Wow this was a very nice problem. Took me a lot of time during the exam and still couldn't solve it. can someone post a solution?

Solution by Rohan Goyal, Anant Mudgal and Daniel Hu We begin by proving a series of lemmas. Lemma 1 Given a non-right angled triangle $ABC$, we can draw the nine-point circle and mark the orthocentre $H$ using only a cyclos. Proof Draw circles $(BC), (CA), (AB)$ and mark their intersections to get the three feet of altitudes $D, E, F$ opposite $A, B, C$. Now draw the circle $(DEF)$ to get the nine-point circle. Draw $(BDF), (CDE), (AEF)$ and they meet at $H$, which we can also mark. Lemma 2 Given points $A, B$, we can mark the midpoint $M$ of $AB$ using only a cyclos. Proof Draw the circle $(AB)$ and choose a point $X$ on it. Draw circles $(XA), (XB)$ and mark their intersection $Y$. Now mark a point $Z$ on the circle $(XA)$ apart from the marked points. Clearly, $Z$ does not lie on $AB$ nor on $(AB)$, hence we can draw $(AZB)$. Mark five points $Z_1, \dots, Z_5$ on this circle, each different from all previous points and verify if either $A$ lies on $(Z_iB)$ or $B$ lies on $(Z_iA)$ for each $1 \le i \le 5$ before marking the new point. By pigeonhole principle, for some three indices $i, j, k$, the three triangles $AZ_iB$ are non-right angled, hence we can draw their ninepoint circles by Lemma 1. All of them pass through $M$, and their centres are not collinear, else homothety at the centre of $(ABZ)$ implies the orthocentres of the three triangles are collinear; but they all lie on the reflection of $(AZB)$ in $AB$, a contradiction! Thus, these three nine-point circles meet at only $M$, and we mark this point. Lemma 3 Given points $A, B, C, D$ on the plane in general position, we can mark the intersection point $E$ of lines $AB$ and $CD$ using only a cyclos. Proof Draw $(AB)$ and mark five points on it, all different from previously marked points. For each marked point $X$; draw $(CX)$ and $(DX)$ and check whether they have an intersection apart from $X$ (i.e., if they are tangent, or if $X$ lies on $CD$). We can find three points $X_1, X_2, X_3$ among them not lying on $CD$. Denote by $Y_i$ the second intersection of $(AX_i), (BX_i)$ and by $Z_i$ the second intersection of $(CX_i), (DX_i)$ and mark them, for each $1 \le i \le 3$. Draw the circles $(X_iY_iZ_i)$ and note that they all pass through $E$ and have diameters $EX_i$ for all $i$; so they are not coaxial as $X_1, X_2, X_3$ are not collinear; all lying on $(AB)$. Thus we mark $E$ as the unique point common to them all. Lemma 4 Given a circle $\Gamma$, we can mark the centre of $\Gamma$ using only a cyclos. Proof Mark points $A, B, C \in \Gamma$ and mark the midpoints of $BC, CA, AB$ to get $A_1, B_1, C_1$ according to Lemma 2. Draw the circles $(AB_1C_1), (BA_1C_1), (CB_1A_1)$ and mark the intersection to get the centre of $\Gamma$. Lemma 5 Given a circle $\Gamma$ and point $A$ on $\Gamma$, we can mark a point $K$ such that line $AK$ is tangent to $\Gamma$ using only a cyclos. Proof Mark points $B_1, B_2, B_3$ and $C$ on $\Gamma$. By Lemma 4, mark the point $O$, the centre of $\Gamma$. Draw $(B_iO)$ and $(AOC)$ and mark the intersection denoted $F_i$; there exists an index $j$ for which $F_j \ne O$; mark the point $K$ which is the intersection of $B_jF_j$ and $OM$ where $M$ is the midpoint of $AC$ (which we mark by Lemma 2) by Lemma 3. Clearly, $K$ lies on $A$ tangent to $\Gamma$. Lemma 6 Given a circle $\Gamma$ and a point $A$ on $\Gamma$, and a point $B$ not on $\Gamma$, we can mark the point $C$ which is the second intersection of line $AB$ and $\Gamma$ using only a cyclos. Proof Mark the foot of perpendicular $M$ from $O$ onto line $AB$ as done in Lemma 1. Mark the intersection of line $OM$ and $AK$ by Lemma 3, where $K$ is a point on the $A$-tangent to $\Gamma$ as constructed in Lemma 5. Draw $(OAK)$ and mark the second intersection with $\Gamma$ to obtain $C$. Lemma 7 Given points $A, B, C$ not all on a line, we can draw the reflection of $A$ in $BC$ using only a cyclos. Proof Draw $(ABC)$ and mark the orthocentre $H$ of $ABC$ by Lemma 1. Mark the intersection $A'$ of line $AH$ with $(BHC)$ using Lemma 6, which is the $A$-reflection in line $BC$. Lemma 8 Given points $A, B$, we can mark the point $C$ which is the reflection of $A$ in $B$ using only a cyclos. Proof Draw $(AB)$, and by Lemma 6, mark a point $K$ such that $BK$ is tangent to $(AB)$. By Lemma 7, mark the reflection $C$ of $A$ in line $BK$ as desired. Thus, a cyclos can do everything a compass can: to draw a circle with given centre $A$ and given radius $B$, we use Lemma 8 to mark the reflection $C$ of $B$ in $A$ and use the cyclos to draw $(BC)$ which has centre $A$ and passes through $B$.\qed Alternative solution (N.V. Tejaswi) We use the first two Lemmas from the existing solution. Let $C$ denote the point such that $A$ is the midpoint of $CB$. Choose a generic point $X$. We can get the midpoint $M$ of $BX$ and the foot of perpendicular $D$ from $X$ to $AB$. Draw the circle passing through $A, D$ and $M$. This is the nine-point of circle of triangle $CBX$. Intersect this circle with the circle whose diameter is $BX$. The intersection point other than $D$ is the foot of perpendicular $E$ from $B$ to $CX$. Note that $|AE| = |AB|$. Similarly, we can construct another point $F$ such that $|AF| = |AB|$. The circle through $B, E$ and $F$ is the required circle. $\square$

