I remembered this question in my TST and was very disappointed. (PS: It was placed as the 3/6 slot) Sketch: 1) $\Delta_1$ and $\Delta_2$ are homothetic 2) Show that the centre of homothety is indeed the point of tangency by directed angle chase and Simson line. (The centre of homothety lies on $(ABC)$.)

This is very trivial for a G5. Could be like 5-10M.

Same idea as 2011 P6 and 2018 G5

I agree with #3, this is definitely not a medium problem and doesn't deserve this position.

I don't know what exactly happened to last year's SL geo problems...

the graph was more difficult than the problem Solution: I'll be quick. Be: $E_i:$ The perpendicular line from $Z_i$ to $AB$ and the perpendicular line from $Y_i$ to $AC$, $i=1,2$ $F_i:$The perpendicular line from $X_i$ to $BC$ and the perpendicular line from $Y_i$ to $AC$, $i=1,2$ $G_i:$The perpendicular line from $Z_i$ to $AB$ and the perpendicular line from $X_i$ to $BC$, $ i=1,2$ (I'll skip the proof of obvious cyclic quadrilaterals and just use them.) $\triangle_1$ and $\triangle ABC$ are paralogical triangles, $\triangle_2$ and $\triangle ABC$ are also. So $\triangle_1\sim\triangle ABC \sim \triangle_2$ Then it is easy to check that $\triangle_1$ and $\triangle_2$ are homothetic, at a point $Y$(center of homothetic). Claim: $B,G_1,G_2$ are collinear Proof: It's easy to see that $Z_1Z_2\cap X_1X_2=B$ $Z_2X_2 //Z_1 X_1$, $Z_2G_2//Z_1G_1$, $G_2X_2//G_1X_1$ $\triangle Z_2 X_2G_2 $ and $\triangle Z_1X_1G_1$ are homothetic, so $B,G_1$ and $G_2$ are collinear. Similarly: $A,E_1$ and $E_2$ are collinear; $C, F_1$ and $F_2$ are collinear Finally $\angle E_2E_1 G_1+\angle G_1G_2E_2=\angle E_1 Y G_2$. therefore $Y$ is the point of tangency of the circles of $\triangle_1$ and $\triangle_2$.

I forgot it $\Omega_i$ is the circumcircle of $\triangle_i$,$(i=1,2)$ Note: also testing that $$T\in \odot ABC$$can be useful to end the problem

Also nice problem! Interesting, who is proposed... Sketch: if this triangles are $P_1Q_1R_1$ and $P_2Q_2R_2$ then 1) they are homothetic at centre $T$ 2) $B, R_1, R_2$ are collinear (by 2 concyclic quadrilaterals $BX_1R_1Z_1$ and $BX_2R_2Z_2$ we have that angles $CBR_1, CBR_2$ are equals) 3) By easy angle chasing $T$ lies on $(ABC)$ 4) By easy angle chasing $T$ lies on $(P_iQ_iR_i) $ 5) By homothetic, circumcircles of $\Delta_1$ and $\Delta_2$ are tangent.

This is just a nerfed 2018 G5. Let $P_1$, $Q_1$, $R_1$ be the points such that $P_1$, $Z_1$, $Q_1$ are collinear, $Q_1$, $X_1$, $R_1$ are collinear, and $R_1$, $Y_1$, $P_1$ are collinear, and define $P_2$, $Q_2$, $R_2$ similarly. Claim: $AP_1$, $BQ_1$, $CR_1$ concur at some point $T$. Proof. Observe that $AZ_1P_1Y_1$, $BX_1Q_1Z_1$, and $CY_1R_1X_1$ are cyclic. Then, angle chasing shows that \[\measuredangle BCR_1 = \measuredangle BAP_1,\quad \measuredangle ACR_1 = \measuredangle ABQ_1, \quad \measuredangle CBQ_1 = \measuredangle CAP_1\]so trig Ceva implies the concurrency. $\blacksquare$ Then, $A$, $P_1$, $P_2$ and the analogous triples are concurrent for homothety reasons, so in fact the quadruples $(T, A, P_1, P_2)$, $(T, B, Q_1, Q_2)$, and $(T, C, R_1, R_2)$ are collinear. To finish, we need to show that $P_1Q_1R_1T$ is cyclic because $T$ is the center of homothety between $\Delta_1$ and $\Delta_2$. First, observe that \[\measuredangle CBT = \measuredangle X_1BQ_1 = \measuredangle X_1Z_1Q_1 = \measuredangle Y_1Z_1P_1 = \measuredangle Y_1AP_1 = \measuredangle CAT\]implies that $A$, $B$, $C$, $T$ are concyclic. Then, we find \[\measuredangle Q_1TR_1 = \measuredangle BTC = \measuredangle BAC = \measuredangle Q_1P_1R_1\]which implies the desired concyclicity.

