$\text {G1}? $ Given Claim- From given $3$ lines cuts each other to form $\triangle ABC,$ we can construct $3$ midlines of $ABC.$ Proof- Draw $3$ perpendicular bisector of $BC,CA,AB.$ They cut $3$ given lines at the midpoints $M,N,P$ of $BC,CA,AB$ respectively, and also concurrent at circumcenter $O$ of $ABC.$ Perpendicular $ON,OP$ cut together at midpoint $O_a$ of $OA.$ Similarly to $O_b, O_c. $ Perpendicular bisector of $NP,PM,NM$ are concurrent at the orthocenter $K$ of $O_aO_bO_c.$ It is clear that $NP$ is the perpendicular bisector of $KO_a,$ so draw $NP.$ Similarly to line $PM,MN. $ Back to main problem Let $3$ lines cut together to form $\triangle ABC.$ $1)$ From the given claim, draw midpoints $A_2,B_2,C_2$ of $BC,CA,AB$ and lines $B_2C_2,C_2A_2,A_2B_2.$ $ 2)$ Using the claim to draw $A_1,A_3,B_1,B_3,C_1,C_3$ which are midpoints of $A_2B,A_2C,B_2C,B_2A,C_2A,C_2B,$ and lines $A_3B_1,B_3C_1,C_3A_1.$ $3)$ Line $A_3B_1$ cuts $C_3A_1$ at $T.$ Use section to draw midline vertex $T$ of $TA_1A_3,$ which cuts $AB,AC,C_3A_1,A_3B_1$ at $X,Y,Z,W$ respectively. $ 4)$ Using the claim to draw the midlines $dB,dC$ of $C_3XZ$ (vertex $X$) and $B_1YW$ (vertex $Y$). It is clear that $dB\parallel CA,dC\parallel AB.$ $5)$ Similarly, draw $dA\parallel BC$ with $A$ lies on $dA.$ Note that, $dA,dB,dC$ cuts each other to form $\triangle A_0B_0C_0.$ $6)$ Draw the perpendicular bisector of $3$ sides of the $\triangle A_0B_0C_0,$ they are concurrent at $H,$ which is the orthocenter of $\triangle ABC. $ $\blacksquare $

Let the three line forms triangle $ABC$ $\textcolor{blue}{CLAIM1:}$ We can draw the centers of circumcircle and ninepoint circle of $ABC$. $\textcolor{red}{proof:}$Draw the perpendicular bisector of three sides to get the circumcentre $O$. Draw the perpendicular bisector of $DE,EF$ where $D,E,F$ are midpoint of $BC,CA,AB$ and where they intersects is the ninepoint centre $O_9$ of $ABC$$\square$ $\textcolor{blue}{CLAIM2:}$Given 2 parallel line $s,t$, if $S,T$ are 2 points $S\in s$,$T\in t$, $ST$ is not perpendicular on $s$ we can draw the line $ST$ $\textcolor{red}{proof:}$Let the perpendicular bisector of $ST$ cut $s,t$ in $M,N$ respectively.Now draw perpendicular bisector of $M,N$.$\square$ $\textcolor{blue}{CLAIM:}$We can get the reflection of $O$ over $BC,CA,AB$ $\textcolor{red}{proof:}$(1) At first draw perpendicular bisector of $BC$ and $EF$ where $E,F$ are midpoint of $CA$ and $AB$.This 2 lines passes through $O$ and $O_9$ which are pointed previously.This 2 lines cut $BC$ at $D,X$ (2)Draw the perpendicular bisector of $X,D$.$R$ be the midpoint of $X,D$.Useing CLAIM2 draw line $OR$ to cut the line $O_9X$ at $L$.Now $XL=OD$ (3)Using CLAIM2 draw the line connecting midpoint of $D,L$ and $X$ AND suppose this line cut $OD$ at $O_A$.Then $|O_AD|=|XL|=|OD|$.Thus $O_A$ is reflection of $O$ over $BC$.Similarly draw $O_C,O_B$$\square$ Now observe that $AO_B=AO_C=OA=r$ where $r$ is circumradius of $ABC$.Also note that $EF$ is midpoint $OO_B$ and $OO_C$.So by midpoint theorem $EF||O_BO_C$.So perpendicular bisector of $O_BO_C$ passes through $A$ and is perpendicular on $EF||BC$.Similarly draw perpendicular from $B$ to $CA$.Their intersection is orthocentre of $ABC$$\blacksquare$

