Circles $\odot AA_1A_2$ and $\odot ABC$ are internally tangent to each other at $A$.And their exsimilicenter is $A$.So the line $AA_3$ cuts $\odot ABC$ at a point $D$ tangent at which to $\odot ABC$ is parallel to $BC$, i.e $D$ is the midpoint of the arc $BC$.Hence $AA_3,BB_3,CC_3$ concur at the incenter of $\triangle ABC$.

Alternatively, just use an angle-chase: By the alternate segment theorem, $\angle CA_3A=\angle AA_1A_3=B+\angle A_2A_1A_3=B+\angle CAA_3$. but also $\angle CA_3A=\angle A_3AB+\angle ABA_3=B+\angle BAA_3$. We conclude that $\angle CAA_3=\angle BAA_3$ and so $AA_3$ passes through the incentre of triangle $ABC$, as do $BB_3$ and $CC_3$.

Or even easier, i think $BA_3^2=BA_1.BA$ and $CA_3^2=CA_2CA$. Suppose $AA_1=xc, A_1b=(1-x)c, AA_2=xb,CA_2=(1-x)b, BA_3=ya, CA_3=(1-y)a$, we have $y^2a^2=(1-x)xc^2, (1-y)^2a^2=(1-x)xb^2$ and dividing these two equations we have $\frac{y^2}{(1-y)^2}=\frac{c^2}{b^2}$. Now, the conclusion is obvious using Ceva's theorem. Furthermore, $A_3$ is the foot of bisector from angle $\widehat A$.

The power of $B$ with respect to $(AA_1A_2)$ is $BA_3^2 = BA_1 \cdot BA$; similarly, the power of $C$ with respect to $(AA_1A_2)$ is $CA_3^2 = CA_2 \cdot CA$. It follows that \[\frac{BA_3^2}{CA_3^2} = \frac{BA_1 \cdot BA}{CA_2 \cdot CA} = \frac{BA_1}{CA_2} \cdot \frac{BA}{CA} = \frac{c^2}{b^2} \implies \frac{BA_3}{CA_3} = \frac{c}{b}\] since $BA_1/CA_2 = BA/CA = c/b$ (because $\triangle{AA_1A_2}$ ~ $\triangle{ABC}$). Then $A_3$ is the point where the bisector of $\angle{A}$ meets $BC$ so $AA_3$ is the bisector of $\angle{A}$. It follows that $AA_3, BB_3, CC_3$ concur at the incenter of $\triangle{ABC}$. Alternatively, $(AA_1A_2)$ is internally tangent to $(ABC)$ since it's the image of $(ABC)$ under the $A$-centered homothety taking $\triangle{ABC}$ to $\triangle{AA_1A_2}$ . It's also tangent to chord $BC$ of $(ABC)$; we can now use the fairly well-known that the line joining $A$ to the contact point with $BC$ is the angle bisector (the above is basically a proof).