We have following properties Property 1 If we have any $3$ marked points $A,B,C$ then we can mark its orthocenter $H$, nine point circle of $(ABC)$,midpoints $M,N,P$ of $BC,CA,AB$ and circumcenter $O$ of $(ABC)$ Proof:- Draw the circles with diameter $BC,CA,AB$ their intersection points $D,E,F$ are the feet of altitudes .$(DEF)$ is the nine point circle of $ABC$.$(BDF),(CED),(AEF)$ concur at $H$. Mark any arbitrary points $P,Q$ on $(ABC)$ and draw the nine point circles of $PBC,QBC,ABC$. The $3$ nine point circles concur at midpoint $M$ of $BC$. We can mark the orthocenter of $MNP$ i.e. $O$ the circumcenter of $ABC$. Property 2We can reflect a drawn circle $(ABC)$ across its chord whose endpoints are marked , reflection of a point across any line joining any $2$ marked points ,mark the antipode of a point on a circle , we can reflect a point across midpoint of segment joining another $2$ points. Proof:- Draw $H$ the orthocenter of $ABC$ then reflection of $(ABC)$ across $BC$ is $(BHC)$.Let $D,E,F$ be the feet of altitudes on $BC,CA,AB$. Reflect the circle with diameter $BA$ across line $B-D-C$ .Note that this circle intersects $(BHC)$ at $B$, reflection of $A$ across $BC$. Mark reflection of $H$ across $OM$ (same labellings as in property $1$) ,say $H'$ and reflection of $H'$ across $BC$ is $A$ antipode. Antipode of $H$ in $(BHC)$ is the reflection of $A$ across $M$. Lemma Using same labellings as in property $1$ .We can reflect $O$ across $A$ Proof:- Reflect $A,B,C$ across $M,N,P$ to get $A',B',C'$ since midpoint of $B'C'$ is $A$ we can reflect $O$ across $A$. Main Proof Let the $2$ marked points be $R,S$. Draw the circle with diameter $RS$. Mark $2$ points $J,K$ on this circle then by property $1$ we can draw circumcenter of $SJK$ i.e. midpoint of $RS$ ,say $O$ is this point. We can reflect $O$ across $R$ to get $V$ by considering triangle $RJK$ and applying the lemma. Similarly considering the circle with diameter $VO$ we can reflect $R$ across $V$ to get $W$. The circle with diameter $SW$ is the circle centered at $R$ with radius $RS$. So, we are done. Alternative sol Let $R,S$ be the given $2$ marked points draw the circle with diameter $RS$. Mark $2$ points $J,K$ on this circle. Mark midpoints of $RJ,RK$ as $M,N$ by property $1$. By property $2$ reflect $S$ across $RM,RN$ let it be $S_1,S_2$. $(S_1SS_2)$ is the circle centered at $R$ with radius $RS$. Edit Note:- The $3$ nine point circles in property $1$ can't be coaxial since there are only $2$ possible circles passing through $2$ given points having same radius .In this case the $3$ circles have same radius $\frac{R}{2}$.Another possible way to mark midpoint of $BC$ is to reflect $H$ across $BC$ by property $2$ say this point is $H'$ then we draw the nine point circle of $H'BC$ .Note that the intersection points of the nine point circles of $ABC,H'BC$ are $D,M$