Let the three lines determine $\triangle ABC$. Then, by drawing the perpendicular bisectors of all three sides of $\triangle ABC$, we can find $O$ (the circumcenter of $\triangle ABC$), and $A_1,B_1,C_1$ (the midpoints of $BC, CA, AB$ respectively). Similarly, we can find the circumcenter of $\triangle A_1B_1C_1$, which is $N$, the nine-point center of $\triangle ABC$. It is well-known that $N$ is the midpoint of $OH$, where $H$ is the orthocenter of $\triangle ABC$. Claim: Given two points $P$ and $Q$ such that a line $l_P$ passes through $P$ and a different line $l_Q$ passes through $Q$, and $l_P\parallel l_Q$, it is possible to construct the reflection of $P$ over $Q$ only with a pencil and Janabel's device. Proof: We will outline a series of steps to follow to reflect $P$ over $Q$. 1. Construct the perpendicular bisector of $PQ$, and let it intersect $l_P$ and $l_Q$ at $A$ and $B$ respectively. Let $M$ be the midpoint of $PQ$, so that $M$ lies on $AB$. Then, $\angle APM=\angle BQM, \angle PMA=\angle QMB$, and $PM=QM$. By Angle-Side-Angle congruence, $\triangle APM\cong\triangle BQM\implies M$ is the midpoint of $AB$, and $PAQB$ is a rhombus. Construct the perpendicular bisector of $AB$; this is line $PQ$ because $PAQB$ is a rhombus. 2. Rename $A,M,Q$ to $X_0,Z_0,Y_0$ respectively. Consider the following algorithm for $i\ge0$: Construct the perpendicular bisector of $X_iZ_i$ (call it $l_{X_iZ_i}$) and the perpendicular bisector of $Y_iZ_i$ (call it $l_{Y_iZ_i}$). Let $l_{X_iZ_i}\cap l_Q=Y_{i+1}, l_{Y_iZ_i}\cap l_P=X_{i+1}$, and $l_{X_iZ_i}\cap l_{Y_iZ_i}=Z_{i+1}$. Claim: $l_{Y_5Z_5}\cap PQ$ is the reflection of $P$ over $Q$. Proof: We use coordinates. Let $P=(0,p),Q=(0,-p),M=(0,0),(A=(0,a),B=(0,-a)$. Then, $l_P$ is the line $y=-\frac{p}{a}x+p$ and $l_Q$ is the line $y=-\frac{p}{a}-p$. $PQ$ and $AB$ are the $y-$ and $x-$ axis respectively. Also, $l_{X_0Z_0}$ is $x=\frac{a}{2}$ and $l_{Y_0Z_0}$ is $y=-\frac{p}{2}$. From here, it is easy to find $X_1=\left(\frac{3a}{2},-\frac{p}{2}\right)$, $Y_1=\left(\frac{a}{2},-\frac{3p}{2}\right)$, and $Z_1=\left(\frac{a}{2},-\frac{p}{2}\right)$. Notice that $\triangle X_1Y_1Z_1$ is $\triangle X_0Y_0Z_0$ translated by $\frac{a}{2}$ units in the $x-$ direction and $-\frac{p}{2}$ units in the $y-$ direction. COntinuing in this fashion, we can see that (trivial by induction, for example) that $\triangle X_kY_kZ_k$ is $\triangle X_{k-1}Y_{k-1}Z_{k-1}$ translated $\frac{a}{2}$ units in the $x-$ direction and $-\frac{p}{2}$ units in the $y-$ direction. Thus, $X_5=\left(\frac{7a}{2},-\frac{5p}{2}\right) , Y_5=\left(\frac{5a}{2},-\frac{7p}{2}\right)$, and $Z_5=\left(\frac{5a}{2},-\frac{5p}{2}\right)$. $l_{Y_5Z_5}$ is therefore the line $y=-3p$, and the intersection of this line with $PQ$ is $(0,-3p)$, which is the reflection of $P$ over $Q$. Going back to $\triangle ABC$, we can apply our claim and algorithm to find the reflection of $O$ over $N$. Note that $A_1B_1\parallel AB$, so the perpendicular bisector of $AB$, which passes through $O$, and the perpendicular bisector of $A_1B_1$, which passes through $N$, are parallel. (This can also be applied to $BC$ and $B_1C_1$ or $CA$ and $C_1A_1$ if the triangle is isosceles.)

Idio-logy wrote:

Please recheck your solution. It doesn't seem to work. @below This is the correct solution

Looking at the above solutions (I need to check more properly), I think people forgot to handle the case where $P \in l_P$, $Q \in l_Q$, $l_P || l_Q$, AND $PQ \bot l_P$. To fix this, we need the following step. First, note that we can draw the midline of any two parallel lines $l$ and $m$. To do this, pick any two points in $l$, use the device to construct a line $n$ perpendicular to both $l$ and $m$, and then use the device on the intersections of $n$ with $l$ and $m$. Going back to the problem, we draw a line perpendicular to $l_P, l_Q$, far away from $P$ and $Q$. Also, we draw the midline $m$ of $l_P$ and $l_Q$. Suppose that $m$ intersects $l_P$ and $l_Q$ respectively at $P_4$ and $Q_4$. Let $P = P_0$ and $Q = Q_0$. Then, for each integer $k$, we denote $P_k$ by the point satisfying $\vec{P_k P_0} = \frac{k}{4} \vec{P_4 P_0}$. We define $Q_k$ similarly. Next, we define $R_k$ as the intersection between $P_k Q_k$ and $m$ for each integer $k$. Recall that we have drawn $l_P$, $l_Q$, $P_0$, $Q_0$, $P_4$, $Q_4$, and the line $P_0 Q_0$. Thus, using the device and line-drawing claim, we can draw the following: $P_2 Q_2$, $P_2$, $Q_2$, $R_2$, $P_3 Q_3$, $P_3$, $Q_3$, $R_3$. Then, $P_6$ since it is the intersection between $Q_0 R_3$ and $l_P$. Next, the line $P_5 Q_5$, and thus $P_5$ and finally the desired line $P_4 Q_4 = PQ$.