Label $\Delta_i$ as $A_iB_iC_i;$ clearly $\Delta_1, \Delta_2$ are homothetic wrt some point $H$. From $$\measuredangle BAA_1=\angle (\ell_1,A_1C_1)=\angle (\ell_2,A_2C_2)=\measuredangle BAA_2$$we deduce $A\in A_1A_2,$ and similarly for $B,C.$ Next, $\overline{AA_1A_2}$ and all analogous lines are concur at $H.$ Then $$\measuredangle HA_1C_1=\angle (AB,\ell_1)=\measuredangle HB_1C_1\implies H\in \odot (A_1B_1C_1),$$so in fact $\odot (A_1B_1C_1),\odot (A_2B_2C_2)$ are tangent at $H.$

This is the same idea as ISL 2018 G5; this one is even easier because the problem already gives you two homothetic triangles for free. Let $\Delta_i$ be labelled as $A_iB_iC_i$. Then $\Delta_1$ and $\Delta_2$ are homothetic, so $\overline{A_1A_2}$, $\overline{B_1B_2}$ and $\overline{C_1C_2}$ are concurrent at a point $T$. Moreover, \[ \measuredangle BAA_1 = \measuredangle Z_1Y_1A_1 = \measuredangle Z_2Y_2A_2 = \measuredangle BAA_2, \]so $A$ lies on $\overline{A_1A_2}$. Similarly, $B$ and $C$ lie on $\overline{B_1B_2}$ and $\overline{C_1C_2}$. In order to show the tangency, we just need to show that $T$ lies on $\Delta_1$ and $\Delta_2$. From now on we can forget $\ell_2$ and redefine $T$ as the concurrency point of $\overline{AA_1}$, $\overline{BB_1}$ and $\overline{CC_1}$. We now claim that $T$ lies on $(ABC)$. This is because \[ \measuredangle BTC = \measuredangle TBC + \measuredangle BCT = 90^\circ - \measuredangle X_1B_1B + 90^\circ - \measuredangle CC_1X_1 = \measuredangle AZ_1Y_1 + \measuredangle Z_1Y_1A = \measuredangle BAC. \]Therefore, $\measuredangle A_1TB_1 = \measuredangle ATB = \measuredangle ACB = \measuredangle A_1C_1B_1$ and hence $T$ lies on $\Delta_1$. Similarly, $T$ lies on $\Delta_2$ so we are done.

Similar to previous sols, let the triangles be $A_1B_1C_1$ and $A_2B_2C_2.$ First, notice $\triangle_1 \sim \triangle_2 \sim \triangle ABC$ using cyclic quadrilaterals formed by the perpendicular lines. To show one of the angle equalities, note that $\measuredangle B_1CA_1C_1=\measuredangle BAC$ from $Z_2B_1$ and other lines being perpendicular. Then, from spiral sim and the many parallel lines, the spiral center sending $B_1C_1$ to $B_2C_2$ is $B_1B_2 \cap C_1 C_2$ and cyclic variants hold so that means that $A_1A_2, B_1B_2, C_1 C_2$ concur, say at $X$. The problem is equivalent to showing $X$ lies on $(\triangle_1)$ and $(\triangle_2)$. First, by Desuarges on $BCX$ at $Z_2A_1Y_2,$ all of the intersections lie on line $B_1C_1X_2$ so $A,A_1,X$ are collinear and similarly $A,A_1,A_2,X$ and cyclic variants are collinear. Then, notice that \[\measuredangle BMC=\measuredangle MBC+\measuredangle BCM=-\measuredangle X_2Z_2B-\measuredangle CY_2Z_2=\measuredangle Y_2Z_2A+\measuredangle AY_2Z_2=\measuredangle BAC\]from parallel lines and $BB_1X_2Z_2$ cyclic so $M$ lies on $(ABC)$. This finishes since $\angle B_1 A_1C_1=\angle A=180-\angle BMC=\angle B_2MC_2$ so $M$ lies on both circumcircles and we are done say by spiral similarity

