My method: Note that you can easily get angle bisectors so you can get incenter and excenter. By Euclidea 9.5 you can get midpoints and by Euclidea 11.5 you can get parallel lines without the distance 1 condition. Then you use midpoint of incenter and excenter and midpoint of a side to get a perpendicular bisector, which you can use parallel line to get to the perpendicular from a vertex to a side and we're done. Alternative without Euclidea 11.5 would be to use perp bisectors to construct circumcenter then use Euler line

I really like this problem! Here are the only two constructions we will need: 1) Note that given two lines $l_1, l_2$, we can construct the lines $1$ away from $l_1$ and the lines $1$ away from $l_2$. These four lines form a rhombus whose diagonals bisect the original angle between $l_1$ and $l_2$. 2) Given a triangle $ABC$, let a line $1$ away from $BC$ intersect $AB, AC$ at $F, E$. Then intersect $CF \cap BE = X$, and intersect $AX \cap BC$ to get the median. This way we can construct the midpoint of any segment. Now note that by Ceva-Menelaus, given collinear $X, Y, Z$ we can always construct $W$ such that $(W, X; Y, Z) = -1$. Hence given any segment $XY$, construct (by two midpoints) a point $N$ on the segment satisfying that $XN = 3YN$. Now by Ceva-Menelaus construct a point $U$ on the ray $XY$ such that $XU = 3YU$, ergo $YU = 2XY$. Now construct the midpoint of $YU$. This allows you to reflect points over other points. Notably Ceva-Menelaus again lets us split a segment into thirds (and really you should be able to construct rational multiples of segments on the same line using this). Anyways, using operation $1$ construct the incenter $I$ and $A$-excenter $I_A$. Note that the external angle bisector of $\angle{BIC}$ and the internal angle bisector of $\angle{BI_AC}$ intersect on the perpendicular bisector of $BC$ by Fact 5. Do the same to construct its antipode on $(BIC)$. This way we can construct the perpendicular bisector of $BC$, and now the circumcenter. Operation $2$ now evidently gives us the centroid. We are now done by applying operation $2$ and noting that $HG = 2OG$.

cute question!!!! We present a series of claims. Claim: We can construct the incenter $I$ and excenters. Proof. By taking lines $1$ away from $AC$ and $BC$, we can get the angle bisectors. Thus, intersect them to get the incenter $I$ and constructing other points of $BC$ we can get exterior angle bisectors. $\blacksquare$ Claim: We can construct the midpoint of any two points. Similarly, we can construct the reflection of a point over another. Proof. The midpoint part is just Ceva: if the points are $X,Y$ take $Z$ and draw a line parallel to $XY$. If it hits $ZX$ and $ZY$ at $E,F$, then $YE\cap XF$ lies on the median by Ceva. To construct $X$ over $Y$, let $Z$ again be any point and $M$ be the midpoint of $XZ$. Let $N$ be the midpoint then of $YM$, and $DN\cap ZY=K$. Then, $MK\cap XY=X'$ is the desired point since $\frac{ZN}{NY}=2$ so $N$ is the centroid $ZXX'$. $\blacksquare$ Now, we can construct midpoints of $ABC$ and medians. But we can also construct all angle bisectors! So by the Iran lemma, we can construct $2$ points on every intouch chord. Thus we can get $D$, the $A$-intouch point, and by reflection over the midpoint, we get $E$, the $A$-extouch point. Then intersect $EI$ and $DI_A$ which is well known to be the midpoint of the $A$-altitude, thus finishing.

