the perpendicular bisector of $M_BM_C$. Finally, point $A,B,C$ are just $M_AI\cap\omega$, $M_BI\cap\omega$ and $M_CI\cap\omega$ so we are done.

Can someone check this solution? Claim 1. It is possible to translate and rotate angles, preserving the measure of the angle. That is, given $\angle ABC$, $\angle B'$, $\overline{DE}$, it is possible to construct $\angle A'B'C'$ such that $\overline{A'B'} \parallel \overline{DE}$. Proof Translating the angle is trivial. For rotation, construct $F$ on circle $\omega$. Let the lines through $F$ parallel to $\overline{BA}$, $\overline{BC}$ intersect the circle again at $A'$, $C'$, respectively. Draw line through $A'$ parallel to $\overline{DE}$, and let the line intersect the circle again at $B'$. Clearly, $\angle A'B'C' = \angle ABC$. $\square$ Claim 2. It is possible bisect a segment. Proof Let the segment be $\overline{AB}$. Chose an arbitrary point $M$ not on $\overline{AB}$. Let the line through $A$ parallel to $BM$ and line through $B$ parallel to $AM$ intersect at point $N$.Let $M = \overline{AB} \cap \overline {MN}$. $M$ is the midpoint of $\overline{AB}$, since $AMBN$ is a parallelogram.$\square$ Claim 3. It is possible construct the perpendicular bisector of a chord. Proof Let the chord be $\overline{AB}$, with $A, B$ on circle $\omega$. Using claim 2, construct the midpoint of $\overline{AB}$, and denote it $M$. Chose arbitrary point $X$ on $\omega$. Using claim 1, construct $Y$ on $\omega$ such that $\angle XMA = \angle YMB$, and $Y$ on arc $AXB$. Let the midpoint of $\overline{XY}$ be $N$. By symmetry, $\angle MNA = \angle MNB$, and $\overline{MN} \perp \overline{AB}$. $\square$ Now, let the lines $l_1$ and $l_2$, $l_1$ and $l_3$, $l_2$ and $l_3$ intersect at $A, B, C$, respectively. Chose random $D$ on circle $\omega$. Draw lines parallel to $l_2$, $l_2$, and have them intersect $\omega$ at $E, F$, respectively. Copy $\angle B$ such that the angle with vertex $E$, edges $\overline{EF}$ and $\overline{EG}$, with $G$ on arc $EDG$, satisfies $\angle FEG = \angle B$. Clearly, $\triangle GEF \sim \triangle ABC$, but $\overline{GE}$ is not parallel to $\overline{AB}$. However, if $G$ is rotated towards $D$ by half of minor arc $DG$, and all other vertices rotated with the same measure in the same direction, then $\overline{GE} \parallel \overline{AB}$. Construct the perpendicular bisectors of $\overline{GE}$, $\overline{FG}$, and have them intersect at circumcenter $O$. Notice that it is possible to rotate each vertex of $\triangle GEF$ by $\frac{1}{2} \angle DOG $, by constructing angles with vertex at $O$ with angles of measure $\angle DEG$. The image $\triangle G'E'F'$ obviously satisfies $\triangle D'E'F' \sim \triangle ABC$, and $G'E' \parallel DE \parallel AB$, so it can be concluded that the three sides of $\triangle G'E'F'$ are parallel to the three liens $l_1, l_2, l_3$. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -9.466571289077443, xmax = 15.155821697684798, ymin = -4.661068709328523, ymax = 10.098887670509952; /* image dimensions */ /* draw figures */ draw(circle((0,0), 4), linewidth(1)); draw((xmin, 0.15555555555555553*xmin + 7.722297674418604)--(xmax, 0.15555555555555553*xmax + 7.722297674418604), linewidth(1)); /* line */ draw((xmin, 11.58333333333334*xmin-82.757)--(xmax, 11.58333333333334*xmax-82.757), linewidth(1)); /* line */ draw((xmin, -0.6607142857142856*xmin + 10.877503322259137)--(xmax, -0.6607142857142856*xmax + 10.877503322259137), linewidth(1)); /* line */ draw((-3.9236011474908685,0.7780450086006198)--(3.1187454701748782,-2.5046809561825793), linewidth(1)); draw((2.787553277104414,2.868718656003834)--(-3.9236011474908685,0.7780450086006198), linewidth(1)); draw((2.787553277104414,2.868718656003834)--(3.1187454701748782,-2.5046809561825793), linewidth(1)); draw((3.1793131647775974,2.427337595035321)--(2.716264595114421,-2.936308336896501), linewidth(1)); draw((2.716264595114421,-2.936308336896501)--(-3.7664153801700992,1.3468909324878997), linewidth(1)); draw((-3.7664153801700992,1.3468909324878997)--(3.1793131647775974,2.427337595035321), linewidth(1)); draw((3.5018647598956445,1.933117483082966)--(-3.9236011474908685,0.7780450086006198), linewidth(1)); draw((3.5018647598956445,1.933117483082966)--(3.1187454701748782,-2.5046809561825793), linewidth(1)); /* dots and labels */ dot((0,0),dotstyle); label("$O$", (0.08134569940244146,0.21398537655419894), NE * labelscalefactor); label("$\omega$", (-2.0079631945472744,2.9547991944237486), NE * labelscalefactor); dot((7.9174883720930245,8.953906976744186),dotstyle); label("$A$", (8.011733221598675,9.17779450220953), NE * labelscalefactor); dot((3.86539534883721,8.323581395348837),dotstyle); label("$B$", (3.9454438688578297,8.548755265321435), NE * labelscalefactor); dot((7.647348837209304,5.824790697674419),dotstyle); label("$C$", (7.742144977218066,6.05506400480078), NE * labelscalefactor); label("$D$", (3.585992876350352,2.168500148313632), NE * labelscalefactor); dot((-3.9236011474908685,0.7780450086006198),dotstyle); label("$E$", (-3.8276838441163816,1.0002844226643157), NE * labelscalefactor); dot((3.1187454701748782,-2.5046809561825793),dotstyle); label("$F$", (3.2040761968111564,-2.279705883966457), NE * labelscalefactor); dot((2.787553277104414,2.868718656003834),dotstyle); label("$G$", (2.867090891335396,3.0895933166140543), NE * labelscalefactor); dot((3.1793131647775974,2.427337595035321),dotstyle); label("$G'$", (3.2714732579063086,2.6627452630114194), NE * labelscalefactor); dot((2.716264595114421,-2.936308336896501),dotstyle); label("$F'$", (2.7996938302402436,-2.7065539375690917), NE * labelscalefactor); dot((-3.7664153801700992,1.3468909324878997),dotstyle); label("$E'$", (-3.67042403489436,1.5619265984572561), NE * labelscalefactor); label("$q$", (2.57503695992307,-0.21286267704843584), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