This configuration is extremely richh!! I didn't even use the fact that $P$ lies on $\odot(ABC)$. There are a ton of cyclicities and collinearities that are not mentioned too! [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings. */ pair A = (-84.71199,74.27384); pair B = (-114.00146,-62.08574); pair C = (35.05358,-62.80351); pair Z = (-106.84610,-28.77336); pair Y = (-22.64184,3.23162); pair X = (-193.48313,-61.70301); pair F = (-193.89094,-146.39008); pair D = (-68.66070,-36.97542); pair E = (-193.23521,-10.21733); pair P = (-59.93071,-97.48183); import graph; size(15.1cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); pen ffxfqq = rgb(1,0.49803,0); draw((-29.77914,11.40060)--(-37.94812,4.26330)--(-30.81082,-3.90567)--Y--cycle, linewidth(0.75) + blue); draw((-96.24027,-31.05144)--(-93.96218,-20.44561)--(-104.56801,-18.16752)--Z--cycle, linewidth(0.75) + blue); draw(circle((-142.75142,-78.54838), 84.95731), linewidth(0.5) + blue); draw(F--Y, linewidth(0.5) + blue); draw(D--E, linewidth(0.5) + blue); draw(E--F, linewidth(0.5) + blue); draw(circle((-76.68635,18.64920), 56.20063), linewidth(0.5) + linetype("4 4") + red); draw(P--A, linewidth(0.5) + ffxfqq); draw(F--C, linewidth(0.5) + ffxfqq); draw(P--E, linewidth(0.5) + ffxfqq); draw(A--B, linewidth(0.5)); draw(X--C, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw((-283.98262,-96.10073)--(85.92973,44.49832), linewidth(0.5) + linetype("3 6 7 6") + blue); dot("$A$", A, N); dot("$B$", B, NW); dot("$C$", C, NE); dot("$Z$", Z, 1.5*dir(120)); dot("$Y$", Y, 1.5*NE); dot("$X$", X, NW); dot("$F$", F, SW); dot("$D$", D, SE); dot("$E$", E, NW); dot("$P$", P, SE); dot("$\ell$", Y, 22*dir(30), linewidth(0pt)+fontsize(14)+blue); [/asy][/asy] We focus on a single line $\ell$ instead of two lines $\ell_1$ and $\ell_2$. So we use the notations $X$, $Y$, $Z$ instead of $X_i$, $Y_i$ and $Z_i$. Now let $D$ be the intersection point of the line passed through $Y$ and perpendicular to $CA$, and the line passed through $Z$ and perpendicular to $AB$. Define $E$ and $F$ similarly. So the triangle $\Delta_i\equiv \triangle DEF$. Now, note that as we have that $\measuredangle DZA=\measuredangle DYA=90^\circ$, we get that $D$ is the $A$-antipode of $A$ in $\odot(AZY)$. Now as $\ell$ varies, note that we have that the $\left\{\ell_i\right\}$ are parallel to each other. This gives us that \[\measuredangle BAD_i=\measuredangle Z_iAD_i=\measuredangle Z_iY_iD_i=90^\circ-\measuredangle AY_iZ_i=90^\circ-\measuredangle (AC,\ell_i),\]which is constant as $\ell_i$ varies. This means that $D_i$ varies along a fixed line through $A$. Similarly, $E_i$ and $F_i$ also vary along some fixed lines through $B$ and $C$ respectively. Now note that we have $\overline{X-Y-Z}$ are collinear, where $X=BC\cap EF$, $Y=CA\cap FD$ and $Z=AB\cap DE$. This means that the triangles $\triangle ABC$ and $\triangle DEF$ are perspective from $\ell$. Now from Desargues' Theorem, we get that $\triangle ABC$ and $\triangle DEF$ are perspective from a point too. This means that $P=AD\cap BE\cap CF$ are concurrent. Now note that this point $P$ is fixed as $\ell$ varies because we had that $D_i$ varies along some fixed line through $A$. This gives us that $\left\{\Delta_i\right\}$ are perspective from $P$. Now note that as $D_iE_i\perp AB$, we get that $D_iE_i\parallel D_jE_j$ and similarly that $E_iF_i\parallel E_jF_j$ and $F_iD_i\parallel F_jD_j$ which means that $\Delta_i$ and $\Delta_j$ are homothetic. Thus the center of perspectivity $P$ is the center of homothety itself. Now to finish, we have that, \begin{align*} \measuredangle DPE&=\measuredangle PDE+\measuredangle DEP\\ &=\measuredangle ADZ+\measuredangle ZEB\\ &=\measuredangle AYZ+\measuredangle ZXB\\ &=\measuredangle CYX+\measuredangle YXC\\ &=\measuredangle YCX\\ &=\measuredangle YFX\\ &=\measuredangle DFE ,\end{align*}where we spammed the angle equalities that we get due to the cyclic quadrilaterals $\odot(AYDZ)$, $\odot(BZEX)$ and $\odot(CXFY)$. This gives us that $P$ lies on all $\odot(\Delta_i)$ and that $P$ is the center of homothety between $\Delta_i$ and $\Delta_j$ which simply means that $\odot(\Delta_i)$ and $\odot(\Delta_j)$ are tangent at $P$ and we are done.