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -4.429663950164643, xmax = 10.25201915337621, ymin = -2.166783588755752, ymax = 6.532066167287861; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); draw((1.100000000000001,5.740000000000006)--(-1.080000000000001,1.540000000000002)--(4.840000000000005,1.160000000000001)--cycle, zzttqq); /* draw figures */ draw((1.100000000000001,5.740000000000006)--(-1.080000000000001,1.540000000000002), zzttqq); draw((-1.080000000000001,1.540000000000002)--(4.840000000000005,1.160000000000001), zzttqq); draw((4.840000000000005,1.160000000000001)--(1.100000000000001,5.740000000000006), zzttqq); draw(circle((1.693203484893987,3.597512302950476), 2.223093319341087)); /* dots and labels */ dot((1.100000000000001,5.740000000000006),dotstyle); label("$A$", (1.173306962570872,5.848313784716748), NE * labelscalefactor); dot((-1.080000000000001,1.540000000000002),dotstyle); label("$B$", (-1.010902037309074,1.650833880599632), NE * labelscalefactor); dot((4.840000000000005,1.160000000000001),dotstyle); label("$C$", (4.914951944973911,1.270971445837903), NE * labelscalefactor); dot((-0.4002837670437572,2.849545035970746),dotstyle); label("$A_1$", (-0.3271496547379604,2.961359280527601), NE * labelscalefactor); dot((3.673881325111770,2.588027682082384),dotstyle); label("$A_2$", (3.756371518950635,2.695455576194390), NE * labelscalefactor); dot((1.551615239817024,1.371078751498233),dotstyle); label("$A_3$", (1.629141884284948,1.479895784956854), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] By PoP, $BA_3^2 = BA_1 \times BA$ and $CA_3^2 = CA_2 \times CA.$ Hence, $\dfrac{BA_3^2}{CA_3^2} = \dfrac{BA_1}{CA_2} \times \dfrac{BA}{CA}.$ Since $A_1A_2 \parallel BC,$ $\dfrac{BA_1}{CA_2} = \dfrac{BA}{CA}.$ It follows that $\dfrac{BA_3^2}{CA_2^2} = \dfrac{BA^2}{CA^2} \implies \dfrac{BA_3}{CA_3} = \dfrac{BA}{CA}.$ Hence, $A_3$ must be the point where the internal angle bisector of $\angle BAC$ meets $BC$. Similarly, the internal angle bisectors of $\angle ABC$ and $\angle BCA$ meet the opposite sides at points $B_3, C_3$ respectively. It follows that the lines in question are concurrent at the incenter of $\triangle ABC.$ $\blacksquare$

Dear Mathlinkers, 1. according to a special case of the Reim's theorem, the circumcircle of ABC is tangent to (AA1A2) ; in consequence AA3 is the A-inner bisector of ABC 2. and we are done... Sincerely Jean-Louis

Easy Solution: Note that since $A_1A_2\parallel BC$, $\angle A_2A_3C=\angle A_3A_2A_1$, and since quadrilateral $AA_1A_2A_3$ is cyclic, $\angle A_3A_2A_1=\angle A_3AA_1$. Also, since $\odot AA_1A_2A_3$ is tangent to $BC$, we have $\angle CAA_3=\angle A_2A_3C$. Hence $\angle CAA_3=\angle A_3AB$, implying that $AA_3, BB_3, CC_3$ are the internal angle bisectors of angles $\angle CAB, \angle ABC, \angle BCA$, respectively. The three lines therefore concur at the incenter of $\Delta ABC$. $\Box$

$(ABC)$ is tangent to $(AA_1A_2)$ so $AA_3$ is an angle bisector.

note that $A_1A_2$ parallel to $BC$ imply that $\frac{AA_1}{AA_2} = \frac{AB}{AC}$ by POP we have $(BA_1)(BA) = BA_3^2$ and $(CA_2)(CA)=CA_3^2$ dividing we get $\frac{BA_1}{CA_2}=\frac{AA_1}{AA_2}=\frac{BA_3^2.AC}{CA_3^2.AB}$ or $\frac{AB^2}{AC^2}=\frac{BA_3^2}{CA_3^2}$ and hence $\frac {BA_3}{CA_3}=\frac {AB}{AC}$ thus $AA_3 $ angle bisector. similarly $BB_3,CC_3$ are also angle bisector and hence all three lines are concurrent at incentre of triangle $ABC$. we are done

Claim:$AA_{3}$ is the angle bisector of $\angle{BAC}$. Proof:We apply POP twice for points $B,C$ WRT $(AA_1A_2)$.We deduce: $CA_3^2=CA_2.CA,BA_3^2=BA_1.BA \Rightarrow \frac{BA_3}{CA_3}=\frac{AB}{AC}$ By angle bisector theorem we are done. By the claim we deduce that these lines are concurrent at the incenter of the triangle.