So nice

I didn't expect a construction problem can be as hard as this one (Well, this type of geometry problem is almost extinct in olympiad) I'm trying to solve this, so I'll post my solution later (Also, maybe every committee hates pure geo nowadays, so I guess this type of geo will replace pure geo for a while

Anecdote: In my notes, I labelled this problem as "toxic geometry" simply because of the number of times one has to say "by pigeonhole principle" to avoid degeneracy issues, like suddenly a right triangle popping up, and orthocentres colliding with vertices. It is quite nice though that the idea of multiple nine-point circles lets you find the common intersection, hence the midpoint of a segment --- this is one of the only times I have seen this idea. All degeneracy issues are quite horrible to write up, and I did this way past bedtime and repeatedly told Daniel how toxic everything in this was while writing it up, hence the name. Originally, Rohan and I decided "find the centre of the circle" is a good enough problem, but once Daniel pointed out the intersection lemma and I found symmedian point, we were excited to see how far this went till we got that cyclos is literally like a compass. Of course, Tejaswi swooped in with a black-magic-Esque nine-point circle finish which did not need any of that, so we thought this might be okay for INMO. An example of the toxicity is as follows: in post #4, the argument used to reflect $(ABC)$ across $BC$ using $(BHC)$ does not carry over to reflecting $(ABD)$ across $BC$, simply because the orthocentre $D$ of $\triangle ABD$ is just a vertex, so something else needs to be done. This can be done, for instance, by choosing another point on $BC$ than $D$, for which you can use the midpoint --- but you must avoid isosceles $ABC$ in your original pick of $A$! See why it is toxic? EDIT: of course, most of these issues are easy to fix, but the point was to highlight that they sneak in somewhat randomly.

anantmudgal09 wrote: Anecdote: In my notes, I labelled this problem as "toxic geometry" simply because of the number of times one has to say "by pigeonhole principle" to avoid degeneracy issues, like suddenly a right triangle popping up, and orthocentres colliding with vertices. It is quite nice though that the idea of multiple nine-point circles lets you find the common intersection, hence the midpoint of a segment --- this is one of the only times I have seen this idea. All degeneracy issues are quite horrible to write up, and I did this way past bedtime and repeatedly told Daniel how toxic everything in this was while writing it up, hence the name. Originally, Rohan and I decided "find the centre of the circle" is a good enough problem, but once Daniel pointed out the intersection lemma and I found symmedian point, we were excited to see how far this went till we got that cyclos is literally like a compass. Of course, Tejaswi swooped in with a black-magic-Esque nine-point circle finish which did not need any of that, so we thought this might be okay for INMO. An example of the toxicity is as follows: in post #4, the argument used to reflect $(ABC)$ across $BC$ using $(BHC)$ does not carry over to reflecting $(ABD)$ across $BC$, simply because the orthocentre $D$ of $\triangle ABD$ is just a vertex, so something else needs to be done. This can be done, for instance, by choosing another point on $BC$ than $D$, for which you can use the midpoint --- but you must avoid isosceles $ABC$ in your original pick of $A$! See why it is toxic? How many points will be deducted for the mistake in post#4?

anantmudgal09 wrote: An example of the toxicity is as follows: in post #4, the argument used to reflect $(ABC)$ across $BC$ using $(BHC)$ does not carry over to reflecting $(ABD)$ across $BC$, simply because the orthocentre $D$ of $\triangle ABD$ is just a vertex, so something else needs to be done. This can be done, for instance, by choosing another point on $BC$ than $D$, for which you can use the midpoint --- but you must avoid isosceles $ABC$ in your original pick of $A$! See why it is toxic? but we can mark any point $P$ on $(ABD)$ such that $PBD$ is non right angled then we can reflect it across $BD$ .It works for $(BDE)$?