Given a conic and $3$ points outside it on the projective plane, construct a triangle inscribed in the given conic such that the lines containing the sides of the triangle pass through the given $3$ points. Indeed, for two lines on the affine plane, to be parallel means to intersect the line at infinity of the projective plane at the same point. We are given with $3$ points where our $3$ given lines intersect the infinite line and we want to draw $3$ lines passing through these $3$ points and intersecting on the circle. The problem we have formulated above can be solved by means of homographies ($=$ linear projective isomorphisms between projective lines $=$ automorphisms of conics preserving the cross-ratio). Namely, realize some naive drawing starting from an arbitrary point $x$ of the circle, that is, draw a line from $x$ to the first given point, write $y$ for the second intersection point of this line with the circle, then draw the line from $y$ to the second given point, etc. Let $x'$ be the return point. We want $x' = x$. But most likely we get $x'$ not equal to $x$. However, the mapping $x \to x'$ is a homography from the circle to itself. There is the standard way to construct the fixed points of a homography (in general there are $2$ such the points) as soon the images of some $3$ points under the homography are known. It is based on Pascal's theorem on inscribed hexagon. See for instance the following notes. http://gorod.bogomolov-lab.ru/ps/stud/projgeom/1718/lec_03.pdf

Proof: Claim 1: From A,B we can draw midpoint, Claim 2: We can construct O' (easily by draw isoceles trapezoid). Now draw O'X, O'Y//AB,AC, OH perp XY, X'Y' cuts OH at M, MN//HX' with N on OX', HN cuts (O') at B notice HXMN isoceles trapezoid thus X'OY' ~ HOB (s.s), draw BA, BC//YZ, ZX and by angle chase, A,C lies on (O') too.

Lemma: Given any two points $A,B$ on $\omega$, we can mark the midpoint of arc $\widehat{AB}$.

For the main problem, we do the following: take any point $A_1$ on $\omega$, and construct $A_2,A_3,A_4,A_5,A_6$ so that $A_1A_2||\ell_1,A_2A_3||\ell_2,A_3A_4||\ell_3$ and $A_4A_5||\ell_1, A_5A_6||\ell_2$. Since then $\widehat{A_1A_4}=\widehat{A_2A_5}=\widehat{A_3A_6}$, we see that $A_6A_1||A_3A_4||\ell_3$. [asy][asy]size(5cm); pair A1=dir(45),A2=dir(135),A5=dir(165),A6=dir(255),A3=dir(285),A4=dir(15),X1=dir(30),X2=dir(150),X3=dir(270); draw(unitcircle,green); draw(MP("A_1",A1,A1)--MP("A_2",A2,A2)--MP("A_3",A3,A3)--MP("A_4",A4,A4)--MP("A_5",A5,A5)--MP("A_6",A6,A6)--cycle,magenta); draw(MP("X_1",X1,X1)--MP("X_2",X2,X2)--MP("X_3",X3,X3)--cycle,blue+dashed); dot(A1^^A2^^A3^^A4^^A5^^A6^^X1^^X2^^X3); [/asy][/asy] Now using the lemma, we can draw $X_1,X_2,X_3$, the arc midpoints of $A_1A_4,A_2A_5,A_3A_6)$. Clearly $X_1X_2X_2$ is the required triangle. $\blacksquare$