Paralogic triangles…

Thank you for your interest! There is an extension for this configuration. It came from my idea on $IG$ line and the comments of Elias M Hagos here. Let $P_1$ and $P_2$ be two points such that they are collinear with centroid $G$ of a triangle $ABC$. Let the tripolar of $P_1$ meet lines $BC$, $CA$, and $AB$ at $X_1$, $Y_1$, and $Z_1$, respectively. Let the tripolar of $P_2$ meet lines $BC$, $CA$, and $AB$ at $X_2$, $Y_2$, and $Z_2$, respectively. Let $\omega_1$ be the circumcircle of the triangle formed by the line passing through $X_1$ and perpendicular to $BC$, the line passing through $Y_1$ and perpendicular to $CA$, and the line passing through $Z_1$ and perpendicular to $AB$. Let $\omega_2$ be the circumcircle of the triangle formed by the line passing through $X_2$ and perpendicular to $BC$, the line passing through $Y_2$ and perpendicular to $CA$, and the line passing through $Z_2$ and perpendicular to $AB$. Prove that two circles $\omega_1$ and $\omega_2$ are tangent.

Bruh what is this!! How is this a G5 ohh gooddd .Seriously no insights!! come on it could be as nice as G3 Aaaaaaaaaaaaaaà ISL 2023 G5 wrote: Let $ABC$ be a triangle and $\ell_1,\ell_2$ be two parallel lines. Let $\ell_i$ intersects line $BC,CA,AB$ at $X_i,Y_i,Z_i$, respectively. Let $\Delta_i$ be the triangle formed by the line passed through $X_i$ and perpendicular to $BC$, the line passed through $Y_i$ and perpendicular to $CA$, and the line passed through $Z_i$ and perpendicular to $AB$. Prove that the circumcircles of $\Delta_1$ and $\Delta_2$ are tangent. Note that $\Delta_1(\triangle IHG) \sim \Delta_2(\triangle JLK) \sim \triangle ABC$.Since $\triangle AY_1Z_1 \sim \triangle AY_2Z_2$ we have $\overline{A-I-J}$ and similarly $\overline{H-L-B}$ and $\overline{G-K-C}$.Since $\Delta_1$ and $\Delta_2$ are homothetic ,$\overline{AIJ},\overline{HLB}$ and $\overline{GKC}$ are concurrent,say they concurr at $M$. Claim : $M \in (ABC)$ Proof : Angle chasing suffices,directed angles to avoid config issues, \begin{align*} \measuredangle BMA &=\measuredangle HBA-\measuredangle BAI \\ &=\measuredangle HX_1Y_1-\measuredangle IY_1X_1 \\ &=\measuredangle HX_1Y_1-\measuredangle GY_1X_1\\ &=\measuredangle HGI=\measuredangle BCA \text{ }\square \end{align*}Claim : $M \in (\Delta_1),M \in (\Delta_2)$ Proof : Another angle chase \begin{align*}\measuredangle AMC &=\measuredangle ABC \\ &=\measuredangle KIJ=\measuredangle GHI \text{ } \square \end{align*}Let $O$ be the centre of $(ABC)$, Note that $$\measuredangle OMI=\measuredangle MAO=\measuredangle BAO-\measuredangle BAM=90-\measuredangle BCA+90-\measuredangle MJL=180-\measuredangle MJL-\measuredangle LKJ=\measuredangle LKM-\measuredangle LKJ=\measuredangle JKM$$So $OM$ is tangent to $(\Delta_1)$ similarly it is also tangent to $(\Delta_2)$ .The End $\blacksquare$

What happened to ISL2022?