Claim 1: Given two distinct parallel lines $l_1, l_2$, and marked points $P_1$ on $l_1$, $P_2$ on $l_2$ such that $P_1P_2 \not\perp l_1$ or $l_2$ we can draw the line $P_1P_2$. Proof: We can use the device to draw the perpendicular bisector of $P_1P_2$. Let it meet $l_1, l_2$ at $Q_1, Q_2$ respectively (these points are defined because $P_1P_2 \not\perp l_1$ or $l_2$). We can mark $Q_1, Q_2$ and then draw the perpendicular bisector of $Q_1Q_2$, which passes through $P_1$ and $P_2$ because as $Q_1Q_2$ is the perpendicular bisector, the midpoint $M$ of $P_1P_2$ is the midpoint of $Q_1Q_2$, (follows from $\triangle{MP_1Q_1} \equiv \triangle{MP_2Q_2}$), and $P_1P_2 \perp Q_1Q_2$. This proves the claim. Let the three lines determine a $\triangle{ABC}$. We can use the device to draw the perpendicular bisectors of $BC, CA, AB$. If one of the perpendicular bisectors does not meet a certain sideline of the triangle, it must be parallel to it. In this case, Janabel can deduce that the triangle is right-angled and mark the orthocenter at the appropriate vertex. So assume that $\triangle{ABC}$ is not right angled, so that each perpendicular bisector meets each of lines $BC, CA, AB$. Let the perpendicular bisector of $BC$ meet $CA, AB$ at $P_{12}, P_{13}$ respectively, the perpendicular bisector of $CA$ meet $AB, BC$ at $P_{23}, P_{21}$ respectively and the perpendicular bisector of $AB$ meet $BC, CA$ at $P_{31}, P_{32}$ respectively. We mark these points with the pencil. We can draw the perpendicular bisectors of $AP_{23}$ and $AP_{32}$. Let them meet at $D$ (which exists as it is the circumcenter of non-degenrate $\triangle{AP_{23}P_{32}}$). Claim 2: $D$ lies on the $A$-altitude Proof: Let $\measuredangle AOB$ denote a directed angle taken modulo $180$. Let $P$ be the foot of the $A$- altitude and $D'$ the point diametrically opposite $A$ on $(AP_{23}P_{32})$. Using $AP_{32}BP_{32}$, $\measuredangle P_{23}BP_{32}= \measuredangle AP_{23}P_{32}$ (collinearity) $=\measuredangle P_{32}AB =\measuredangle CAB$ (collinearity), and in the same way, using $CP_{23}=AP_{23}$, $\measuredangle P_{23}CP_{32}= \measuredangle P_{23}CA$ (collinearity) $=\measuredangle CAP_{23} =\measuredangle CAB$ (collinearity) $=\measuredangle P_{23}BP_{32}$, so $P_{23}, B, C, P_{32}$ are concyclic. This means $\measuredangle P_{23}P_{32}A= \measuredangle P_{23}P_{32}C$ (collinearity) $=\measuredangle P_{23}BC$ (cyclic) $=\measuredangle ABC$ (collinearity). Now, $\measuredangle BP_{23}D = \measuredangle AP_{23}D$ (collinear) $=\measuredangle AP_{23}D'-\measuredangle DP_{23}D' =90-\measuredangle DP_{23}D'$ (Thales) $=90-\measuredangle P_{23}D'D$ $$(DP_{23}=DD)$$$=90-\measuredangle P_{23}D'A=90-\measuredangle P_23D'A=90-\measuredangle P_{23}P_{32}A$ (cyclic) $=90-\measuredangle ABC$ (from above). So $\measuredangle PAB=\measuredangle DAP_{23}=\measuredangle AP_{23}D$ ($DA=DP_{23}$) $=90-\measuredangle ABC=90-\measuredangle ABP$. It follows by triangle sum that $\measuredangle PBA=90$ which proves the claim. Because we have drawn the perpendicular bisectors of $AP_{32}$ and $AP_{23}$, we can mark the midpoints $M_1, M_2$ of $AP_{32}, AP_{23}$ respectively. We can now draw the perpendicular bisectors of $AM_1, AM_2$. Upon dilating from $A$ with scale factor $\frac{1}{2}$, $P_{32}, P_{23}$ are taken to $M_1, M_2$ respectively so the perpendicular bisectors of $AP_{32}$ and $AP_{23}$ are taken to those of $AM_1, AM_2$ respectively. It follows that the perpendicular bisectors of $AM_1, AM_2$ meet at a point $E$ on the $A$-altitude, The perpendicular bisectors of $AM_2, AP_{23}$ are parallel and are both perpendicular to $AB$, so we have $D$, $E$ lying on two distinct parallel lines. Suppose for contradiction that $DE$ is perpendicular to the perpendicular bisectors of $AM_2, AP_{23}$. Then it follows that the $A$-altitude is perpendicular to the perpendicular bisector of $AB$ which implies that $\triangle{ABC}$ is right-angled at $B$, which violates our assumption from earlier that $\triangle{ABC}$ is not right-angled. Therefore by Claim 1, we can draw line $DE$, which is the $A$-altitude. In the same way, we can draw the $B$ and $C$ altitudes and then mark the orthocenter at the concurrency point. We have shown that we can determine whether the triangle is right angled or not, and that in both cases we can mark the orthocenter so we are done.