Has anybody done this using homothety

$\textbf{Lemma.}$ Given two points $X$ and $Y$ in the plane, we can construct the midpoint of segment $\overline{XY}$. $\textbf{Proof:}$ The line $\overline{XY}$ is not parallel to at least two of the directions $l_1,l_2,l_3,$ w.l.o.g. the last two of them. Build two line through $X$, one parallel to $l_2$ and one parallel to $l_3$ and two lines through $Y$, one parallel to $l_2$ and one parallel to $l_3$. Let the intersection points be $U$ and $V$. Now consider the point $M\in \overline{XY}\cap \overline{UV}$. Since $XUYV$ is a parallelogram, $M$ is the midpoint of $\overline{XY}$, as required. Consider now on $\omega$ two distinct very close points $A_0$ and $A_1$. For $i\in \{0,1\}$, define $B_i,C_i,D_i,E_i$ and $F_i$ as follows: all these points lie on $\omega$ and $\overline{A_iB_i}\parallel l_1$, $\overline{B_iC_i}\parallel l_2$, $\overline{C_iD_i}\parallel l_3$, $\overline{D_iE_i}\parallel l_1$, $\overline{E_iF_i}\parallel l_2$. We claim that $\overline{A_iF_i}\parallel l_3$. Indeed, $\overline{A_iB_i}\parallel \overline{D_iE_i}$ gives $A_iD_i=B_iE_i$ and $\overline{B_iC_i} \parallel \overline{E_iF_i}$ gives $B_iE_i=C_iF_i$. These two equalities give further that $A_iD_i=C_iF_i$, so $\overline{A_iF_i}\parallel \overline{C_iD_i}$ or $\overline{A_iF_i}\parallel l_3$, as required. Define $M_i,N_i,P_i$ the midpoints of $\overline{A_iD_i}$, $\overline{B_iE_i}$ and $\overline{C_iF_i}$ for both $i$. $A_iD_iE_iF_i$ is an isosceles trapezoid with $\overline{D_iE_i}\parallel \overline{A_iB_i}$, so $\overline{M_iN_i}\parallel \overline{A_iB_i}\parallel l_1$. Similarly $\overline{M_iP_i}\parallel l_3$ and $\overline{N_iP_i}\parallel l_2$. Now, again by parallelisms, $A_0A_1=B_0B_1=C_0C_1=D_0D_1=E_0E_1=F_0F_1$, so $A_0D_0D_1A_1$, $E_0B_0B_1E_1$ and $C_0F_0F_1C_1$ are all isosceles trapezoids. Therefore, if we let $\overline{M_0M_1}$ cut the smaller arc of the four arcs ${A_iD_i}$, in point $M$, $M$ will be the midpoint of the arc ${A_0D_0}$ (and of the arc ${A_1D_1}$, too). Similarly define $N\in \overline{N_0N_1}\cap \omega$ and $P\in \overline{P_0P_1}\cap \omega$, both on the smaller arcs. Now, conclude: $M$ is the midpoint of arc ${A_0D_0}$, $N$ is the midpoint of arc ${B_0E_0}$, $P$ is the midpoint of arc ${C_0F_0}$ so $\overline{MN}\parallel l_1$, $\overline{NP}\parallel l_2$, $\overline{PM}\parallel l_3$. Since $M,N,P\in \omega$, we are done. $\textbf{Remark.}$ Obviously all these operations were possible, thanks to te special straightedge, the normal straightedge and to the lemma. $\textbf{Further remark.}$ Lines $\overline{M_0M_1}$, $\overline{N_0N_1}$ and $\overline{P_0P_1}$ cut each other at the center of $\omega$, so we can obtain the center of $\omega$, too.

We show how to do the following constructions first: 1) Constructing the midpoint of a line segment $AB$. Construct a triangle $ABC$ with any arbitrary point $C$ not on line $AB$. Then construct the parallelogram $ACBX$ with $AB$ and $CX$ as two diagonals. The intersection of $AB$ and $CX$ is the midpoint of $AB$. $\square$ 2) Constructing the centre of a circle $\Omega$. Draw two parallel chords of $\Omega$. Construct their midpoints, $M,N$ and let the line passing through $M,N$ hit $\Omega$ at $X$ and $Y$. So $\overline{XY}$ is a diameter and the midpoint of $\overline{XY}$ is the centre of $\Omega$. $\square$ 3) Constructing a perpendicular to a line $l$ through a point $X$ on $l$. Note that the circle $\omega$ is given to us. So we may use it as we wish to. Construct two distinct chords of $\omega$ parallel to $l$. Let the midpoints of the two chords be $M_1, M_2$. We know that $M_1M_2$ is perpendicular to the chords and hence to $l$. Now just draw the line parallel to $M_1M_2$ from $X$. This is our desired line. $\square$ 4) Constructing the circumcentre of a triangle $ABC$. Construct the midpoints of the sides, draw the perpendiculars to the sides, from their midpoints, and they meet at $O$, the circumcentre. $\square$ Now we move to the main problem at hand. If the lines $l_1, l_2,l_3$ intersect at a point $P$, then draw another line $l_4$ parallel to $l_3$, not passing through $P$. The pairwise intersections of $l_1,l_2,l_4$ gives a non-degenerate triangle $PQR$ with sides parallel to $l_1,l_2,l_3$ in some order. In case $l_1,l_2,l_3$ do not meet at a point, we simply take the pairwise intersections of $l_1,l_2,l_3$. Now construct $O$, the circumcentre of $\triangle PQR$ and the centre $O'$ of $\omega$. Also, WLOG suppose that $P$ is not on $OO'$. Construct $P'$ on the circumference of $\omega$ such that $P'$ is on the same side of $OO'$ as $P$, and $OP\parallel O'P'$. Now construct $Q', R'$ on the circumference of $\omega$ such that $P'Q'\parallel PQ$ and $P'R'\parallel PR$. We can see that the homothety that maps $(PQR)$ to $\omega$, also maps $\triangle PQR $ to $\triangle P'Q'R'$, so all the sides of $\triangle P'Q'R'$ are parallel to the sides of $\triangle PQR$, and hence to $l_1,l_2,l_3$ in some order. Finally draw in line segment $Q'R'$, and so $\triangle P'Q'R'$ is our desired triangle. $\blacksquare$

amaanmathbuddy_2006 wrote: Has anybody done this using homothety See the first solution.