I have a naive doubt. Since the question doesn’t restrict one’s means to mark a point on a drawn circle: By a simple coordinate geometry calculation one can show that the circle $\Gamma$ with $AB$ as diameter (where $A$ and $B$ are the given points) is tangent to the required circle with centre $A$. And any circle with $BC$ as diameter for a point $C$ on $\Gamma$ will intersect the required circle at a point other than $B$. Mark those!

We start with the following lemmas. Observe first that given any non-right nondegenerate triangle $ABC$: Draw $(AB)$, $(BC)$, $(CA)$ to get feet $D$, $E$, $F$. Draw $(AEF)$, $(BFD)$, $(CDE)$ to get the orthocenter $H$. Note also that $(BCH)$ is the reflection of $(ABC)$ across $\overline{BC}$. So, we may reflect a circle over a chord. Suppose we're given to start a circle $\gamma$ with diameter $\overline{AD}$. Pick points $B$ and $C$ on $\gamma$ so that $ABC$ is acute. Claim: We may construct get the reflection $B'$ of $A$ across $B$. Proof. Let $\delta_B$ be any circle through $A$ and $B$ other than $\gamma$ or $(AB)$ (for example one may let the third vertex be the foot from the orthocenter of $ABC$ to line $BD$). Reflect both $\delta_B$ and $\gamma$ about $\overline{BD}$. $\blacksquare$ [asy][asy] pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = dir(290); filldraw(unitcircle, invisible, red+1.4); draw(A--D, red); pair B_prime = 2*B-A; pair C_prime = 2*C-A; draw(circumcircle(A, B_prime, C_prime), deepgreen); draw(circumcircle(B_prime, B, D), red+dashed); pair V = foot(orthocenter(A, B, C), B, D); draw(circumcircle(A, B, V), blue); draw(circumcircle(B_prime, B, V), blue+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$B'$", B_prime, dir(B_prime)); dot("$C'$", C_prime, dir(C_prime)); dot(V); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ A = dir 110 B = dir 210 C = dir 330 D = dir 290 unitcircle / 0.1 orange / red+1.4 A--D / red B' = 2*B-A C' = 2*C-A circumcircle A B' C' / deepgreen circumcircle B' B D / red dashed V .= foot (orthocenter A B C) B D circumcircle A B V / blue circumcircle B' B V / blue dashed */ [/asy][/asy] Define $C'$ similarly. Then $(AB'C')$ is the desired circle by homothety.

Nice problem! Sketch of solution: Claim 1: Given points $A,B,C,D$ and $AC\cap BD$, we can mark $AB\cap CD$. Proof. Use the Miquel point. Claim 2: Given points $A,B,C$, we can mark the orthic triangle $DEF$, orthocenter $H$, $A$-humpty point $P$, $A$-queue point $Q$, and $BC\cap AQ\cap PH=X$. Proof. $(AB),(BC),(CA)$ gives $DEF$, $(AEF)\cap (BFD)$ gives $H$, $(ABC)\cap (AEF)$ gives $Q$, $(AEF)\cap (BHC)$ gives $P$, $(QFB)\cap(QEC)$ gives $X$. Claim 3: Given $B,C,$ we can mark the midpoint $M$. Proof. Take random point $A$ and do the stuff in claim 2. By claim 1, we can mark $AP\cap XD=M$. Claim 4: Given points $B,C$, we can mark the reflection of $B$ across $C$, $B'$. Proof. Take random point $A$ and mark $M_{AB}$ the midpoint of $AB$, $T$ on segment $AC$ such that $TA=3TC$ (midpoint twice), and $K=M_{AB}T\cap BC$ (possible by claim 1). Now mark $M_{BC}$. By Menelaus, we have $M_{BC}K=BC$. Repeat the previous steps again to get $B'$. Finally, draw the circle with diameter $BB'$. This gives the desired circle. Note: As noted by many posts above, the config issues are a mess. I will gladly handwave all of them.