Really beautiful question imo the best 'weird tool' construction question in any ELMO so far. Solution (with Rg230403) We present the solution with RAINBOW diagrams$^{\text{TM}}$!! Steps of construction will be presented in order of the colours green, yellow, orange and red. [since violet and indigo are basically just fancy synonyms for blue] The prerequisite points/lines for any lemma will be drawn in blue and the end result which we want will be black in colour. Lemma 1: Suppose $X,Y$ lie on a drawn line $l$. Suppose lines $m,n$ are perpendicular to $X,Y$ and are drawn. Then we can draw the reflection of $X$ across $Y$ Proof: Mark $Z$, an arbitrary point on $m$ Let the perpendicular bisector of $XZ$ meet $n$ at $P$ Let the perpendicular bisector of $ZP$ meet $m,n$ at $U,V$ The perpendicular bisector of $UV$ is the line $ZP$ and it intersects $l$ at the desired point [asy][asy] size(9cm); defaultpen(fontsize(10pt)); import olympiad; pair A = (-6,0), B = (0,0), C = (-6,5), D = (0,2.5), E = (6,0), U1, V1; pair P = extension(circumcenter(C,D,A),circumcenter(C,D,B),A,C); pair Q = extension(circumcenter(C,D,A),circumcenter(C,D,B),B,D); U1 = P+(6,0); V1 = Q-(6,0); draw(A--E, royalblue); draw(P--V1, royalblue); draw(Q--U1, royalblue); dot("$X$",A,dir(-45), purple); dot("$Y$",B,dir(225), purple); dot("$Z$",C,dir(30),springgreen); draw((-6,2.5)--D, lightolive); dot("$P$",D,dir(30),olive); draw(P--Q, orange); dot("$U$",P,fuchsia); dot("$V$",Q,fuchsia); draw(C--E, red); dot("$E$",E); markscalefactor=.06; draw(rightanglemark(C,A,B),heavyblue); draw(rightanglemark(D,B,A),heavyblue); [/asy][/asy] Lemma 2: Suppose $S_0,S_1, \cdots S_5$ are equally spaced points on that order on a line $l$. We can draw in the reflection of $S_0$ across $S_4$, if $S_0,S_4$ and $l$ are marked. Proof: For any point $S_i$ denote $l_i$ to be a line through $S_i$ perpendicular to $l$ Draw $S_2$ and $l_2$ Draw $S_1,S_3,l_1,l_3$ Using lemma 1, draw $S_5$ or $S_1$ reflected across $S_3$ Draw in $l_4$ Similarly, we can draw $l_0$. Now, using lemma 1, we can reflect $S_0$ across $S_4$ Corollary: given a point $X$ on a line $l$, we can draw the perpendicular to $l$ through $X$ [asy][asy] size(9cm); defaultpen(fontsize(12pt)); import olympiad; pair S0 = (-2,0), S1 = (-1,0), S2 = (0,0), S3 = (1,0), S4 = (2,0), S8 = (6,0), S11 = (-3,0), P = (0,3), S5 = (3,0); draw(S0--S8, royalblue); dot("$S_0$",S0,dir(225),purple); dot("$S_4$",S4,dir(225),purple); draw((S2-P)--(S2+P), chartreuse); dot("$S_2$",S2,dir(225),springgreen); draw((S3-P)--(S3+P), lightolive); draw((S1-P)--(S1+P), lightolive); dot("$S_1$",S1, dir(225),olive); dot("$S_3$",S3, dir(225),olive); dot("$S_5$",S5, dir(225),orange); draw((S0-P)--(S0+P), red); draw((S4-P)--(S4+P), red); dot("$S_8$",S8, dir(225),black); [/asy][/asy] While I've worded this slightly weirdly, all lemma 2 is saying, is that it is possible to reflect a point $X$ across another point $Y$ if the line $XY$ is drawn. Problem statement proof: Suppose the initial triangle is $ABC$ Let the midpoints of the sides $BC,AC,AB$ be $D,E,F$. Mark these points. Let the perpendicular bisector of $EF$ meet $BC$ at $Q$ Let $R$ be the reflection of $D$ across $Q$. Observe that $AR \perp BC$ and we can draw $AR$ Corollary of proof: We can draw the altitudes given the sides of a triangle [asy][asy] size(12cm); defaultpen(fontsize(12pt)); import olympiad; pair A,B,C,D,E,F,R,R2,H; A = dir(110); B = dir(210); C = dir(330); R2 = foot(B,A,C); draw(A--B--C--A, royalblue); dot("$A$",A,dir(A),purple); dot("$B$",B,dir(B),purple); dot("$C$",C,dir(C),purple); D = (B+C)/2; E = (A+C)/2; F = (A+B)/2; dot("$D$",D,dir(-90),chartreuse); dot("$E$",E,chartreuse); dot("$F$",F,dir(180),chartreuse); pair Q = extension(circumcenter(E,F,C),circumcenter(E,F,B),B,C); pair Q4 = (E+F)/2; draw(Q--Q4, lightolive); dot("$Q$",Q,dir(-90), olive); R = 2*Q - D; dot("$R$",R,dir(-90),orange); draw(A--R, red); draw(B--R2, red+dashed); H = extension(A,R,B,R2); dot("$H$",H); draw(D--E--F--D, deepgreen+dotted); draw(A--D, dotted+lightolive); markscalefactor=.006; draw(rightanglemark(A,R,B),magenta); [/asy][/asy] Now, a few claims for fun. Claim 1: If $A$ is a point on line $l$ and $B$ is an arbitrary point, we can draw $AB$ given $A,B$ and $l$ Proof: Let the line through $A$ perpendicular to $l$ be called $n$. Draw $n$ Let the perpendicular bisector of $AB$ be called $m$. Draw $m$. Call the triangle formed by lines $l,m,n$, triangle $CAT$ Draw the perpendicular from $A$ to $CT$. It passes through $B$ [asy][asy] size(8cm); defaultpen(fontsize(10pt)); import olympiad; pair A = (0,0), B = (3, 5), S = (7,8), D = (1.5,2.5), U = (0,3), V = (3,0); draw((-1,0)--(7,0), royalblue); dot("$A$",A,dir(-90),purple); dot("$B$",B,purple); pair C = extension(circumcenter(A,B,S),D,A,U); pair T = extension(circumcenter(A,B,S),D,A,V); draw(A--C, chartreuse); draw(C--T, orange); dot("$T$",T,dir(-90),fuchsia); dot("$C$",C,dir(90),fuchsia); draw(A--B); [/asy][/asy] Claim 2: Janabel's tool is superior to the straightedge Proof: Suppose $A,B$ are two arbitrary points Draw an arbitrary line $l$ Pick an arbitrary point $C$ on $l$ Draw $AC$ using Claim 1 Draw $AB$ using Claim 1 [asy][asy] size(8cm); defaultpen(fontsize(10pt)); import olympiad; pair A = (7,-1), B = (3, 5), C = (0,0); dot("$A$",A,royalblue); dot("$B$",B,dir(-150),royalblue); draw((-4,0)--(10,0), chartreuse); dot("$C$",C,dir(90),olive); draw(A--C, fuchsia); draw(B--A); [/asy][/asy] The last claim makes finding the centroid easy. Does anyone know whether it is possible to find the incentre of a triangle? I've a feeling it's not possible to even construct the angle $\frac{\pi}{4}$, but I have no idea how to prove this. Edit; As @below proves, not possible @below, Nice! Pretty cool to see that trying to bisect $\frac{\pi}{4}$ leads to a contradiction, and not trying to bisect $\frac{\pi}{2}$ which is much more natural Note: The last claim also means that for the original question, we do not require the lines forming the triangle; only having the vertices suffices

I think it is impossible to bisect a given angle. In particular, we will prove that $\frac{\pi}{4}$ cannot be bisected. Suppose we are given two lines forming an angle of $\frac{\pi}{4}$. Take the origin at their point of intersection and take one of the lines as the X axis. Call a line good is it passes through at least $2$ rational points. Note that a good line has a rational (or infinite) slope. Conversely, if a line has a rational (or infinite) slope, and passes through at least one rational point, then it is good. Note that both the initially given lines are good, but the required angle bisector is not good. We will prove that all constructible lines are good. This follows from the following 3 claims: Claim 1: Intersection of two good lines is a rational point. Claim 2: Perpendicular bisector of a line segment with rational endpoints is good. Claim 3: We can assume any "randomly chosen" point to be rational. Claims 1 and 2 are easy to prove, and for claim 3 we can inductively prove that every line drawn is good, so the set of rational points is dense, and so we can assume the "randomly chosen" point is rational.