nice

Solved with MathLuis and kingu Claim: We can construct the midpoint of a segment $XY$. Proof. Let $T$ be a point at a distance further than $1$ from $XY$. Construct a line parallel to $XY$ intersecting segments $TX$ and $TY$ at $D$ and $E$. By Ceva, the line connecting $T$ and $XE \cap YD$ is a median of $TXY$, which intersects $XY$ at its midpoint. Claim: We can construct the reflection of a point $B$ over a point $A$. Proof. Let $C$ be an arbitrary point on a line $\ell$ at a distance $1$ from $AB$ which is also parallel to $AB$. Construct the midpoint of $AC$ and let the line through this point and $B$ intersect $\ell$ at $D$. Then, $ABCD$ is a parallelogram. Now, construct the midpoint of $AD$ and let the line through this point and $C$ intersect $AB$ at $E$. This forms parallelogram $ACDE$. Therefore, $E$ is the reflection of $B$ over $A$. Claim: We can construct the line through $X$ parallel to another line $YZ$ at an arbitrary distance. Proof. Reflect $X$ over $Y$ and over $Z$ to $Y'$ and $Z'$. Now, reflect $Y'$ over $Z$ to $Z''$ and $Z'$ over $Y$ to $Y''$. We can easily check that $Y''Z''$ is parallel to $YZ$ and passes through $X$. Claim: We can construct an angle bisector given two lines $\ell_1$ and $\ell_2$. Proof. Construct parallel lines at a distance $1$ away from each line. This forms a rhombus, and connecting opposite vertices creates an angle bisector. Now, we can construct the incenter and excenters of $ABC$. If $I$ is the incenter and $I_A$ is the $A$-excenter, we can take the midpoint, which is the midpoint of minor arc $BC$. We can also construct the midpoint of $BC$, so we can construct the perpendicular bisector of $BC$. Doing this repeatedly gives us the circumcenter $O$. Now, we can construct midpoints midpoints, so we also have medians and therefore the centroid $G$. We can now finish using the Euler Line. Specifically, reflect $O$ over $G$ to $O'$, then reflect $G$ over $O'$ to the orthocenter, and we are done.

This solution doesn't make use of the incenter construction, which is unfortunate as it's the first and nicest one I found Rhombus with given diagonal: Just use our double edged ruler, where we assume the diagonal has length greater than 1 (patched in the next construction). Integer Homotheties: We first use this to scale up the diagram to where all necessary lengths later in the solution are greater than 1. Choose a far away point $X$. For each point $K$, construct one rhombus with diagonal $XK$, and another by shifting the ruler. The opposite vertex in this second rhombus is the reflection of $X$ across $K$, and we can induct to get all integers. Other than trivial degenerate cases, this allows us to scale the diagram by powers of 2, as well as constructing homotheties in our scaled diagram. Midpoints and Perpendicular Bisectors: Simply use the other diagonal of a constructed rhombus. Thus we get $G$, $O$. Thus, Euler line gives us $H$, from which we scale back (if necessary) using midpoints to get the desired orthocenter. $\blacksquare$

Wow. Step 1: Given $X,Y$ can mark the midpoint of $XY$ Proof: Mark an arbitrary point $P$, and intersect a line parallel to $XY$ with $PX$ and $PY$. The intersection of the diagonals of the resulting trapezoid lies on the $P-$ median of $\triangle PXY$. Step 2: Given $\triangle XYZ$ we can mark the angle bisectors of $\angle YXZ$. Proof: Draw the distance $1$ lines from $XY$ and $XZ$. Their intersection lies on the internal or external bisector of $\angle YXZ$, depending on which lines are chosen. Step 3: Given $X,Y$ we can draw the perpendicular bisector of $XY$. Proof: Mark an arbitrary point $P$. The midpoint of the incenter and the $P-$ excenter of $\triangle PXY$ is the arc midpoint of $XY$, so we can draw the line through this point and the midpoint of $XY$. Step 4: We can mark the Symmedian point of $\triangle ABC$. Proof: Let the perpendicular bisectors of $AB$ and $AC$ intersect the $A-$ median at $D$ and $E$. By 2008 USAMO $2$, $BD\cap CE$ is the $A-$ Dumpty point, so we can draw the $A-$ Symmedian which suffices. Finish: The line through the midpoint of $BC$ and the Symmedian point is the $A-$ Schwatt line, which intersects the $A-$ midline at the midpoint of the $A-$ altitude. Hence, we can draw the altitudes, which finishes.