Severus wrote: We show how to do the following constructions first: 1) Constructing the midpoint of a line segment $AB$. Construct a triangle $ABC$ with any arbitrary point $C$ not on line $AB$. Then construct the parallelogram $ACBX$ with $AB$ and $CX$ as two diagonals. The intersection of $AB$ and $CX$ is the midpoint of $AB$. $\square$ 2) Constructing the centre of a circle $\Omega$. Draw two parallel chords of $\Omega$. Construct their midpoints, $M,N$ and let the line passing through $M,N$ hit $\Omega$ at $X$ and $Y$. So $\overline{XY}$ is a diameter and the midpoint of $\overline{XY}$ is the centre of $\Omega$. $\square$ 3) Constructing a perpendicular to a line $l$ through a point $X$ on $l$. Note that the circle $\omega$ is given to us. So we may use it as we wish to. Construct two distinct chords of $\omega$ parallel to $l$. Let the midpoints of the two chords be $M_1, M_2$. We know that $M_1M_2$ is perpendicular to the chords and hence to $l$. Now just draw the line parallel to $M_1M_2$ from $X$. This is our desired line. $\square$ 4) Constructing the circumcentre of a triangle $ABC$. Construct the midpoints of the sides, draw the perpendiculars to the sides, from their midpoints, and they meet at $O$, the circumcentre. $\square$ Now we move to the main problem at hand. If the lines $l_1, l_2,l_3$ intersect at a point $P$, then draw another line $l_4$ parallel to $l_3$, not passing through $P$. The pairwise intersections of $l_1,l_2,l_4$ gives a non-degenerate triangle $PQR$ with sides parallel to $l_1,l_2,l_3$ in some order. In case $l_1,l_2,l_3$ do not meet at a point, we simply take the pairwise intersections of $l_1,l_2,l_3$. Now construct $O$, the circumcentre of $\triangle PQR$ and the centre $O'$ of $\omega$. Also, WLOG suppose that $P$ is not on $OO'$. Construct $P'$ on the circumference of $\omega$ such that $P'$ is on the same side of $OO'$ as $P$, and $OP\parallel O'P'$. Now construct $Q', R'$ on the circumference of $\omega$ such that $P'Q'\parallel PQ$ and $P'R'\parallel PR$. We can see that the homothety that maps $(PQR)$ to $\omega$, also maps $\triangle PQR $ to $\triangle P'Q'R'$, so all the sides of $\triangle P'Q'R'$ are parallel to the sides of $\triangle PQR$, and hence to $l_1,l_2,l_3$ in some order. Finally draw in line segment $Q'R'$, and so $\triangle P'Q'R'$ is our desired triangle. $\blacksquare$ thank you

We start by a lemma: Lemma: Given any segment $XY,$ we can construct its midpoint. Proof: $XY$ can be parallel to at most one side of $\triangle ABC.$ Assume wlog that it is not parallel to $AB,AC.$ Let the parallels through $X,Y$ to $AB,AC$ meet at $Z,$ while the parallels through $X,Y$ to $AC,AB$ meet at $W.$ Clearly, $ZXWY$ is a parallelogram. So $ZW \cap XY$ gives the midpoint of $XY.$ $\square$ We now give an algorithm: Draw an arbitrary line parallel to $BC,$ say $EF,$ such that $E,F \in \omega.$ Let $K,L \in \omega$ be points such that $KE \parallel AB, LF \parallel CA.$ Let $KE \cap LF=P.$ Construct the midpoint $M$ of $KL$ as in the lemma. Draw the parallel to $KL$ through $F,$ and let it meet $\omega$ again at $G.$ Let $N$ be the midpoint of $FG.$ Construct $N$ as in the lemma. Let $MN$ intersect the arc $EKLF$ of $\omega$ at $X.$ Draw through $X$ the lines parallel to $AB,AC$ and let them meet $\omega$ in $Y,Z$ respectively. We claim that $XYZ$ is the desired triangle. Why does it work? Here's the reason: $KLFG$ is an isosceles trapezium and so $XK=XL \implies X$ is the midpoint of arc $KL.$ It suffices to show $YZ \parallel EF$ for which it is sufficient to show $\measuredangle PEF=\measuredangle XYZ.$ Now \begin{align*} \measuredangle XYZ &=\measuredangle XYL+\measuredangle LYZ \\ &=\measuredangle KYX+\measuredangle LXZ \quad (\text{by Fact 5 on } \triangle KYL) \\ &=\measuredangle KYX+\measuredangle XZF \quad (\text{as }LF \parallel XZ) \\ &=\measuredangle KEX+\measuredangle XEF \\ &= \measuredangle KEF \end{align*}and so we are done. $\blacksquare$ [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 = -17.107256337000432, xmax = 26.907824407442693, ymin = -13.21600794459077, ymax = 14.714241592959675; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((10.8333700137642,6.4954226812781215)--(10.75087712372571,-1.423894762417018)--(18.17523722718988,1.1058871987633738)--cycle, linewidth(0.4) + rvwvcq); draw((-7.305571861205497,4.280859560734392)--(-7.439487977078006,-8.575087563026436)--(4.612962451447728,-4.468326676269502)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw(circle((-2.8879473626380943,-2.193828402776268), 7.838175198040212), linewidth(0.4) + wrwrwr); draw((10.8333700137642,6.4954226812781215)--(10.75087712372571,-1.423894762417018), linewidth(0.4) + rvwvcq); draw((10.75087712372571,-1.423894762417018)--(18.17523722718988,1.1058871987633738), linewidth(0.4) + rvwvcq); draw((18.17523722718988,1.1058871987633738)--(10.8333700137642,6.4954226812781215), linewidth(0.4) + rvwvcq); draw((-7.305571861205497,4.280859560734392)--(-7.439487977078006,-8.575087563026436), linewidth(0.4) + rvwvcq); draw((-7.439487977078006,-8.575087563026436)--(4.612962451447728,-4.468326676269502), linewidth(0.4) + rvwvcq); draw((4.612962451447728,-4.468326676269502)--(-7.305571861205497,4.280859560734392), linewidth(0.4) + rvwvcq); draw((-10.592087736443204,-0.7504949222947814)--(-1.4221568838126235,5.50607086327633), linewidth(0.4) + wrwrwr); draw((-7.305571861205497,4.280859560734392)--(-10.592087736443204,-0.7504949222947814), linewidth(0.4) + wrwrwr); draw((-7.305571861205497,4.280859560734392)--(-1.4221568838126235,5.50607086327633), linewidth(0.4) + wrwrwr); draw((-10.620482197493962,-3.4763631831675283)--(4.018914916854186,1.511875833573339), linewidth(0.4) + wrwrwr); draw((-8.857363735136426,-7.273499514858382)--(4.018914916854186,1.511875833573339), linewidth(0.4) + wrwrwr); draw((-7.305571861205497,4.280859560734392)--(-2.41922440914112,-2.8808118406425214), linewidth(0.4) + wrwrwr); draw((-10.45782222955676,12.138993738803883)--(-10.620482197493962,-3.4763631831675283), linewidth(0.4) + wrwrwr); draw((-10.45782222955676,12.138993738803883)--(4.018914916854186,1.511875833573339), linewidth(0.4) + wrwrwr); draw((-10.592087736443204,-0.7504949222947814)--(-7.439487977078006,-8.575087563026436), linewidth(0.4) + wrwrwr); draw((-7.439487977078006,-8.575087563026436)--(-1.4221568838126235,5.50607086327633), linewidth(0.4) + wrwrwr); draw((-1.4221568838126235,5.50607086327633)--(4.612962451447728,-4.468326676269502), linewidth(0.4) + wrwrwr); draw((4.018914916854186,1.511875833573339)--(4.612962451447728,-4.468326676269502), linewidth(0.4) + wrwrwr); /* dots and labels */ dot((10.8333700137642,6.4954226812781215),linewidth(3pt) + dotstyle); label("$A$", (10.490490229864756,6.817296039187973), NE * labelscalefactor); dot((10.75087712372571,-1.423894762417018),linewidth(3pt) + dotstyle); label("$B$", (9.86704715983015,-2.1602841693103847), NE * labelscalefactor); dot((18.17523722718988,1.1058871987633738),linewidth(3pt) + dotstyle); label("$C$", (18.51212439764334,0.956931180862656), NE * labelscalefactor); dot((-7.305571861205497,4.280859560734392),linewidth(3pt) + dotstyle); label("$X$", (-8.088113257166478,4.822278215077227), NE * labelscalefactor); dot((4.612962451447728,-4.468326676269502),linewidth(3pt) + dotstyle); label("$Z$", (5.4198199269166345,-4.944996548798302), NE * labelscalefactor); dot((-7.439487977078006,-8.575087563026436),linewidth(3pt) + dotstyle); label("$Y$", (-8.129676128502117,-9.683163881061324), NE * labelscalefactor); dot((-10.592087736443204,-0.7504949222947814),linewidth(3pt) + dotstyle); label("$K$", (-11.579394449360267,-1.5784039706114172), NE * labelscalefactor); dot((-10.620482197493962,-3.4763631831675283),linewidth(3pt) + dotstyle); label("$E$", (-11.371580092682064,-4.695619320784458), NE * labelscalefactor); dot((4.018914916854186,1.511875833573339),linewidth(3pt) + dotstyle); label("$F$", (4.089808044176143,1.9128772215823886), NE * labelscalefactor); dot((-1.4221568838126235,5.50607086327633),linewidth(3pt) + dotstyle); label("$L$", (-1.022425130107619,5.736661384461319), NE * labelscalefactor); dot((-10.45782222955676,12.138993738803883),linewidth(3pt) + dotstyle); label("$P$", (-11.205328607339503,12.80234951152021), NE * labelscalefactor); dot((-6.007122310127914,2.377787970490774),linewidth(3pt) + dotstyle); label("$M$", (-6.176221175727021,0.956931180862656), NE * labelscalefactor); dot((-8.857363735136426,-7.273499514858382),linewidth(3pt) + dotstyle); label("$G$", (-9.70906523925645,-8.145337641642623), NE * labelscalefactor); dot((-2.41922440914112,-2.8808118406425214),linewidth(3pt) + dotstyle); label("$N$", (-2.1861855275055486,-3.698110408729085), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Lemma 1: Given any point $A$ on $\omega$, we can construct the tangent at $A$ to $\omega$.