I missed on a lot Claim 1 : Given two parallel lines $\ell _1, \ell _2$ and points $A, B$ on $\ell _1, \ell _2$ respectively. We can draw $A^\prime = \mathrm{reflection of}$A $\mathrm{over}$ $B$. Proof : Consider a point $C \neq A$ on $\ell _1$. Observe that if perpendicular bisector of $CB$ intersects $\ell _1, \ell _2$ at $D, E$ respectively then perpendicular bisector of $DE$ intersects line $AB$ at $A^\prime$. Now, as many above has stated that we can draw line $AB$ and as BlazingMuddy has also made a construction for when $AB \perp \ell _1, \ell_2$, we see that we can get the point $A^\prime$. Now, see that $M$ the midpoint of $BC$ is the reflection of the foot of the perpendicular from $A$ to $BC$ over the intersection of $BC$ with perpendicular bisector of $EF$, where $E, F$ are vertices of the medial triangle of $\triangle ABC$, and since $M$ lies on perpendicular bisector of $BC$ which is parallel to perpendicular bisector of $EF$ (this is because $EF \mathbin\Vert BC$ which is a result due to similarity), we can draw the foot of the perpendicular from $A$ to $BC$. Now, I use the corollary p_square has used above that we can draw the perpendicular to a line from a point $P$ on the line only using Janabel's tools (which is trivial to prove) and this completes my proof

I know lots of people have posted something similar though this is what I did which someone already did just edited a bit though there aren't that many solutions anyway.

See attached

Assume that we have a device that$,$ when given a line $l$ and a point $P$ in the plane$,$ draw the line passing through point $P$ and perpendicular to $l.$ Can we draw a square using this device$,$ a pencil$,$ and no other tools$?$

I claim that we can actually draw each of the individual altitudes. Mark the midpoints $D, E, F$ of $BC, AC, AB$. Let $\ell_1$ denote the perpendicular bisector of $EF$ and $\ell_2$ denote the perpendicular bisector of $BC$. We will mark $\ell_1 \cap BC = D'$. Now choose a point $P \in \ell_2$. Let the perpendicular bisector of $PD$ intersect $\ell_1$ at $Q$, which we mark. We can actually draw the line connecting $P$ and $Q$ - first let the perpendicular bisector of $PQ$ hit $\ell_1$ and $\ell_2$ at $Y$ and $Z$, which we mark. Since $PYQZ$ is a parallelogram, we can now draw the perpendicular bisector of $YZ$, which is just the line $PQ$. Indeed, mark $PQ \cap BC = H_A$ which is the reflection of $D$ over $D'$, the foot of the perpendicular from $A$ to $BC$. It remains to be able to connect $AH_A$. Mark $D_1$ which is the midpoint of $D'H_A$ and let $\ell_3$ be the line through $D_1$ perpendicular to $BC$, also the perpendicular bisector of $D'H_A$. Using a similar process to previously, we may mark $D_2$, the reflection of $D$ over $D_1$. Clearly $D_2H_A = D'D = D'H_A$ so now we may draw the perpendicular bisector of $D_2D'$ which is the $A$-altitude. Similarly we may draw altitudes and we mark the orthocenter $H$, and we are done. $\blacksquare$

Fun Problem We first deal with the general cases and deal with the annoying degenerate cases later on. Claim. Let $X \in \ell$ and $Y \in \ell'$ where $\ell \parallel \ell'$. If $XY$ is not perpendicular to $\ell$. Then we can construct line $XY$.

Proof. To prove this, let the perpendicular bisector of $XY$ intersects $\ell$ and $\ell'$ at $X'$ and $Y'$ respectively. This perpendicular bisector passes through the midpoint of $XY$, which we denote as $M$. Now, notice that $XX'YY'$ is a parallelogram. Now, we know that since $X'M = MY'$, then the perpendicular bisector of $X'Y'$ is $XY$, and we have constructed $XY$. We denote the midpoint of $BC, CA, AB$ as $D,E,F$ respectively. Claim. We can construct the feet of altitudes of $\triangle DEF$. Proof. To prove this, we first use our tool to construct the perpendicular bisector of $AB, BC, CA$, which must concur at the circumcenter of $\triangle ABC$, which we denote as $O$. Now, these perpendicular bisectors must pass through the midpoints. Well Known Fact. $O$ is the orthocenter of $\triangle DEF$. Proof. To prove this, notice that $OD \perp BC \parallel EF$. Similarly, we have $OE \perp AC \parallel DF$. Therefore, we are done. Now, from our first claim, since $AF \parallel DE$ and $DF$ is not perpendicular to $AF$. Then, we can draw $DF$. Similarly, we can draw $EF$ and $DE$. Now, since we can construct line $OD$. By our previous claim, we can construct $OD \perp EF$ which is the feet of $D$-altitude in $\triangle DEF$. Similarly, we can construct the other feets of altitude which we will name $H_D, H_E, H_F$.

Claim. We can construct the centroid $G$ of $\triangle ABC$. Proof. To prove this, Notice that $AC \parallel DF$ and $AC$ is not perpendicular to $AD$. Therefore, by our previous claim, we can construct line $AD$. Therefore using the same reason, we can construct the three medians of the triangle. Mark their intersection as the centroid: $G$.