Easy.... ELMO Shortlist 2010 G1 wrote: Let $ABC$ be a triangle. Let $A_1$, $A_2$ be points on $AB$ and $AC$ respectively such that $A_1A_2 \parallel BC$ and the circumcircle of $\triangle AA_1A_2$ is tangent to $BC$ at $A_3$. Define $B_3$, $C_3$ similarly. Prove that $AA_3$, $BB_3$, and $CC_3$ are concurrent. Carl Lian. Solution:- Introduce $\odot(ABC)$, the next steps are trivial. Notice that $A_1A_2\|BC$. So, $\odot(ABC)$ and $\odot(AA_1A_2)$ are interally tangent at $A$. Now it is well known that $A,A_3$ and the midpoint of $\widehat{BC}$ are collinear. Reason- Homothety centered at $A$, maps $A_3$ to the midpoint of $\widehat{BC}$. So, $AA_3$ is the internal bisector of $\angle BAC$. Similarly you get that $BB_3,CC_3$ are the internal bisectors of $\angle ABC$ and $\angle ACB$. So, $AA_3,BB_3,CC_3$ concur at the incenter of $\triangle ABC$. $\blacksquare$.

Probably same as others $\angle AA_2A_1=\angle ACB$ ,thus $(ABC)$ and $(AA_1A_2)$ are internally tangent . By PoP , $BA_1.BA=BA_3^2$ and $CA_2.CA=CA_3^2$ $\Rightarrow{\frac{BA_1.BA} {CA_2.CA}=\frac{BA_1.AA_1}{CA_2.AA_2}}$ $(\xi)$ And since $\frac{AA_1}{AA_2}=\frac{A_1B}{A_2C}$ so, $(\xi)=(\frac{AA_1}{AA_2})^2=(\frac{BA_3}{CA_3})^2\rightarrow{\frac{AA_1}{AA_2}=\frac{AB}{AC}=\frac{BA_3}{CA_3}}$ .Done .

ELMO Shortlist 2010, G1 wrote: Let $ABC$ be a triangle. Let $A_1$, $A_2$ be points on $AB$ and $AC$ respectively such that $A_1A_2 \parallel BC$ and the circumcircle of $\triangle AA_1A_2$ is tangent to $BC$ at $A_3$. Define $B_3$, $C_3$ similarly. Prove that $AA_3$, $BB_3$, and $CC_3$ are concurrent. Carl Lian. Here are two solutions which I got for this problem .

How is a non-marathon thread still alive after 8 years?!?

LucaTu1 wrote: How is a non-marathon thread still alive after 8 years?!? Even my reply on your reply about alive thread is keeping this thread alive

$\overline{A_1A_2} \parallel \overline{BC}$ implies $\triangle AA_1A_2 \sim \triangle ABC$. From PoP, we have $BA_3^2=BA_1\cdot BA$ and $CA_3^2=CA_1\cdot CA$. Hence $$\frac{BA_3}{CA_3}=\sqrt{\frac{BA_1\cdot BA}{CA_1\cdot CA}}=\sqrt{\frac{BA^2}{CA^2}}=\frac{BA}{CA}.$$Hence $$\frac{BA_3}{CA_3}\cdot \frac{CB_3}{AB_3}\cdot \frac{AC_3}{BC_3}=\frac{BA}{CA}\cdot \frac{CB}{AB}\cdot\frac{AC}{BC}=1,$$and we're done by Ceva. $\blacksquare$

Even though , this is very simple, Still I will post They are just the angle bisectors.

It is easy to see that \[\angle A_1AA_3 = \angle A_1A_2A_3=\angle A_2A_1A_3=\angle A_2AA_3\]as $A_3$ is the midpoint of arc $A_1A_3A_2$. Hence, $\sin \angle A_1AA_3=\sin \angle A_2AA_3$. Similarly, $\sin \angle B_1BB_3=\sin \angle B_2BB_3$ and $\sin \angle C_1CC_3=\sin \angle C_2CC_3$. This yields \[\frac{\sin \angle A_1AA_3}{\sin \angle A_2AA_3}\cdot \frac{\sin \angle B_1BB_3}{\sin \angle B_1BB_3}\cdot \frac{\sin \angle C_1CC_3}{\sin \angle C_1CC_3}=\frac{\sin \angle BAA_3}{\sin \angle CAA_3}\cdot \frac{\sin \angle CBB_3}{\sin \angle ABB_3}\cdot \frac{\sin \angle ACC_3}{\sin \angle BCC_3}=1\]which by the converse of Ceva's Theorem, these three cevians $AA_3,BB_3,CC_3$ are concurrent.