shendrew7 wrote: This solution doesn't make use of the incenter construction, which is unfortunate as it's the first and nicest one I found Rhombus with given diagonal: Just use our double edged ruler, where we assume the diagonal has length greater than 1 (patched in the next construction). Integer Homotheties: We first use this to scale up the diagram to where all necessary lengths later in the solution are greater than 1. Choose a far away point $X$. For each point $K$, construct one rhombus with diagonal $XK$, and another by shifting the ruler. The opposite vertex in this second rhombus is the reflection of $X$ across $K$, and we can induct to get all integers. Other than trivial degenerate cases, this allows us to scale the diagram by powers of 2, as well as constructing homotheties in our scaled diagram. Midpoints and Perpendicular Bisectors: Simply use the other diagonal of a constructed rhombus. Thus we get $G$, $O$. Thus, Euler line gives us $H$, from which we scale back (if necessary) using midpoints to get the desired orthocenter. $\blacksquare$ How do you do it given diagonal???

What is hapening with this manufacturing problems? A similar one

Claim 1: we can construct angle bisectors. Proof: Given angle $\angle XYZ$, place your ruler along the interior of $XY$, then along the interior of $YZ$. Their intersection $W$ is such that $WXYZ$ is a rhombus, and so $YW$ is the angle bisector. Claim 2: we can construct midpoints. Proof: Given $\overline{XY}$, mark some point $T$ on the plane. Draw $l$ parallel to $XY$ and let it intersect $TX, TY$ at $D, E$ respectively. Then, let $DY, EX$ intersect at $K$. It's clear now by Ceva/projection/homothety that $TK$ is the median in $\triangle TXY$, so it intersects $XY$ at its midpoint $M$. Claim 3: give two points $P, Q$, we can construct the reflection $P'$ of $P$ across $Q$. Proof: draw $l$ parallel to $PQ$, pick two points $p,p'$ on $l$ and construct its midpoint $q$. Let $Pp$ intersect $Qq$ at $T$. It's clear now that $Tp'$ intersects $PQ$ at $P'$ as defined. In fact, we can easily generalize this claim to find the point $Q+n(Q-P)$. Using the angle-bisector tool, we can construct triangle $\triangle I_AI_BI_C$ where the vertices are the excenters of $\triangle ABC$. If we change our reference triangle to the excentral triangle, we notice that $\triangle ABC$ is the orthic triangle. We also find the incenter $I$ of $\triangle ABC$ which is the orthocenter of $\triangle I_AI_BI_C$. Now, mark the midpoints of $\triangle I_AI_BI_C$, as well as the midpoints of $II_A, II_B, II_C$. Note that these points form the 9 point circle of $\triangle I_AI_BI_C$, which is just the circumcircle of $\triangle ABC$! A diameter of this circle is formed by connecting the midpoint of $II_A$ and the midpoint of $I_BI_C$. We can mark its midpoint to find the circumcenter $O$. Now, by finding the medians in $\triangle ABC$, we can find the centroid $G$. We now use the general form of claim $3$ to reflect $O$ across $G$ to $G+ 2(G-O) = 3G = H$, as desired.

Tool 1: Given two lines $l_1$ and $l_2$ we can construct both angle bisectors Construct all four lines that are a distance one away from $l_1$ or $l_2$. The diagonals of the rhombus formed by these lines are the desired bisectors. Tool 2: Given two points $X$ and $Y$ we can construct their midpoint Let $l$ be a line a distance one away from $XY$. Chose arbitrary $E$ and $F$ along $l$. Then the line passing through $XE\cap YF$ and $XF\cap YE$ bisects $XY$. Tool 3: Given three points $X$, $Y$, and $P$ we can construct the line through $P$ parallel to $XY$ Construct the midpoint $M$ of $XY$ and chose a point $Q$ along $XP$. Define $Z=QM\cap PY$. Then we can take the line passing through $P$ and $XZ\cap QY$. Now we can construct the incenter $I$ and the excenter $I_A$. We can also construct the midpoint, $M$, of $BC$. Notice that the midpoint of $II_A$ is the arc midpoint, $N$. Now draw the line passing through $A$ parallel to $MN$. This is the $A$-altitude so we can easily finish.