Suppose we are given points $A$ and $B.$ Invert around $A,$ essentially we want to construct the perpendicular bisector of segment $AB.$ After inversion, we can draw circle between any 3 points, lines between any 2 points, as long as this objects don't contain $A.$ We also can draw the perpendicular to $AX$ passing through $X,$ for any $X.$ Now, to the solution - Add a random point $C,$ and call the circumcircle of triangle $ABC$ as $\omega.$ now mark the intersections of the circle with diameter $BC$ and lines perpendicular to $AB$ and $AC$ (at points $B$ and $C,$ respectively) as $B_1$ and $C_1.$ By connecting $B, C, B_1$ and $C_1$ obtain $A_1,$ the antipode of $A$ in $\omega$ and $A',$ the reflection of $A$ in $M,$ the midpoint of $BC.$ Now, intersect the circle of diameter $A'A_1$ with $\omega$ to obtain $A_2, $ that lies on line $A-M-A',$ reconstructing $M$ as such by taking the intersection of $A'A_2$ and $BC.$ Next, reconstruct the orthocenter $H$ of $\triangle ABC$ by intersecting the ine $MA_1$ with the circumcircle of $A'BC.$ Draw in the center of diameter $HM$ and intersect it with $BC$ to get $A_3,$ the feet of the perpendicular of $A$ on $BC.$ Now, draw-in the 9-point circle of $A'BC,$ say $\gamma,$ from these you get the midpoint triangle of $A'BC$ and the midlines of $\triangle ABC,$ without the one parallel to $BC.$ Now, take the midpoint of $A'A_1,$ which lies on $\gamma,$ and connect it to $M$ to form a line that cuts $A_3H$ at the midpoint of $AH.$ Now, we can reconstruct the O-9 circle of $\triangle ABC$, intersect it with the midline of $\triangle ABC$ parallel to $AC,$ get $N$ the midpoint of $AB$ and draw a line perpendicular to $AN$ through $N,$ which is precisely the perpendicular bisector of $AB.$

Given two points $A,B$ we can construct the midpoint of $AB$ through the nine-point circles --- I'm not gonna describe it as others already did. Now, given two points $A,B$ let $C$ be the midpoint of $AB$, $D$ be the midpoint of $AC$, $E$ be the midpoint of $AD$, $F$ be the midpoint of $CD$. Draw the circles with diameters $AD$, $EF$ and $CD$ and mark the intersections $G,H,I,J$ as in the attached picture. Then draw the circles $BIG$, $BJH$. They intersect at a point $K$ satisfying $2\vec{AK}=\vec{BA}$. Applying the above construction to the points $K$ and $C$ we find the point $L$ with $2\vec{KL}=\vec{CK}$, i.e. $L$ is the reflection of $B$ in $A$. It remains to draw a circle with diameter $BL$.