Not doing a complete writeup because this is just an exercise in linear motion. 1. Claim that Simson line $\ell$ creates a point $L$ which is the point of homothety and point of tangency. 2. Claim that if $\triangle = A'B'C'$ then $(LA'B'C')$ is similarly to $(L'ABC)$ where $L'$ is the antipode of $L$. 3. Suffices to show speed of motions is proportional and angles are equal when moving line $\ell$. The angles are a simple chase. 4. Speed of $C'$ relative to $L'C$ is proportional to $\frac{1}{AL' \cdot CL' \cdot \sin \angle (\ell, AC)}$. 5. By sixfold symmetry this finishes immediately. 5 (remark). To confirm this is cyclically equal, it suffices to show $\frac{CL'}{BL'} = \frac{\sin \angle(\ell, AB)}{\sin \angle (\ell, AC)}$. Let $X, Y$ be $AB, AC \cap \ell$. Then by LoS we have $RHS = \frac{AY}{AX} = \frac{\cos LAC}{\cos LAB} = \frac{\sin L'AC}{\sin L'AB} = \frac{CL'}{BL'}$, finishing.

Cute. Let the lines perpendicular to $AC$ and $AB$ through $Y_1$ and $Z_1$ meet at $P_1$. Define $Q_1, R_1, P_2, Q_2, R_2$ similarly. First, note that $P_1Q_1 \parallel P_2Q_2$ as both are perpendicular to $AB$. By making similar observations we conclude triangles $P_1Q_1R_1$ and $P_2Q_2R_2$ are similar with sides parallel to one another. Next, we claim $A,P_1, P_2$ collinear. This is evident, as triangles $AY_1Z_1$ and $AY_2Z_2$ are similar, and $P_1, P_2$ are constructed analogously. Similarly $B,Q_1,Q_2$ and $C,R_1, R_2$ are collinear, and they must all concur at $T$, the center of homothety between the aforementioned similar triangles. Note now that quadrilaterals $CX_2R_2Y_2$ and $AY_1Z_1P_1$ are cyclic, so $\angle R_2CX_2 = \angle R_2Y_2X_2 = \angle P_1Y_1Z_1 = \angle P_1AZ_1$ as lines $Y_1Z_1$ and $Y_2X_2$ are collinear. It follows that $T$ lies on $(ABC)$. Thus $\angle P_1Q_1R_1 = \angle ABC = \angle ATC$ as lines $P_1Q_1$ and $Q_1R_1$ are just a rotation of lines $AB$ and $BC$. It follows that $T$ lies on both $(P_1Q_1R_1)$ and $(P_2Q_2R_2)$ and thus is the desired point of tangency. We are done.

Inconsistent wrote: Not doing a complete writeup because this is just an exercise in linear motion. 1. Claim that Simson line $\ell$ creates a point $L$ which is the point of homothety and point of tangency. 2. Claim that if $\triangle = A'B'C'$ then $(LA'B'C')$ is similarly to $(L'ABC)$ where $L'$ is the antipode of $L$. 3. Suffices to show speed of motions is proportional and angles are equal when moving line $\ell$. The angles are a simple chase. 4. Speed of $C'$ relative to $L'C$ is proportional to $\frac{1}{AL' \cdot CL' \cdot \sin \angle (\ell, AC)}$. 5. By sixfold symmetry this finishes immediately. 5 (remark). To confirm this is cyclically equal, it suffices to show $\frac{CL'}{BL'} = \frac{\sin \angle(\ell, AB)}{\sin \angle (\ell, AC)}$. Let $X, Y$ be $AB, AC \cap \ell$. Then by LoS we have $RHS = \frac{AY}{AX} = \frac{\cos LAC}{\cos LAB} = \frac{\sin L'AC}{\sin L'AB} = \frac{CL'}{BL'}$, finishing. is this basically the concept of moving points?