A different solution as in we don't really need to be able to construct arbitrary pairs of parallel lines to be able to construct the orthocenter. We start off with our preliminary loci. Construction : Given two points $X$ and $Y$ we can mark their midpoint $M_{XY}$. Proof : Draw line $XY$ and the two lines $\ell_1$ and $\ell_2$ on either side of $\overline{XY}$ each a distance one away from $XY$ and parallel to it. Now, it is fairly intuitive that there exists two parallel lines $m_1$ and $m_2$ which are a distance of one apart where $m_1$ passes through $X$ and $m_2$ passes through $Y$. Consider the quadrilateral formed by $\ell_1$ , $\ell_2$ , $m_1$ and $m_2$, it is not hard to see that it is a parallelogram so intersecting the diagonals of this quadrilateral yields the midpoint of segment $XY$, as desired. Construction : Given two intersecting lines $\ell_1$ and $\ell_2$ we can construct their internal and external angle bisectors. Proof : Draw the lines $m_1$ and $m_2$ parallel to $\ell_1$ and $\ell_2$ and a distance of 1 away from each respectively (on the side such that they intersect these lines). This then forms a parallelogram. We can further note that it is also a rhombus since the distances between the parallel side pairs is the same. Thus, if we draw the diagonal of this parallelogram which passes through the intersection of $\ell_1$ and $\ell_2$ which must be the internal angle bisector of $\ell_1$ and $\ell_2$. Repeating this process of the obtuse angle formed by $\ell_1$ and $\ell_2$ gives us the external angle bisector. Now, we use our configuration geometry knowledge. We first prove the following claim. Claim : Given a triangle $\triangle ABC$ it is possible to construct its intouch points. Proof : Since we can construct midpoints, it is possible to construct the midlines of $\triangle ABC$ as well. Now, by Iran Lemma we know that the intersections $K$ and $J$ of the $C$-angle bisector and $B$-midline and the $B$-angle bisector and $C$-midline lie on the line $\overline{EF}$ where $E$ and $F$ are the $B$ and $C$-intouch points. Then, drawing the line $\overline{KJ}$ and taking its intersections with the sides $AB$ and $AC$ allows us to mark the $B$ and $C$-intouch points. Similarly, we can also construct the $A$-intouch point which proves the claim. Now, since we can construct internal and external angle bisectors it is immediate that we can construct the incenter $I$ and $A-$excenter $I_A$. Now, we draw line $\overline{I_AD}$ and the $A-$midline. By the Midpoint of the Altitude Lemma, it is well known that their intersection is the midpoint of the $A-$altitude, $H_A$. It simply remains to construct the perpendicular from $A$ to $BC$ by drawing line $\overline{AH_A}$. Repeating this process on all 3 sides, we can construct all three perpendiculars and in particular mark the orthocenter of $\triangle ABC$ as desired.

original solution hopefully not fakesolve. Construction: Given $X,Y$ we can construct midpoint $M$. Construct $\ell$ distance $1$ from $XY$ parallel to it. Pick a point $P$, and let $XP\cap \ell=X', YP\cap \ell=Y'$ Consider $XY'\cap X'Y=Q$, by Ceva clearly $PQ$ bisects $XY$. Construction: Given $\angle XPY$ we can construct its bisectors. Construct $\ell$ parallel to $XP$ with distance $1$, and construct $\ell'$ parallel to $PY$ with distance $1$ such $XP,PY,\ell,\ell'$ form a rhombus. Taking $P\ell\cap\ell'$ is the internal bisector from symmetry. The external bisector is the same but with the rhombus on the outside. Construction: Given $XY$, and a point $P$ we can construct the parallel line to $XY$ through $P$. Construct the midpoint of $XY=M$. Let $D$ be a point on $XP$. Let $DM\cap PY=N$. Let $DY$ intersect $XN$ at $Z$. $PZ$ is the desired parallel line by Ceva. Construction: Given $XY$, we can construct its perpendicular bisector.. Now, we can construct the incenter and $P$-excenter of $(PXY)$ for any point $P$. The midpoint of the two points is by Fact 5, the midpoint of minor arc $XY$ in $(PXY)$. Taking the line through that and the midpoint of $XY$ finishes. Hence, we return to our problem. From constructing the midpoints, we may get the medians of $ABC$ and hence the centroid $G$. The perpendicular bisector of $BC$ intersects $AB$ at $K$. Then $KC$ intersects the parallel line to $BC$ through $A$ at the point $A'$, such $ABCA'$ is an isosceles trapezium. Now, intersect $GA'$ with $BC$. This gives the foot from $A$ to $BC$ by a well-known Why-Point /2011 G4 Point property. Hence, repeating for all three sides, and intersecting we have the orthocentre.