rmtf1111 wrote: Suppose we are given points $A$ and $B.$ Invert around $A,$ essentially we want to construct the perpendicular bisector of segment $AB.$ After inversion, we can draw circle between any 3 points, lines between any 2 points, as long as this objects don't contain $A.$ We also can draw the perpendicular to $AX$ passing through $X,$ for any $X.$ Now, to the solution - Add a random point $C,$ and call the circumcircle of triangle $ABC$ as $\omega.$ now mark the intersections of the circle with diameter $BC$ and lines perpendicular to $AB$ and $AC$ (at points $B$ and $C,$ respectively) as $B_1$ and $C_1.$ By connecting $B, C, B_1$ and $C_1$ obtain $A_1,$ the antipode of $A$ in $\omega$ and $A',$ the reflection of $A$ in $M,$ the midpoint of $BC.$ Now, intersect the circle of diameter $A'A_1$ with $\omega$ to obtain $A_2, $ that lies on line $A-M-A',$ reconstructing $M$ as such by taking the intersection of $A'A_2$ and $BC.$ Next, reconstruct the orthocenter $H$ of $\triangle ABC$ by intersecting the ine $MA_1$ with the circumcircle of $A'BC.$ Draw in the center of diameter $HM$ and intersect it with $BC$ to get $A_3,$ the feet of the perpendicular of $A$ on $BC.$ Now, draw-in the 9-point circle of $A'BC,$ say $\gamma,$ from these you get the midpoint triangle of $A'BC$ and the midlines of $\triangle ABC,$ without the one parallel to $BC.$ Now, take the midpoint of $A'A_1,$ which lies on $\gamma,$ and connect it to $M$ to form a line that cuts $A_3H$ at the midpoint of $AH.$ Now, we can reconstruct the O-9 circle of $\triangle ABC$, intersect it with the midline of $\triangle ABC$ parallel to $AC,$ get $N$ the midpoint of $AB$ and draw a line perpendicular to $AN$ through $N,$ which is precisely the perpendicular bisector of $AB.$ You are not allowed to draw any circles through A as I understand it and circles with BC diameter, you will need to show how to draw I think. Maybe I'm mistaken but you seem to be using these things.

2 fakesolves (+First line of geoking's sol) later..... Note that it is possible to construct a perpendicular from $X$ to $YZ$ just by marking $W=(XY) \cap (YZ)$. This is the only thing we will be using in this solution. Also we will call the intersection of $YZ$ with this line as $X(YZ)$ (Just to make my life easier ) First construct the circle $(AB)$ and choose a point $E$ on it ($E$ is closer to $B$ than $A$, again to make my life easier). Now mark $C=E(AB)$ and $H=C(AE)$ Now $(AB) \cap (HA)=F$ and $H(AB)=D$. Notice that $DEF$ is the orthic triangle of $\triangle ABX$ where $X=AE \cap BF$ which we can therefore mark using $(ADE) \cap (BDF)$. This means we have constructed a triangle with $H$ as it's orthocenter. Now by constructing $(DEF)$ and varying $E$ we can construct $M$, the midpoint of $AB$. Now using $M(AB)=H_1$, we again construct another triangle such that $H_1$ is the orthocenter of the new triangle. Also notice this triangle has it's third vertex $X_1$ on line $BE$ (as $H_1E \perp BE + HE \perp BE$). This is crucial because now we can finish the problem by noticing that $(H_1BX_1) \cap (HBX)=A'$ as the reflection of $A$ about $XB \equiv X_1B$ must lie on on both $(H_1BX_1)$ and $(HBX)$. Now repeat this for another $E$ and drawing the circle about $A,A',A''$ will give us the desired circle $\blacksquare$ Note: Unlike the other solutions, we can completely avoid using midpoints at all by just constructing $G=H(AB)$ and then $H_1=G(AE)$. This works because we just need 2 positions of the orthocenter on the line $AE$ for $(H_1BX_1)$ to include the reflection of $A$ about $BE$ This took me a lot of time to come up with, even after the 2 other fakesolves Atleast I hope this weird sol is correct

Huh.

Now construct equilateral triangles $ABC$, $BCD$, $CDE$ such that $A,B,C,D,E$ are distinct. Then $B$ is the midpoint of $AE$, and so the circle $AE$ is what we want.

bruh Given triangle $ABC$ we construct its orthocenter by constructing the feet $D,E,F$ from $A,B,C$ respectively by drawing circles with diameters $AB,BC,CA,$ then considering the circumcircles of $AEF,BDF,CDE.$ Construct an arbitrary point $P$ and construct a point $Q$ on $AP$ by drawing the foot from an arbitrary point on $(AP).$ If $H,H'$ are the orthocenters of $APB,AQB$ then $(AHP)$ is the reflection of $(APB)$ across $AP$ so it passes through the reflection of $B$ over $AP.$ Similarly $(AH'Q)$ does as well so we may construct the reflection of $B$ over $AP$ which is equidistant from $A$ as $B.$ Repeat this to get another equidistant point and construct their circumcircle.