Let $D_1$ be intersection of the perpendiculars from $Y_1$ and $Z_1$, $E_1$ be intersection of the perpendiculars from $Z_1$ and $X_1$, and $F_1$ be intersection of the perpendiculars from $X_1$ and $Y_1$. Define $D_2$, $E_2$, and $F_2$ similarly. Let $P$ be the point on $(ABC)$ such that its Simson line, which intersects $BC$ at $X$, $CA$ at $Y$, and $AB$ at $Z$ is parallel to $\ell_1$ and $\ell_2$. Note that $AYZ$, $AY_1Z_1$ and $AY_2Z_2$ are homothetic through $A$ and thus so are $D_2$, $P$, and $D_1$ so $D_1,P,D_2$ are collinear. Similarly, $E_1,P,E_2$ are collinear and $F_1,P,F_2$ collinear which means that $\Delta_1$ and $\Delta_2$ are homothetic about $P$. Note that $BX_1E_1Z_1$ is cyclic so \[\measuredangle D_1PF_1=\measuredangle APC = \measuredangle ABC=\measuredangle Z_1BX_1=\measuredangle Z_1E_1X_1=\measuredangle D_1E_1F_1\]so $P$ lies on the circumcircle of $\Delta_1$ which concludes the proof.

...... Let $\ell$ vary while being parallel to a fixed line and have it the sides (extended to lines) of $\triangle ABC$ at $X,Y,Z$. Let the lines through $Y,Z$ perpendicular to $\overline{CA},\overline{AB}$ intersect at $P$ and define $Q,R$ similarly. It suffices to show that all the $\triangle PQR$ that result as $\ell$ varies share a common homothetic center $T$, and that $T$ lies on $(PQR)$, so the circumcircles of $\Delta_1$ and $\Delta_2$ are tangent at $T$. As $\ell$ varies, all triangles $AYZ$ are related by a homothety at $A$, hence $P$ varies along a fixed line $\ell_A$ passing through $A$, and cyclic statements hold for $Q,R$. On the other hand, since a $90^\circ$ rotation of $\triangle PQR$ produces a triangle homothetic to $\triangle ABC$, it follows that all $\triangle PQR$ are pairwise related to each other by a homothety. This homothety must be centered at $\ell_A \cap \ell_B \cap \ell_C$, which means that these three lines concur at some point $T$, and hence $T$ is the center of all homotheties between all $\triangle PQR$. Pick $\ell$ such that $X$ is the foot of the perpendicular from $T$ to $\overline{BC}$, so $Y,Z$ are the feet from $T$ to $\overline{CA},\overline{AB}$. Since these are collinear, Simson lines imply that $T \in (ABC)$. To finish, note that $(AYZP)$ is clearly cyclic, so $$\measuredangle QPR=\measuredangle ZPY=\measuredangle ZAY=\measuredangle BAC=\measuredangle BTC=\measuredangle QTR,$$so $(PQRT)$ is indeed cyclic and we're done. $\blacksquare$

This is for sure straightforward for G5, though not as trivial as people claim it to be. I think it could still be a good 2/5. We will look at a reduced picture, where there is only one such line $\ell$. Let $DEF$ be the vertices of the triangle $\triangle$. Obviously $DEF$ is similar to $ABC$. Mark the point $P = \overline{AD} \cap \overline{BE}$. I claim that $P = (DEF) \cap (ABC)$. To show this, note that $BXEZ$ is cyclic, so $\measuredangle PED = \measuredangle BXZ = \measuredangle(\overline{BC}, \ell)$. Similarly, $\measuredangle EDP = \measuredangle(\ell, \overline{AC})$, so $\measuredangle EPD = \measuredangle(\overline{BC}, \overline{AC})$. Both claims follow from here. In particular, $P$ lies on $\overline{CF}$ too. Now consider any $\ell' \parallel \ell$ and let $D'E'F'$ be the corresponding triangle for $\ell'$. As the dimensions of $PDE$ rely only upon the directionality of $\ell$, $PDE$ and $PD'E'$ are similar. As $\overline{DE}\parallel \overline{D'E'}$, they are homothetic with center at $P$ (as they are similarly oriented), hence $D'$ lies on $\overline{PAD}$ and cyclic permutations. But then $DEF$ and $D'E'F'$ are homothetic at $P$, which lies on both their circumcircles. The result follows.