Solution from Twitch Solves ISL: We prove the following series of claims. Claim: Given two intersecting lines $\ell_1$ and $\ell_2$, we get the two angle bisectors. Proof. Make a parallelogram two of whose sides are $\ell_1$ and $\ell_2$ and where the distance to opposite sides is $1$. This is actually a rhombus. $\blacksquare$ Claim: Given a line segment $\overline{XY}$ we can find its midpoint. Proof. Draw two lines parallel to $\overline{XY}$ distance $1$ away, say $\ell$ and $\ell'$. Let $Z$ be on $\ell'$. Let rays $ZX$ and $ZY$ meet $\ell$ again and $X'$ and $Y'$. Then we can get the centroid of $\triangle X'Y'Z$. $\blacksquare$ From now on fix triangle $ABC$. Our goal is to draw the $A$-altitude; see the figure below. [asy][asy] pair A = dir(130); pair B = dir(210); pair C = dir(330); draw(unitcircle, dotted+blue); draw(A--B--C--cycle, blue); pair N = dir(270); pair M = midpoint(B--C); pair Q = 0.6*C+0.4*A; pair P = extension(B, Q, A, M); pair R = extension(B, P, A, A+B-C); draw(B--N--C, grey); draw(B--R, red); pair S = extension(A, R, M, N); pair L = midpoint(A--M); pair D = foot(A, B, C); draw(A--D, dashed+deepgreen); draw(A--M, blue); draw(A--R, blue); draw(S--N, blue); draw(D--S, deepgreen+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$N$", N, dir(N)); dot("$M$", M, dir(45)); dot("$Q$", Q, dir(85)); dot("$P$", P, dir(250)); dot("$R$", R, dir(R)); dot("$S$", S, dir(S)); dot("$L$", L, dir(L)); dot(D); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ A = dir 130 B = dir 210 C = dir 330 unitcircle / dotted blue A--B--C--cycle / blue N = dir 270 M 45 = midpoint B--C Q 85 = 0.6*C+0.4*A P 250 = extension B Q A M R = extension B P A A+B-C B--N--C / grey B--R / red S = extension A R M N L = midpoint A--M D .= foot A B C A--D / dashed deepgreen A--M / blue A--R / blue S--N / blue D--S / deepgreen dashed */ [/asy][/asy] Claim: We can construct the perpendicular bisectors of side $\overline{BC}$, which meets $\overline{BC}$ at its midpoint $M$. Proof. Bisect all the angles of $ABC$ to get the incenter $I$ and excenters $I_A$. Let $N$ be the midpoint of $\overline{II_A}$ which is equidistant from $B$ and $C$. Then the internal bisector of $\angle BNC$ is the perpendicular bisector of side $BC$. $\blacksquare$ To finish, an arbitrary line $g$ through $B$ inside the triangle, and let it meet $\overline{AM}$ at $P$ and $\overline{AC}$ at $Q$. Using a straightedge alone we can construct the harmonic conjugate $R$ of $P$ with respect to $\overline{BQ}$. Then from $(BC;M\infty) = -1$, where $\infty$ is the point at infinity along line $BC$, we conclude that $\overline{AR} \parallel \overline{BC}$. Bisect $\overline{AM}$ to get $L$. Let line $AR$ meet the perpendicular bisector of $BC$ at $S$. Then $\overline{SL}$ passes through the foot of the $A$-altitude and we're done.

Great Solution above