Here is a ridiculously inefficient solution that nonetheless does its job. Note that given some points $A, B$, we can always mark some $C$ such that $\angle ACB \neq 90^\circ$. Claim. Given a segment $\overline{BC}$, we can construct its midpoint. Proof. Mark any point $A$ with $\angle BAC \neq 90^\circ$. Through the circles $(AB), (AC), (BC)$ we can construct the feet of altitudes $D, E, F$ from $A, B, C$. Then the circles $(AEF), (BDF), (CDE)$ concur at $H$. Then, construct the $A$-HM point $G = (BHC) \cap (AH)$ and the midpoint $M = (HQD) \cap (DEF)$ using the nine-point circle. $\blacksquare$ Claim. Given four points $A, B, C, D$ that form a convex quadrilateral, and the point $E = \overline{AB} \cap\overline{CD}$ we may construct the points $\overline{AC} \cap \overline{BD}$ and $\overline{AB} \cap \overline{CD}$. Proof. Use the Miquel point $M = (EAC) \cap (EBD)$. Then $\overline{AC} \cap \overline{BD} = (MAD) \cap (MBC)$. The other point follows similarly. $\blacksquare$ Now, suppose we are initially given points $X$ and $Y$. Mark any two points $A$ and $B$ such that triangle $AYB$ is not right. We may construct the midpoints $M, N, P$ of $\overline{AB}, \overline{AY}, \overline{BY}$. By the second claim on complete quadrilateral $ANPB$, we obtain the centroid $G$ of triangle $AYB$. In particular, $YG = 2YM$. Now, construct the midpoint $M_1$ of $\overline{XM}$; by the second claim on quadrilateral $XM_1GY$, we can construct the point $\overline{M_1G} \cap \overline{XY}$. By Menelaus, this point $Y'$ is the reflection of $Y$ over $X$. Hence the circle $(YY')$ is our desired circle.

Given three points $A$, $B$, $C$, we start by constructing the foot of the altitude from $A$ to $BC$ as $(AB) \cap (AC)$. We mark the orthocenter a the Miquel point of the feet of the altitude, from which the reflecting the orthocenter lemma allows us to construct the reflection of any circle over a marked chord. Suppose the starting points are $X$ and $Y$. We draw $(XY)$ and mark 2 points $P$, $Q$ on $(XY)$, and mark a point $R$ as the foot from $Q$ to $PY$. Note the intersection of the reflections of $(XY)$ and $(XPR)$ over $PY$, say $X_1$, is simply the reflection of $X$ over $P$. Repeating, we can mark $X_2$ as the reflection of $X$ over $Q$, giving $(XX_1X_2)$ as our desired circle. $\blacksquare$

Solved with leejae019 and asbodke: Claim: Given any triangle $ABC$, we may construct the foot from each point to the opposite side. Proof: The circles with diameters $AB$ and $AC$ meet again at the foot from $A$ to $BC$; we may repeat this construction twice more to get the other two feet. Claim: We can construct the midpoint of any two points $X$ and $Y$. Proof: Pick two points $Z_1$ and $Z_2$ not on line $XY$. Then, draw the feet of $\triangle XYZ_1$ and $\triangle XYZ_2$ and the nine-point circles. These two circles meet on line $XY$ at the midpoint of $X$ and $Y$, so our claim is proved. Claim: We can reflect a point $A$ across a line $\ell$. Proof: Select distinct points $B_1$ and $C_1$ on $\ell$ and construct the feet $D,E_1$ and $F_1$ of $\triangle AB_1C_1$. We can construct the orthocenter $H_1$ of $\triangle AB_1C_1$ by intersecting $(B_1DF_1)$ and $(C_1DE_1)$. It's well known that $(B_1H_1C_1)$ passes through the reflection of $A$ across $\ell$, so now we select points $B_2$ and $C_2$ on $\ell$ and repeat this construction a second time. Intersecting the two circles obtained gives us the reflection of $A$ across $\ell$, as desired. Now, we describe a process for which, given points $J$ and $L$, constructs the circle centered at $J$ passing through $L$: Construct the midpoint $K_2$ of $J$ and $L$. Then, construct the midpoints $K_1$ and $K_3$ of $\overline{JK_2}$ and $\overline{K_2L}$ respectively. (So, $J, K_1, K_2, K_3$ and $L$ are collineqr and equally spaced.) Let $K_{1.5}$ be one of the intersections between the circles with diameter $\overline{JK_2}$ and $\overline{K_1K_3}$. Note that $\triangle K_1 K_{1.5} K_2$ is equilateral. Reflect each vertex of equilateral $\triangle K_1 K_{1.5} K_2$ over the opposite side to make three more equilateral triangles. Repeat this process arbitrarily to form a lattice of equilateral triangles; note that $J$ and $L$ are part of this lattice, so the reflection of $L$ over $J$ (call this point $L'$) is also part of the lattice. The circle with diameter $\overline{LL'}$ is the desired circle.