Claim. Let the feet of $A$-altitude in $\triangle ABC$ be $H_A$. Then $H_A, G, H_D$ collinear Proof. Consider the homothety at point $G: \triangle ABC \mapsto \triangle DEF$. This homothety maps $H_A$ to point $H_D$, which is enough to prove that $H_A, G, H_D$ is collinear. Claim. We can construct $H_A$. Proof. First, mark the midpoint of $EF$ as $M_{EF}$. Now, take the perpendicular bisector of $EF$, let it intersect $BC$ at $J$. Now suppose that the midpoint of $M_{EF} J$ is $K$, we claim that $K,H_A H_D$ is collinear. Consider the homothety centered at point $G$ mapping $\triangle DEF$ to its medial triangle. Since $J$ is in the medial line and $M_{EF} J$ is perpendicular to the medial line, we are hence done as this homothety will map $H_D \mapsto J$. Now, we just need to use our claim on $K \in M_{EF} J$ and $H_D \in OD$ and connect the two points to get the line.

Claim. We can connect $AH_A$.

From the previous claims, we can construct $H_A, H_B, H_C$.Therefore, construct the circumcenter of $AH_B H_C$. Now, we know that $AHH_B H_C$ is cyclic, where $H$ is the orthocenter of $\triangle ABC$, and therefore, the circumcenter of $AH_B H_C$ is the midpoint of $AH$, let it be point $O_H$, which we have constructed. Now, suppose that the midpoint of $AH_B$ is $X$. We use the tool to construct the circumcenter of $AXO_H$, which must be the midpoint of $AO_H$, let it be $M_A$ (as $\angle AXO_H = 90^{\circ}$.) Furthermore, we define line $\ell$ to be the perpendicular bisector of $AX$ and $\ell'$ to be the perpendicular bisector of $AH_B$. Now, we know that $M_A \in \ell$ and $O_H \in \ell'$. Therefore, by our previous claim, since $M_A X$ is not perpendicular to $\ell$, we can construct $M_A O_H \equiv AH_A$ itself.

Since we have constructed the altitude, just mark the intersection of the three altitudes and we are done. Remark. The main difficulty of the problem is to figure out how to extract the orthocenter with using perpendicular bisectors without using the tool invented in the first claim. After deriving the first claim, and we construct any possible line and points possible with the available tool (once you got the centroid, and circumcenter), the homothety argument naturally flows in. Remark 02. Too many edge cases (degenerate cases) to consider. I might miss some along the way.

Solved with Kevinmathz Claim 1: If two parallel lines $\ell_{1}$ and $\ell_{2}$ exist such that points $X, Y$ exist such that $X \in \ell_{1}$, $Y \in \ell_{2}$ and $XY$ is not perpendicular to $\ell_{1}$, then we can draw $XY$. Proof: We see that we can draw the perpendicular bisector of $XY$. Then, letting that perpendicular bisector hit $\ell_{1}$ at $X'$ and $\ell_{2}$ at $Y'$, we see that the perpendicular bisector of $X'Y'$ is $XY$ and so $XY$ is constructable. Claim 2: If two parallel lines $\ell_{1}$ and $\ell_{2}$ exist such that $A \in\ell_{1}$ and $B, C\in\ell_{2}$, then if $\angle ABC = 90$ we can draw $D$ on $\ell_{1}$ such that $ABCD$ is a rectangle. Proof: The midpoint of $AC$ is constructable because you are able to make $AC$ a line and its perpendicular bisector meets $AC$ at the midpoint. Now, draw the perpendicular bisector with respect to $AB$ and we note that $M$ is on that, meaning with respect to that perpendicular bisector and $BC$, then $BM$ is constructable by Claim 1. Extending $BM$ to hit $\ell_{1}$ at $D$, our claim is proven. Claim 3: Given two lines $\ell_{1} || \ell_{2}$, and $A,D\in\ell_{1}, B,C\in\ell_{2}$ such that $ABCD$ is a rectangle with $AB \perp BC$, I claim we can construct the reflection of $B$ over $C$ (and similarly for the other pairs). Proof: By Claim 1, we can draw $AC$ and $BD$, and let $E = AC\cap BD$. Furthermore, we can draw the line parallel to $\ell_{1}$ that goes through $E$ (this is the perpendicular bisector of $AB$, and let's call this $\ell$. We can also draw the midpoints of $AD$ and $BC$ (draw perpendicular bisector of $AD$ and mark intersection with $\ell_{1}$, same for $BC$). Let $M$ and $N$ be the midpoints of $BC$ and $AD$, respectively. Next, by Claim $2$, since $\angle END = 90$, we can draw the projection of $D$ onto $\ell$, which we call $R$. Finally, by claim $1$, since $\ell || \ell_{1}$, and $A\in \ell_{1}, R\in\ell$, we can draw $AR$. Let $AR$ intersect $\ell_{2}$ at $X$; since $R$ is the midpoint of $DC$, we have $X$ is the reflection of $B$ over $C$, which gives the construction. Finally, let's solve the problem! Let $M, N, P$ be the midpoints of $BC, AC, AB$ respectively, we can draw these as the intersection of the perpendicular bisectors with the original line. Let those three perpendicular bisectors intersect at $O$ (which is the circumcenter) Next, draw in the perpendicular bisectors of $NP, NM, MP$, these three lines intersect at the circumcenter of $\triangle NPM$, which is the nine point center. Let's denote the nine point center as $N_{9}$, and let the perpendicular bisector of $NP$ intersect $BC$ at $D$. Finally, denote $H$ as the orthocenter (which we wish to construct) Now, since $OM || N_{9}D$, we can let $Y$ be the projection of $O$ onto $N_{9}D$, and $X$ be the projection of $N_{9}$ onto $OM$. Next, we can let $X', Y'$ be the reflections of $O$ over $X$ and $Y$ over $N_{9}$ respectively. Observe that, since $N_{9}$ is the midpoint of $OH$, $X$ is the midpoint of $X'O$, by homothety, $\angle HX'O = 90$. Finally, let $T$ be the reflection of $O$ over $X'$ (again constructable by claim 2). Since $X'$ is the midpoint of $OT$, and $\angle HX'T = 90$, we have $H$ lies on the perpendicular bisector of $OT$. Now, do the same setup symmetrically for the other three sides (using $AC$ and $AB$ as the perpendicular lines instead of $AC$). The intersection of the three perpendicular bisectors is the orthocenter.

wrong

First of all construct the Circumcenter and then using midpoints construct the Nine point center , and then use some of the above lemma's to prove that we can reflect any point over another point. Just reflect $O$ over $N_9$ to get the desired orthocenter $H$..