Easy G5... Let $\Delta_i$ be $\triangle M_iN_iL_i$ Claim 1: $\Delta_1$ and $\Delta_2$ are homothetic. Just note that $M_1N_1 \parallel M_2N_2$ as they are perpendiculars to $AB$. Similarly do for all sides. Let $P$ be center of homothety. Claim 2: $P,M_1,N_1,L_1$ are cyclic. Note $X_1Y_1M_1$ and $X_2Y_2M_2$ are also homothetic. Hence $\overline{A-M_1-M_2}$ and similarly $\overline{B-N_1-N_2}$ and $\overline{c-L_1-L_2}$ As $L_1,Y_1,Z_1,C$ are cyclic we get $\angle CY_1Z_1 = \angle CL_1Z_1$. And from $A,M_1,Y_1,X_1$ cyclic we get $\angle AY_1X_1 = \angle AM_1X_1 = \angle PM_1N_1$ hence $\angle PM_1N_1=\angle PL_1N_1$ which prove our claim. Now as $\Delta_1$ and $\Delta_2$ are homothetic we get circumcircles of both triangle are tangent at $P$.

Let the triangle $\Delta_i$ be $A_iB_iC_i$. Since $ABC$ and $A_iB_iC_i$ are perspective from $l_i$, they are perspective from some point $P_i$ by Desargue's theorem. Also, note that $\Delta ABC \stackrel{+}{\sim} \Delta A_iB_iC_i$. Let $Q_i$ be the centre of spiral similarity of the two triangles. Due to spiral similarity, we have that $A,B,P_i,Q_i$ are concyclic, $A_i,B_i,P_i,Q_i$are concyclic, and its cyclic variants. Thus, $P_i$ and $Q_i$ lie on the circles $(ABC)$ and $(A_iB_iC_i)$. Next we show that $P_1=P_2$. It suffices to show that $\measuredangle BAA_1 = \measuredangle BAA_2$, since $P_i$ is the second intersection between line $AA_i$ and $(ABC)$. Note that $\measuredangle AY_iA_i = 90^{\circ} = \measuredangle AZ_iA_i \implies A,A_i,Y_i,Z_i$ are concyclic. We have $\measuredangle BAA_i=\measuredangle Z_iAA_i = \measuredangle Z_iY_iA_i = 90^{\circ} - \measuredangle AY_iZ_i = 90^{\circ} - \measuredangle (\overline{AC}, l_i)$. But $\measuredangle (\overline{AC}, l_1)=\measuredangle (\overline{AC}, l_2)$ because $l_1,l_2$ are parallel. Now rename $P_1,P_2$ as $P$. Since $A,P,A_1,A_2$ are collinear, $A_1B_1 // A_2B_2$, and all cyclic variants, $\Delta_1$ and $\Delta_2$ are homothetic with centre $P$ on their circumcircles. This implies that their circumcircles are tangent.

Label triangles $\Delta_1$ and $\Delta_2$ with $A_1B_1C_1$ and $A_2B_2C_2$. These triangle both have parallel sides and are similar to $ABC$ so they must be homothetic about some point $P$. Notice that $A$, $A_1$, and $A_2$ are collinear as they are the diameters of tangential circles $AY_1A_1Z_1$ and $AY_2A_2Z_2$. Similar collinearities also hold. Now it is sufficient to show that $PA_1B_1C_1$ is concyclic. $$\measuredangle (A_1P,A_1C)=\measuredangle (l_1,AB)=90^{\circ}+\measuredangle(BB_1,BC)=\measuredangle(B_1C_1,B_1P)$$

let the line perpendicular to $BC$ through $X_1$ intersect the line perpendicular to $AC$ through $Y_1$ at $P_1$, let the line perpendicular to $AC$ through $Y_1$ intersect the line perpendicular to $AB$ through $Z_1$ at $Q_1$, and let the line perpendicular to $AB$ through $Z_1$ intersect the line perpendicular to $BC$ through $X_1$ at $R_1$ similarly, define $P_2, Q_2$ and $R_2$ we easily get that $P_1Q_1$ is parallel to $P_2Q_2$, $Q_1R_1$ is parallel to $Q_2R_2$, and $P_1R_1$ is parallel to $P_2R_2$, so $P_1Q_1R_1$ and $P_2Q_2R_2$ are similar also, these parallel lines let us easily prove that $P_1P_2$ intersects $X_1X_2$ and $Y_1Y_2$ at $C$, $Q_1Q_2$ intersects $Y_1Y_2$ and $Z_1Z_2$ at $A$, and that $R_1R_2$ intersects $X_1X_2$ and $Z_1Z_2$ at $B$ $CPX_1Y_1$, $CY_2X_2P_2$, $Z_1Q_1Y_1A$, $AY_2Q_2Z_2$ are all cyclic, so angle chasing yields $OQ_1R_1P_1$ and $OQ_2R_2P_2$ are cyclic and since $P_1Q_1R_1$ and $P_2Q_2R_2$ are similar, they are homothetic over $O$ thus, circles $OQ_1R_1P_1$ and $OQ_2R_2P_2$ are tangent