Lemma 2: Given any points $M$ and $N$, we can construct the midpoint of $MN$.

We claim we can construct lines $k_1, k_2, k_3$ tangent to $\omega$ parallel to $\ell_1, \ell_2, \ell_3$ in that order.

Let $k_2 \cap k_3 = A$, $k_1 \cap k_3 = B$, $k_1 \cap k_2 = C$. The problem can be restated as follows: New problem wrote: Given a triangle $ABC$ and its incircle $\omega$ centered at $I$ intersecting $BC, AC, AB$ at $D,E,F$, construct a triangle $A'B'C'$ with circumcircle $\omega$ and with sides parallel to those of $\Delta ABC$, using the same tools as in the original problem. We claim we can construct the isogonal conjugate $N^*$ of the Nagel Point $N$ of $\Delta ABC$.

Lemma 3: $N^*$ is the exsimilicenter of $\omega$ and the circumcircle $\Omega$ of $\Delta ABC$.

Thus we can let $A'$ be the intersection of segment $AN^*$ and $\omega$, and similarly define $B'$ and $C'$. We know that $A'$ is the image of $A$ under the homothety at $N^*$ sending $\Omega$ to $\omega$, and similarly for $B'$ and $C'$. Since the image of a line under a homothety is parallel to the original one, we have that the sides of $\Delta A'B'C'$ are parallel to $\Delta ABC$ and we're done.

Alright, the Poncelet-Steiner theorem says we can do anything a compass and straightedge can do, if only we can construct the center of the given circle. That turns out to be fairly easy. Given any line, we can construct a diameter of the fixed circle that is perpendicular to the line. Now that I think of it (it's not in the diagram) we might as well make that first diameter, then simply make the diameter perpendicular to that. The intersection of the two (perpendicular) diameters is the center of the circle. See attached diagram. The fixed circle is in red, so is the initial line being used. Given two diameters that are not identical, the center is just where the diameters intersect. So, the original problem ( inscribed triangle) can be solved. We know that by Poncelet-Steiner theorem

tastymath75025 wrote: Carl is given three distinct non-parallel lines $\ell_1, \ell_2, \ell_3$ and a circle $\omega$ in the plane. In addition to a normal straightedge, Carl has a special straightedge which, given a line $\ell$ and a point $P$, constructs a new line passing through $P$ parallel to $\ell$. (Carl does not have a compass.) Show that Carl can construct a triangle with circumcircle $\omega$ whose sides are parallel to $\ell_1,\ell_2,\ell_3$ in some order. Proposed by Vincent Huang It is well known that (straightedge+circle with center) is equivalent to (straight edge + compass) with which we can obviously solve this problem. Claim- Given $A,B$, we can find there midpoint $M$. Let $C$ be any point not on $AB$ and let $E$ be a point on $BC$ other than $B$ or $C$ , let line through $C$ and parallel to $AB$ intersect $AE$ at $F$ , using straight edge construct the fourth harmonic of $A,E,F$ call it $G$ and let $CG$ intersect $AB$ at $M$. Then $(AB,M\infty_{AB})=(AE,GF)=-1$ and hence $M$ is mid point of $AB$ Now just construct two sets of parallel chords on the circle and the intersection of the lines joining midpt of parallel pair of chords is the center of the circle Hence proved.