By sinus Law $AA_3$ bisector of angle $BAC$.

let $O$ the center of $\odot AA_1A_2$ $BC\parallel A_1A_2$$\longrightarrow$$OA_3 \perp A_1A_2$ then $A_3$ is the midpoint of arc$ A_1A_2$not contain $A$ so $AA_3$ is angle bisector of $\angle A$ and at the same way we can found that $BB_3$ $CC_3$ are angle besictors of $\angle B$ $\angle C$ respectively and angle besictors in traingle are concurrent so we are done

Notice $$\angle A_1AA_3=\angle A_1A_2A_3=\angle CA_3A_2=\angle A_3AA_2$$so $I,$ the incenter of $\triangle ABC,$ lies on $\overline{AA_3}.$ Similarly, $I$ lies on $\overline{BB_3}$ and $\overline{CC_3}.$ $\square$

Draw the tangent line from A to (O) cut BC at A_4. It's clear that the center of (A_1A_2A_3) lies on the bisector of <AA_4A_3. Call A_5 is the symmetry point of A3 to A_4. According to Newton formula we will have that (A_5,A_3;A_4,C) =-1. Also, AA_5 $\perp$ AA_3 then AA_3 is the bisector of <BAC. Similarly, BB_3 and CC_3 are the bisectors of <ABC and <ACB. Nevertheless, AA_3,BB_3,CC_3 are concurrent at the incenter of $\triangle$ABC

Let $\omega$ be the circumcircle of $ABC$. Let $\omega_A$ be the circumcircle of $\triangle AA_1A_2,$ and define $\omega_B,\omega_C$ similarly. Let $M_a$ be the arc midpoint of arc $BC$ on $\omega$. Let the tangent to $\omega$ at $M_a$ intersect $AB$ and $AC$ at $B'$ and $C'$. Take a homothety centered at $A$ with ratio $AB/AA_1.$ Then, $\omega_A$ gets sent to $\omega$, so $BC$ gets sent to $B'C'$ and consequently $A_3$ gets sent to $M_a$. Therefore, line $AA_3$ is the same as line $AM_a$, which is also the bisector of angle $A$. Similarly, the other two lines are also angle bisectors, so they concur at the incenter, QED.

Note, that $\triangle A A_1 A_2 \sim ABC,$ using PoP with respect to $B,$ we have $(BA_3)^2=BA_1 \cdot BA,$ and with PoP with respect to $C,$ we have $(CA_3)^2=CA_1 \cdot CA,$ so $\frac{BA_3}{CA_3}=\sqrt{\frac{BA_1 \cdot BA}{CA_1 \cdot CA}=\sqrt{\frac{BA^2}{CA^2}}=\frac{BA}{CA}.$ So Ceva's finishes $\blacksquare$

Proposed by me (Just kidding ) $\color{red} \textbf{Geo Marabot Solve 9}$ $\bold {\textbf{One Liner :}}$ $\angle BAA_3 = \angle A_1A_2A_3 = \angle CA_3A_2 =\angle CAA_3 \Longrightarrow AA_3$ is the angle bisector of $\triangle ABC$ Hence our desired concurrency happens at incenter

Lmfao I solved this without a diagram There is a homothety taking circumcircle A A1 A2 to circumcircle ABC at A, since A1A2 parallel to BC. Due to the tangency condition with BC the tangency point A3 must be at the "bottom" point of A A1 A2, so after the homothety the bottom pt would just be the arc midpoint of BC. Since A3,arc midpoint are homothetic at A, this line is just angle bisector, and we win by googling the word "incenter".

Use standard notation. Let $AA_1 = \lambda c, AA_2 = \lambda b$. From PoP, $$BA_3^2 = BA_1 \cdot BA = (1-\lambda)c^2.$$Similarly, $$CA_3^2 = (1-\lambda)b^2.$$Dividing, we get $$\left(\frac{BA_3}{A_3C}\right)^2 = \left(\frac{c}{b}\right)^2 \implies \frac{BA_3}{A_3C} = \frac{c}{b}.$$Taking the cyclic product wrt $A, B, C$, we get $\prod_{cyc} \frac{BA_3}{A_3C} = 1$, so from the converse of Ceva's theorem, the result follows. $\square$