Let $\ell_{X_1}$ be the line through $X_1$ perpendicular to $BC$, and define the other perpendicular lines analogously. Let $A_1 = \ell_{Y_1} \cap \ell_{Z_1}$ , $B_1 = \ell_{X_1} \cap \ell_{Z_1}$, and $C_1 = \ell_{X_1} \cap \ell_{Y_1}$. Define $A_2$, $B_2$, and $C_2$ similarly. Easy angle chasing shows that $\triangle ABC \sim \triangle A_1B_1C_1 \sim \triangle A_2B_2C_2$, and $\triangle A_1B_1C_1$ and $\triangle A_2B_2C_2$ are homothetic, so it suffices to prove that the center of homothety $G = A_1A_2 \cap B_1B_2$ lies on the circumcircle of both triangles. Claim: $A_1, A_2, A$ are collinear and $B_1, B_2, B$ are collinear. Proof. Note that $A_1AY_1Z_1$ and $A_2AY_2Z_2$ are cyclic, so \[ \measuredangle A_1AC = \measuredangle A_1ZY = \measuredangle A_2Z_2Y_2 = \measuredangle A_2AC. \]Therefore, $A_1, A_2, A$ are collinear. Similarly, $B_1, B_2, B$ are collinear. $\square$ Claim: $G$ lies on $(ABC)$. Proof. Let $\alpha = \measuredangle AZ_1Y_1$. Then from $A_1AY_1Z_1$ we get $\measuredangle CAG = 90^\circ - \alpha$ and from $BX_1B_1Z_1$ we get $\measuredangle CBG = 90^\circ + \alpha$. Therefore, $A, B, C, G$ are concyclic. $\square$ Finally, $\measuredangle A_1GB_1 = \angle ACB = \angle A_1C_1B_1$, so $G$ lies on $(A_1B_1C_1)$. Similarly, $G$ lies on $(A_2B_2C_2)$, as desired.

Let $x_i, y_i, z_i$ denote the lines determining $\Delta_i$. Let $f(\Delta, \ell)$ denote the triangle determined by the reflections of $\ell$ in the sides of $\Delta$. Lemma: For any line $\ell$, the incenter of $f(\triangle ABC, \ell)$ is the anti-Steiner point of the line through $H$ parallel to $\ell$. Proof: Let $m$ be the line through $H$ parallel to $\ell$. The anti-Steiner point $I$ is the concurrence point of the reflection of $m$ across the three sides of the triangle. As such, if $r$ is the distance between $\ell$ and $m$, it is also the distance from $I$ to each of the three reflections of $\ell$. In this problem, let $I$ be the anti-Steiner point of the line through $H$ parallel to $\ell_1$ and $\ell_2$. From our previous claim, $I$ is the incenter of $f(\triangle ABC, \ell_1)$. Since \[f(\triangle ABC, \ell_1) = f(\Delta_1, \ell_1),\]it follows that $I$ lies on the circumcircle of $\Delta_1$ – similar reasoning shows that it lies on the circumcircle of $\Delta_2$.

so $I$ is the center of homothety between the two triangles. This finishes.

Let $d$ be the line passes through orthocenter of $\triangle ABC$ and parallels to $\ell_1$ and $S$ be it's Anti - Steiner point WRT $\triangle ABC$. Suppose that $\triangle_1$ is $\triangle U_1V_1W_1$ and $\triangle_2$ is $\triangle U_2V_2W_2$. We can see that $A, U_1, U_2, S$ are collinear. Similarly, we have $\triangle U_1V_1W_1$ and $\triangle U_2V_2W_2$ are homothetic with homothety center $S$. Note that $\triangle U_1V_1W_1 \stackrel{+}{\sim} \triangle U_2V_2W_2 \stackrel{+}{\sim} \triangle ABC,$ then $(SV_1, SW_1) \equiv (SB, SC) \equiv (AB, AC) \equiv (U_1V_1, U_1W_1) \equiv (U_2V_2, U_2W_2) \pmod \pi$. This means $S$ lies on both $(U_1V_1W_1)$ and $(U_2V_2W_2)$. Hence $(U_1V_1W_1)$ tangents $(U_2V_2W_2)$ at $S$