Cute tastymath75025 wrote: Carl is given three distinct non-parallel lines $\ell_1, \ell_2, \ell_3$ and a circle $\omega$ in the plane. In addition to a normal straightedge, Carl has a special straightedge which, given a line $\ell$ and a point $P$, constructs a new line passing through $P$ parallel to $\ell$. (Carl does not have a compass.) Show that Carl can construct a triangle with circumcircle $\omega$ whose sides are parallel to $\ell_1,\ell_2,\ell_3$ in some order. Proposed by Vincent Huang Recall that if we are given $\triangle ABC$ then knowing either the midpoint of $\overline{BC}$ or the line through $A$ parallel to $\overline{BC}$ is enough to construct the other, using straight-edge only. Lemma. Given a point $P$ and a line $\ell$, Carl can draw a perpendicular from $P$ to $\ell$. Proof. Pick $A, B, C \in \omega$ such that $\overline{AB} \parallel \ell$. Then draw parallel from $C$ to $\ell$ to meet $\omega$ again at $D$. Since $\overline{AB} \parallel \overline{CD}$ we can construct both their midpoints and join them by a line to meet $\omega$ at $C_1, C_2$; the arc midpoints. Repeat this for another triangle with base not parallel to $\overline{AB}$ and we obtain the centre $O$ of $\omega$. Thus we can draw parallel line from $P$ to $\overline{C_1C_2}$. $\blacksquare$ Notice that the above lemma also constructs the centre of $O$. Thus we draw perpendiculars from $O$ to $\ell_i$ to meet the circle $\omega$ at $A_1, B_1, C_1$. These are the arc-midpoints. Draw perpendiculars from $A_1$ to $\overline{B_1C_1}$ to meet $\omega$ again at $A$. Likewise, draw $B$ and $C$. So $\triangle ABC$ is our desired triangle.

Cool problem though i missed it last year during the contest. We begin with a claim firstly. Claim 1: Carl can draw the perpendicular bisector of any chord $\overline{AB}$ of $\omega$ [asy][asy] size(6cm); pair A=(7.76,1.48); pair B=(15.196429620360057,1.5990461124301159); pair C=(6.3685098294922105,-0.9114229107566338); pair D=(10.05444993479033,2.8299068927268523); pair D1=(12.859946485499774,2.8748186951439227); pair C1=(16.663753362237856,-0.7466114647181894); draw(circumcircle(A,B,D),orange); draw(A--B,green); draw(C--C1,green); draw(D--D1,green); draw(B--D,green); draw(A--D1,green); draw(A--C1,green); draw(B--C,green); dot("$A$",A,W); dot("$B$",B,E); dot("$C$",C,SW); dot("$C'$",C1,SW); dot("$D'$",D1,NE); dot("$D$",D,NE); [/asy][/asy] Proof: Choose and 2 random points on $\omega$ $C,D$ on opposite sides of $\overline{AB}$. Let $C',D'$ be the intersections of the line through $C,D$ parallel to $\overline{AB}$. Then clearly the line joining $\overline{CB} \cap \overline{C'A}$ and $\overline{DB} \cap \overline{D'A}$ is the perpendicular bisector (in particular take the intersection of the diagnols). A corllary of this is that you can also draw a line through any point $P$ perpendicular to a chord $\overline{AB}$ and infact it can be easily seen that you could remove the "chord" condition and can extend this to any line $\ell$ in the plane $\qquad \square$ Now we state another claim. Claim 2: Given any segment $\overline{XY}$ you can draw the midpoint of $\overline{XY}$ [asy][asy] pair X=(-8,2); pair Y=(2,2); pair P=(-2.38114,6.36038); pair Q=(-3.61886,-2.36038); pair M=(-3,2); draw(X--P--Y--Q--X,green); draw(X--Y,orange); draw(P--Q,orange); dot("$P$",P,N); dot("$Q$",Q,S); dot("$X$",X,W); dot("$Y$",Y,E); dot("$M$",M,S); [/asy][/asy] Proof: Take any random point $P$ and draw the lines thorugh $X,Y$ parallel to $\overline{PY},\overline{PX}$ respectively. Let these lines meet at $Q$. Then notice that $PQXY$ is a parallelogram and hence $\overline{PQ} \cap \overline{XY}$ is the desired midpoint $\qquad \square$ Now back to the main problem. Let $A,B,C$ be the pairwise intersection of the lines $\ell_1,\ell_2,\ell_3$. Notice that the first claim toghether with the second claim implies that we can draw the perpendicular bisectors of $\overline{BC} ,\overline{CA},\overline{AB}$ respectively. To finish notice that there intersection $O$ is nothing but the circumcenter of $\triangle{ABC}$. By claim 1 we already have the center of $\omega$. Now let the desired triangle be $\triangle{XYZ}$. Then notice that if $\ell_O$ be the line through the circumcenter of $\omega$ parallel to $\overline{AO}$ then $\ell_O \cap \omega=X$ and now we can definitely construct the other points $\qquad \blacksquare$

Very fun problem, although the solution took me way longer than $4.5$ hours to find