Notice that by drawing $(AB)$ and $(AC)$, we can mark the foot of the altitude from $A$ to $\overline{BC}$. Given arbitrary points $A$ and $B$, we construct a point on the line through $A$ perpendicular to $\overline{AB}$ as follows. Let $C$ be an arbitrary point, and let $D$ be the foot of the altitude from $C$ to $\overline{AB}$. Let $E$ be an arbitrary point, and let $F$ be the foot of the altitude from $E$ to $\overline{CD}$. Let $G$ be the foot of the altitude from $A$ to $\overline{EF}$. We can see that $G$ is on the line through $A$ perpendicular to $\overline{AB}$. Now, construct another point $H$ on $\overline{AB}$ by dropping a perpendicular from an arbitrary point. Let $I$ and $J$ be the feet of the perpendiculars from $B$ to $\overline{GH}$ and from $H$ to $\overline{BG}$, respectively. Mark the intersection $K$ of $(ABJ)$ and $(AHI)$, which is the orthocenter of $AGH$. Notice that $(GHK)$ passes through the reflection $B'$ of $B$ over $A$. We can repeat this construction to obtain another circle passing through $B'$. Thus, we can find $B'$ and draw $(BB')$, which is the circle through $B$ centered at $A$, as desired. $\blacksquare$

Claim: Given any points $A, B, C$ we can construct the projection $p(A, BC)$ of $A$ onto $BC$. Proof: Take $(AC) \cap (BC)$ $\blacksquare$. Claim: Given any points $A, B$ we can construct a point $P \ne B$ such that $BP \perp BA$. Proof: Take general $C$ on $(AB)$ and general $D$ on $(BC)$. Then project $C_1 = p(C, AB), D = p(D, CC_1)$ and $P = p(B, DD_1)$ $\blacksquare$. Claim: We can construct the antipode $C_1$ in $(AB)$ and reflection $C'$ of $C$ over $AB$. Proof: Take $P$ such that $AC \parallel PB$, then $C' = p(A, PB)$ is the antipode of $C$. Then take $Q$ such that $QC' \parallel AB$ and consider $p(C, QC')$ $\blacksquare$. Claim: Given right triangle $ABC$ with right angle at $C$, we can construct $D = AC \cap BB$ where $BB$ is the tangent to $(AB)$. Proof: Take $P$ such that $BP \perp AB$. Let $Q$ be the common foot of $C_1, C'$ to $BP$. Let $F$ be the foot from $C_1$ to $AC$ respectively. Then $AC \cap BB$ is $(AC'Q) \cap (TC_1Q)$ $\blacksquare$. Now, we can construct $D = AC \cap BB$ as above, and $D' = AC' \cap BB$ the same way. Then we can construct the reflection $A'$ of $A$ over $DD'$ since $A$ lies on $(DD')$, and finish by considering $(AA')$ $\blacksquare$.

Call the points $B,C$ pick a point $P$ on $(BC)$ and pick a point $A$ on $(BP)$. Now we perform a few operations. $(AB)\cap (AC),(AB)\cap (BC)$ and $(BC)\cap (AC)$ gives the feet of the altitudes Doing miquel on altitudes gives orthocentre $H$ of $\triangle ABC$ Drawing $(BCH)$ means we can reflect a circle over a chord. Now call the altitudes $D,E,F$ (as in regular convention).Pick a point $Q$ on $(CEB)$ other than $C$.Reflect $(QEB)$ over $EB$ and intersect it with $(BHA)$ to get $C'$ .$C'$ is the reflection of $C$ over $BA$ , so $BC=BC'$ .Repeating this process we can construct another point $C''$ such that $BC'=BC''$ and we finish by drawing $(CC'C'')$ $\blacksquare$