One obvious construction we'll use throughout the proof is adding in the circumcenter of any three points. Additionally, we need to make the following distinction: we can easily find the midpoint of any two points if there's a line drawn between them, but it's nontrivial to draw the midpoint of two points that don't have a line drawn between them. We'll begin by establishing some constructions that we can do with the given tool: Construction 1: Given parallel lines $x$, $y$ and points on them $X$, $Y$ such that $\overline{XY}$ is not perpendicular to them, we can draw the line through $X$ and $Y$. Construct the perpendicular bisector of $XY$, and let it intersect $x$ at $A$ and $y$ at $B$. This generates a rhombus $AXBY$, at which point we can construct the perpendicular bisector of $AB$ which is precisely $\overline{XY}$. $\square$ Construction 2: Given a triangle $ABC$ with all of its sidelines drawn in with circumcenter $O$, we can draw cevian $\overline{AO}$. Construct the perpendicular bisector of $AO$, and let it intersect $\overline{AB}$ and $\overline{AC}$ at $D$ and $E$, respectively. Let $N$ and $P$ denote the midpoints of $AC$ and $AB$, respectively, and let $X$ be the intersection of $DE$ and the perpendicular bisector of $PN$. We claim that $X$ lies on $\overline{AO}$. To see this, we can redefine $X'$ to be the midpoint of $AO$, which obviously lies on $\overline{DE}$. Recall that quadrilateral $ANOP$ is cyclic with diameter $AO$. Then, $X'$ is the center of $(ANOP)$, which implies $X'$ lies on the perpendicular bisector of $PN$ and so $X' = X$. To finish this construction, draw in the perpendicular bisector of $BC$. Noting that it's parallel to the perpendicular bisector of $NP$, we can apply the first move to draw $\overline{XO}$. $\square$ At this point, we can finish. Let $O$, $\triangle ABC$, and $D$ and $E$ be defined as in the creation of move $2$. Add in $O'$ to be the circumcenter of $\triangle ADE$, and let $H$ be the desired orthocenter. Since $O$ lies on the alttiude from $A$ to $DE$, we see that $\overline{AO'}$ and $\overline{AO}$ are isogonal. However, $\overline{AO}$ and $\overline{AH}$ are isogonal too, implying $\overline{AO'H}$ is collinear. To finish, observe that all the sidelines of $\triangle ADE$ are drawn in, which means we can apply the cevian construction and add in $\overline{AH}$. Repeating this procedure for $\overline{BH}$(or $CH$) and marking $H$ finishes.

Try to guess which half of the solution took 120x as much as the other: https://yu-dylan.github.io/Writeups/ELMO-2020-3.pdf