Great problem Claim. Given a segment $\overline{AB}$, we can construct the midpoint of segment $\overline{AB}$. Proof. Take a random point $P$, now construct line parallel to $\overline{AB}$ passing through $P$, let it be $\ell$. Take any two random points in $\ell$, let it be $C$ and $D$. Now, consider trapezoid $ABCD$ (we could easily ensure that $ |AB| < |CD|$). Now, consider $AC \cap BD = E$ and $AD \cap BC = F$. We claim that the line $EF$ bisect $\overline{AB}$. Consider the line parallel to $AB$ passing through $F$. Let it intersect $AC$ and $BD$ at $M$ and $N$ respectively. Now, we have \[ |MF| = \frac{|CF|}{|BC|} \cdot |AB| = \frac{|DF|}{|AD|} \cdot |AB| = |FN| \]By homothety from point $E$, we get that $EF$ bisects $\overline{AB}$ easily. Claim. Given a line $\ell$ and a point $P$, we can construct line $\ell' \perp \ell$ passing through $P$. Proof. Take two point $X,Y \in \omega$ such that $XY \nparallel \ell$. Now, construct point $X', Y' \in \omega$ such that $XX' \parallel YY' \parallel \ell$. Now, notice that we can construct midpoint of $XX'$ and $YY'$ by the previous claim. Connecting these two points form a line that passes through the center of the circle, which therefore forces this line, name it $k$ to be perpendicular to both of these lines, and therefore these line is perpendicular to $\ell$ as well. Use our tool to construct line $\ell'$ by taking a line parallel to $k$ passing through $P$. By our proof, we even prove that: 1. For any chord $k$ on circle $\omega$, we can construct a perpendicular bisector of $k$. Just do the same exact procedure as before. 2. We can construct the center of the circle $\omega$: construct two perpendicular bisector of the chords of different gradient, which is possible since we are given $\ell_1, \ell_2, \ell_3$. Now, to deal with the problem. First, we construct the center of $\omega$, let it be $O$. Construct points $A, B \in \omega$ such that $AB \parallel \ell_1$ and $AB$ passes through $O$ with our tool. Now, let $C,D$ be the point in $\omega$ such that $BC \parallel \ell_2$ and $AD \parallel \ell_3$. If $C = D$, we are done, since we have constructed $\triangle ABC$ that satisfies the condition. Otherwise, connect segment $CD$. Construct the perpendicular bisector of segment $CD$. Let it intersects the minor arc of $CD$ at $X$. Now, construct line parallel to $\ell_2$ and $\ell_3$ passing through $X$ which intersects $\omega$ at $Y$ and $Z$ respectively. Main Claim. $YZ \parallel \ell_1$. Proof. It suffices to prove that $\widehat{BY} = \widehat{AZ}$. But to prove this, we just need to observe that \[ \widehat{BY} \overset{XY \parallel BC}{=} \widehat{CX} \overset{X \ \text{midarc } CD}{=} \widehat{XD} \overset{XZ \parallel AD}{=} \widehat{AZ} \]Thus, triangle $XYZ$ satisfies the given criteria, and we are done.

We split the problem into tasks and corresponding constructions. Task 1: Prove that given any two points $A$ and $B$ in the plane, one can find their midpoint. Task 2: Find the circumcenter $O$ of $\omega$. Task 3: Find the diameter of $\omega$ perpendicular to $\ell_1$. Task 4: Let the desired triangle be $\triangle XYZ$ with incenter $I$, so we have eight candidates for the triangle formed by the second intersections of $XI, YI, ZI$ with $\omega$ determined by the diameters found by iterating Task 3 on each line. Find $I$ given such a triangle, then note that finding $I$ determines the possible candidate of $\triangle XYZ$ and check if it works by going through all cases for sides corresponding to $\ell_1, \ell_2, \ell_3$. Construction 1: Choose a point $C$ not on line $AB$. Use the parallel line feature to generate the point $C’$ such that $CAC’B$ is a parallelogram. Then the midpoint of $AB$ is $CC’\cap AB$. Construction 2: Pick a chord of $\omega$. Draw a parallel chord to it which is not the first chord. Then the line through the midpoints of the chords passes through $O$. Choose a chord not parallel to the first and iterate the process, then consider the intersection of the two lines passing through $O$. Construction 3: Draw a parallel chord to $\ell_1$ on $\omega$. Take the midpoint of this chord. If the chord does not pass through $O$, then the line through $O$ and the midpoint of the chord works. Otherwise, choose a distinct parallel chord, which suffices. Construction 4: Let the triangle of second intersections be $\triangle MNP$. Then let the second intersection of $OM$ and $\omega$ be $M’$. The point $I$ is such that $INM’P$ is a parallelogram, so we can generate $I$ this way.

A difficult thing to write up without asy Let $\Delta$ be the desired triangle. We claim we can find any pair of arc midpoints as well as the line perpendicular to a chord $AB$. Indeed, assume we have two points $U$ and $V$, by marking an arbitrary point $X$ not on $AB$ we can construct the line parallel to $UX$ passing through $V$ and the line parallel to $VX$ passing through $U$. Let them intersect at $Y$. Notice that $UXVY$ is a parallelogram, thus $XY$ bisects $UV$, yielding the midpoint of $UV$. Now consider chord a $AB$ on $\omega$, marking another chord $BC$ which is parallel to $AB$ then marking both midpoints and drawing a line between them gives a line perpendicular to $AB$ passing through the arc midpoints of $AB$. Now, consider an arbitrary point $P$ on $\omega$ then let the parallels of $\ell_1,\ell_2,$ and $\ell_3$ passing through $P$ intersect $\omega$ at $A$, $B$, and $C$. Mark the arc midpoints of $PA, PB$ and $PC$. This gives six points, choose an arbitrary set of $3$ alternating points, say $D$, $E$, and $F$. Notice that $D$, $E$, and $F$ are the arc midpoints of $\Delta$, so drawing the three altitudes of $DEF$ may give $\Delta$, it is easy to check whether or not a desired triangle on $\omega$ is $\Delta$. Thus, if $DEF$ is not $\Delta$ then choose the other triple of alternating points and apply the same process. This yields $\Delta$.

here's a sketch Claim: we can construct the perpendicular bisector of any two points $A,B$ Proof: for their midpoint, draw a parallelogram $ABCD$, identify $AB\cap CD$ and draw a line parallel to $AD$ through it. It intersects $AB$ at its midpoint To identify a perpendicular line, just take two chords in the circle parallel to $AB$ and connect their midpoints. Now, we can identify the circumcenter $O$ of the triangle formed by the intersections of $\l_1,l_2,l_3$, draw an arbitrary triangle inscribed by $\omega$ to identify its center $O'$, then draw parallels to the lines connecting $O$ to pairwise intersections of $l_1,l_2,l_3$ that go through $O'$.

