A line $\ell$ passes through the vertex $B$ of a regular triangle $ABC$. A circle $\omega_a$ centered at $I_a$ is tangent to $BC$ at point $A_1$, and is also tangent to the lines $\ell$ and $AC$. A circle $\omega_c$ centered at $I_c$ is tangent to $BA$ at point $C_1$, and is also tangent to the lines $\ell$ and $AC$. Prove that the orthocenter of triangle $A_1BC_1$ lies on the line $I_aI_c$.
Problem
Source: Sharygin 2013 Final 9.6
Tags: geometry, orthocenter, circles
GeoMetrix
27.12.2019 12:54
Another beautiful problem. Here is my solution. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.522569310450763, xmax = 5.875176892321606, ymin = -2.880379393830543, ymax = 9.92969066345483; /* image dimensions */ /* draw figures */ draw((-7.656053437790895,0.8782420183861012)--(-5.310378663295351,5.293874459191753), linewidth(2.8)); draw((-5.310378663295351,5.293874459191753)--(-2.6591661830307425,1.054644295059452), linewidth(2.8)); draw((-2.6591661830307425,1.054644295059452)--(-7.656053437790895,0.8782420183861012), linewidth(2.8)); draw((-7.330337885668523,1.4913876147255478)--(-3.6166305365757125,2.58560863641023), linewidth(1.2)); draw((-8.392337210965565,2.055543499816846)--(-7.330337885668523,1.4913876147255478), linewidth(2)); draw((-0.9649225127799422,4.243985543186208)--(-3.6166305365757125,2.58560863641023), linewidth(1.2)); draw((-8.392337210965565,2.055543499816846)--(-0.9649225127799422,4.243985543186208), linewidth(1.2)); draw(circle((-5.789358611833416,3.1105530944130018), 2.2352436048450905), linewidth(1.6) + linetype("2 2")); draw((-13.032144129531794,0.6884529378495974)--(-8.392337210965565,2.055543499816846), linewidth(1.2)); draw((-6.268338560371481,0.9272317296342512)--(-3.6166305365757125,2.58560863641023), linewidth(1.2)); draw((-6.268338560371481,0.9272317296342512)--(-7.330337885668523,1.4913876147255478), linewidth(1.2)); draw((-5.310378663295351,5.293874459191753)--(-8.392337210965565,2.055543499816846), linewidth(1.2)); draw((-0.9649225127799422,4.243985543186208)--(-5.310378663295351,5.293874459191753), linewidth(1.2)); draw((-0.9649225127799422,4.243985543186208)--(-2.6591661830307425,1.054644295059452), linewidth(1.2)); draw((-8.392337210965565,2.055543499816846)--(-7.656053437790895,0.8782420183861012), linewidth(1.2)); draw((-13.032144129531794,0.6884529378495974)--(-5.310378663295351,5.293874459191753), linewidth(1.2)); draw((-13.032144129531794,0.6884529378495974)--(-7.656053437790895,0.8782420183861012), linewidth(1.2)); draw((-6.777963493692568,2.5312087831157224)--(-6.268338560371481,0.9272317296342512), linewidth(1.2)); draw((-4.079799832913284,3.326206105209937)--(-6.268338560371481,0.9272317296342512), linewidth(1.2)); /* dots and labels */ dot((-7.656053437790895,0.8782420183861012),dotstyle); label("C", (-7.715841388711762,0.5583119784790999), NE * labelscalefactor); dot((-2.6591661830307425,1.054644295059452),dotstyle); label("$B$", (-2.5766982388863546,0.8480185776114141), NE * labelscalefactor); dot((-5.310378663295351,5.293874459191753),linewidth(4pt) + dotstyle); label("A", (-5.259633265633442,5.395152590079476), NE * labelscalefactor); dot((-13.032144129531794,0.6884529378495974),dotstyle); label("$X$", (-13.069115503113228,0.3945647702738787), NE * labelscalefactor); dot((-8.392337210965565,2.055543499816846),linewidth(4pt) + dotstyle); label("$I_A$", (-8.50938555155245,1.8053099486573219), NE * labelscalefactor); dot((-0.9649225127799422,4.243985543186208),linewidth(4pt) + dotstyle); label("$I_B$", (-0.9140342786487231,4.349689645384603), NE * labelscalefactor); dot((-7.330337885668523,1.4913876147255478),linewidth(4pt) + dotstyle); label("$A_1$", (-7.375751033208609,1.2133008112999841), NE * labelscalefactor); dot((-3.6166305365757125,2.58560863641023),linewidth(4pt) + dotstyle); label("$C_1$", (-3.5087977317468453,2.498086598756334), NE * labelscalefactor); dot((-4.678629861872753,3.1497645215015253),linewidth(4pt) + dotstyle); label("$H$", (-4.629836310997976,3.253842944318893), NE * labelscalefactor); dot((-6.268338560371482,0.9272317296342522),linewidth(4pt) + dotstyle); label("$N$", (-6.330288088513735,0.6338876130353557), NE * labelscalefactor); dot((-6.777963493692568,2.5312087831157224),linewidth(4pt) + dotstyle); label("$P$", (-6.657782504924178,2.334339390551113), NE * labelscalefactor); dot((-4.079799832913284,3.326206105209937),linewidth(4pt) + dotstyle); label("$F$", (-4.0252312345479275,3.4301860916168234), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Solution :Firstly we'll define a few points. Instead of $B$ let $A$ be the apex of $\Delta ABC$ and restate the problem accordingly. Let $X=I_AI_B \cap BC$. Let $I_AI_B \cap AC=P$ and let $I_AI_B \cap AB=F$. Let $H$ denote the orthocentre of $AA_1C_1$. Let $I_AA_1 \cap I_BC_1=N$. Note that $I_A$ is the incentre of $\Delta XAC$ while $I_B$ is the $X$-excentre of $\Delta XAB$. Claim 1: $N \in BC$. Proof: Reflect $AI_A$ about $AC$ and $AI_B$ about $AB$. Observe that there reflections are he same line since $\angle I_AAC+\angle I_BAB=60^\circ =\angle ABC$. Let $I_AA_1 \cap BC=N_1$ and let $I_BC_1 \cap BC=N_2$. Observe that $N_1,N_2$ are the reflections of $I_A,I_B$ about $AC,AB$ respectively but as we argued earlier the reflection of $AI_A,AI_B$ under $AC,AB$ are the same lines $\implies N \in BC$. Now back to the main problem. Observe that since $\overline{I_AA_1}=\overline {A_1N}$ and $AC \perp I_AN \implies PN$ is the reflection of $I_AI_B$ in $AC$ . Similiarly $NF$ is the reflection of $I_AI_B$ in $AB \implies N$ is the Anti-stiener point of the line $I_AI_B$ w.r.t $\Delta AA_1C_1$ . But since $N \in \odot(AA_1C_1) \implies H \in I_AI_B$. Done $\blacksquare$.