We begin with the following claim Claim: If we have two lines $l$ and $l'$ which are parallel and have a perpendicular transversal, we can construct the reflection of $P$ across $P'$ where $P$ and $P'$ are the two intersections of the transversal with the parallel lines. Proof: We pick a random point $X$ on $l'$ and draw the perpendicular bisector of $P'X$, let that be $k$. Now let $k \cap l = X'$. We draw the perpendicular bisector $n$ of $XX'$. Let $n \cap l' = S$ and $n \cap l = T$. The intersection of the perpendicular bisector $f$ of $ST$ with $m$ is the reflection of $P'$ over $P$. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(17cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.82466825336453, xmax = 13.833344712248609, ymin = -8.19117751842921, ymax = 6.331333185244802; /* image dimensions */ pen ccwwff = rgb(0.8,0.4,1); pen qqzzff = rgb(0,0.6,1); pen ttffqq = rgb(0.2,1,0); /* draw figures */ draw((-6.072874491262288,2.0998141577720775)--(6,2), linewidth(0.8) + ccwwff); draw((5.959025362196399,0)--(-6.149120142621427,0), linewidth(0.8) + ccwwff); draw((-3.988826687445849,3.980540224630813)--(-4,-5), linewidth(0.8) + qqzzff); draw((xmin, 120.9535276431472*xmin + 197.4543801430809)--(xmax, 120.9535276431472*xmax + 197.4543801430809), linewidth(0.8) + linetype("2 2") + ttffqq); /* line */ draw((xmin, -1.1710445444730258*xmin + 0.5109998371537062)--(xmax, -1.1710445444730258*xmax + 0.5109998371537062), linewidth(0.8) + linetype("2 2") + blue); /* line */ draw((xmin, 0.8539384814349685*xmin + 1.4089324127252953)--(xmax, 0.8539384814349685*xmax + 1.4089324127252953), linewidth(0.8) + linetype("2 2") + red); /* line */ /* dots and labels */ dot((-3.988826687445849,3.980540224630813),linewidth(3pt) + dotstyle); dot((-4,-5),linewidth(3pt) + dotstyle); dot((-6.072874491262288,2.0998141577720775),linewidth(3pt) + dotstyle); dot((6,2),linewidth(3pt) + dotstyle); dot((-6.149120142621427,0),linewidth(3pt) + dotstyle); dot((5.959025362196399,0),linewidth(3pt) + dotstyle); label("$l'$", (-0.05884868657842482,2.8267689702956913), NE * labelscalefactor,ccwwff); label("$l$", (-0.3866136850988449,-0.5517317836840268), NE * labelscalefactor,ccwwff); label("$m$", (-4.471069820507156,-1.1820490885309891), NE * labelscalefactor,qqzzff); dot((-3.9911880420409016,2.0826035274598875),linewidth(3pt) + dotstyle); label("$P'$", (-4.521495204894913,1.591347052795645), NE * labelscalefactor); dot((-3.9937791534391742,0),linewidth(3pt) + dotstyle); label("$P$", (-4.471069820507156,-0.4256683227146343), NE * labelscalefactor); dot((0.7603369479456799,2.0433196381631227),linewidth(3pt) + dotstyle); label("$X$", (0.7983828480134431,2.3729405108058783), NE * labelscalefactor); label("$k$", (-1.2942706040784697,5.196762036520269), NE * labelscalefactor,ttffqq); dot((-1.632481366939842,0),linewidth(3pt) + dotstyle); label("$X'$", (-1.470759449435619,-0.4508810149085128), NE * labelscalefactor); label("$n$", (-3.109584442037719,4.5664447316733074), NE * labelscalefactor,blue); dot((-1.3232168470225445,2.060545707014255),linewidth(3pt) + dotstyle); label("$S$", (-1.2186325274968344,2.221664357642607), NE * labelscalefactor); dot((0.436362424952552,0),linewidth(3pt) + dotstyle); label("$T$", (0.899233616788957,-0.37524293832687733), NE * labelscalefactor); label("$f$", (-8.984141723211403,-6.3506509882760795), NE * labelscalefactor,red); dot((-3.9962720184087885,-2.003638046075762),linewidth(3pt) + dotstyle); label("$P''$", (-4.471069820507156,-2.720023312357577), NE * labelscalefactor); dot((-0.44342721103499605,1.0302728535071273),linewidth(3pt) + dotstyle); label("$Z$", (0.016789390003210582,0.9106043635609258), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Now, we first draw all the perpendicular bisectors of $\bigtriangleup ABC$ to get the midpoints of the $BC$, $CA$ and $AB$, suppose they are $M, N$ and $L$ respectively. Now, let the intersection of the perpendicular bisector of $NL$ with $BC$ be $K$. By our above claim, we perform a reflection of the point $M$ on $K$ to get $D$. This is the foot of the $A$-altitude. Now we draw the perpendicular bisector of $DK$. Suppose the intersection of it with $BC$ is $Z$. Now we take the reflection of $M$ over $Z$ and name it $Z'$ and the perpendicular bisector of $Z'K$ is the the $A$-altitude. We can similarly draw the other altitudes, and get their concurrency point to be the orthocenter. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(11cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.7034372707775916, xmax = 7.104534133364298, ymin = -3.9633425136180174, ymax = 4.75791671826921; /* image dimensions */ pen fuqqzz = rgb(0.9568627450980393,0,0.6); /* draw figures */ draw((-3.62,2.8007482798682477)--(-5.3,-2.24), linewidth(0.8) + blue); draw((-5.3,-2.24)--(3.3,-2.1992517201317523), linewidth(0.8) + blue); draw((-3.62,2.8007482798682477)--(3.3,-2.1992517201317523), linewidth(0.8) + blue); draw((-4.46,0.28037413993412375)--(-0.16,0.3007482798682477), linewidth(0.8) + linetype("2 2")); draw((-3.62524301452334,3.907296213830642)--(-3.5918717956643698,-3.135761385535315), linewidth(0.8) + linetype("2 2") + fuqqzz); draw((-2.9762044347600733,3.910371470306629)--(-2.9428332159011004,-3.132686129059328), linewidth(0.8) + linetype("2 2") + fuqqzz); draw((-2.327165854996803,3.913446726782616)--(-2.293794636137832,-3.129610872583342), linewidth(0.8) + linetype("2 2") + fuqqzz); draw((-1.0290886954702654,3.9195972397345886)--(-0.9957174766112938,-3.1234603596313684), linewidth(0.8) + linetype("2 2") + fuqqzz); /* dots and labels */ dot((-3.62,2.8007482798682477),linewidth(3pt) + dotstyle); label("$A$", (-3.554782705609005,2.895564486459959), NE * labelscalefactor); dot((-5.3,-2.24),linewidth(3pt) + dotstyle); label("$B$", (-5.6442510632486576,-2.2372599573070038), NE * labelscalefactor); dot((3.3,-2.1992517201317523),linewidth(3pt) + dotstyle); label("$C$", (3.3646885946904264,-2.1009902818087656), NE * labelscalefactor); dot((-1,-2.219625860065876),linewidth(3pt) + dotstyle); label("$M$", (-0.9353767210317584,-2.131272431919485), NE * labelscalefactor); dot((-0.16,0.3007482798682477),linewidth(3pt) + dotstyle); label("$N$", (-0.10261759298696906,0.39728710232559655), NE * labelscalefactor); dot((-4.46,0.28037413993412375),linewidth(3pt) + dotstyle); label("$L$", (-4.402682908709154,0.367004952214877), NE * labelscalefactor); dot((-2.298077159526538,-2.2257763730178493),linewidth(3pt) + dotstyle); label("$K$", (-2.2375091757927015,-2.131272431919485), NE * labelscalefactor); dot((-3.596154319053076,-2.2319268859698225),linewidth(3pt) + dotstyle); label("$D$", (-3.539641630553645,-2.146413506974845), NE * labelscalefactor); dot((-2.947115739289807,-2.2288516294938354),linewidth(3pt) + dotstyle); label("$Z$", (-2.8885754031731734,-2.131272431919485), NE * labelscalefactor); dot((-4.894231478579614,-2.238077398921795),linewidth(3pt) + dotstyle); label("$Z'$", (-4.826633010259228,-2.146413506974845), NE * labelscalefactor); dot((-3.6080771595265375,0.2844106969492128),linewidth(3pt) + dotstyle); dot((-2.959038579763269,0.2874859534251992),linewidth(3pt) + dotstyle); dot((-2.31,0.2905612099011857),linewidth(3pt) + dotstyle); dot((-1.0119228404734624,0.2967117228531588),linewidth(3pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Really fun problem.

I doubt anyone here has posted this solution yet; can someone plz check it?

solved with adrian_042