The main idea is that we can construct the midpoint of any given arc in any circle. Firstly, we can construct the midpoint of any segment $AB$; to do this, we pick an arbritary point $C$ not on line $AB$, and then use the parallel line thing to construct the point $D$ such that $ACBD$ is a parallelogram. Then, the intersection of the two diagonals is the midpoint of segment $AB$. Then, to construct the midpoint of an arc $AB$ in a circle $\Omega$, we draw a line parallel to $\overline{AB}$ that intersects $\Omega$ at points $C$ and $D$; then, we construct the midpoint $M$ of $\overline{AB}$ and the midpoint $N$ of $\overline{CD}$; the intersection of $\overline{MN}$ and $\Omega$ is clearly the midpoint of $AB$. Now we solve the problem: denote the initial triangle by $XYZ$, and then, Let $\overline{MN}$ be some chord of $\omega$ satisfying $\overline{MN} \parallel \overline{YZ}$. Let $O$ and $P$ be the points on $\omega$ such that $\overline{ON} \parallel \overline{XZ}$ and $\overline{MP} \parallel \overline{YX}$. Let $X'$ be the midpoint of arc $OP$. Major/minor doesn't actually matter, either one produces a valid triangle. Let $Y'$ and $Z'$ be the points on $\omega$ such that $\overline{X' Y'} \parallel \overline{MP}$ and $\overline{X' Z'} \parallel \overline{ON}$. We claim that $\triangle X' Y' Z'$ is our desired triangle. We have $\overline{X' Y'} \parallel \overline{MP} \parallel \overline{YX}$ and $\overline{X' Z'} \parallel \overline{ON} \parallel \overline{XZ}$. Now it remains to show that $\overline{Y'Z'} \parallel \overline{YZ}$. By the definition of where we placed $X'$, we have that $OX' = X'P$; since $\overline{X' Y'} \parallel \overline{MP}$, we have that $X' P = YM$; since $\overline{X' Z'} \parallel \overline{ON}$, we have $OX' = NZ'$. Therefore, we have $Y'M = NZ'$, hence $\overline{Y'Z'} \parallel \overline{MN} \parallel \overline{YZ}$.

the only nonbash geo ive done in the past month Nice problem! Call the pairwise intersections of $l_1,l_2,l_3$ $A,B,C$ respectively going clockwise. It remains to show only one criterion: Claim: Given any two points $X,Y$, we can construct the perpendicular bisector of $XY$. Proof: Draw paralells through two other points $W,Z$ such that $WX || YZ$ and $WY || XZ$; then $WXYZ$ is a parallellogram and thus when we intersect its diagonals we get the midpoint. The perpendicular condition is done by taking any two chords on $\omega$ and connecting their midpoints. Then all we have to do is construct the perpendicular bisectors of $AB,BC,CA$ and find $O$, the circumcenter of $ABC$. We can then form a mapping to $\omega$ by drawing parallel lines through an arbritrary triangle $DEF$ with vertices on $\omega$ to $AB,BC,CA$ respectively after finding the center of $\omega$ in a similar fashion. Thus we are done. $\blacksquare$

Claim: Given two points $A$ and $B$ on $\omega$ we can construct the midpoint of $\widehat{AB}$. Proof: Pick two points $A_1,A_2$ on $\omega$ such that $\overline{A_1A_2}$ and $\overline{AB}$ are not parallel. Then draw points $B_1,B_2$ on $\omega$ such that $\overline{A_1B_1} \parallel \overline{A_2B_2} \parallel \overline{AB}$. Then draw the line between $\overline{AB_1} \cap \overline{BA_1}$ and $\overline{AB_2} \cap \overline{BA_2}$ and mark its intersections with $\omega$. Now pick an arbitrary point $P_1$ on $\omega$ and draw the lines through it parallel to $\ell_1$ and $\ell_2$, letting them intersect $\omega$ again at $P_2$ and $P_3$ respectively. Then draw the line parallel to $\ell_3$ through $P_2$ and let it intersect $\omega$ again at $P_4$. By the above claim, mark the midpoint $M$ of $\widehat{P_3P_4}$. Then draw lines parallel to $\ell_2$ and $\ell_3$ through $M$, mark one of the second intersections with $\omega$, and draw the line parallel to $\ell_1$ through this second intersection. It's not hard to see that this works. $\blacksquare$ Redacted for profanity.

It suffices to construct the centers of $\omega$ and $(XYZ)$, from which we can use homothety. We only require the following two constructions: Center of $\omega$: Draw two sets of three parallel chords. In each set, the line which goes through the pairwise intersections of corresponding points is the perpendicular bisector of the three chords, so it must also pass through the center. Thus we take the intersection of these two lines. Perpendicular bisector of segment $\ell$: Take a chord of $\omega$ parallel to $\ell$ and draw its perpendicular bisector as described perviously. Notice the midpoint of $\ell$ is constructed by intersecting the diagonals of a parallelogram with $\ell$ as a diagonal, so we draw the line parallel to the previous perpendicular bisector through this midpoint. We finish by intersecting the perpendicular bisectors of $XY$, $YZ$, $ZX$. $\blacksquare$