Consider two circles $k_1,k_2$ touching at point $T$. A line touches $k_2$ at point $X$ and intersects $k_1$ at points $A,B$ where $B$ lies between $A$ and $X$.Let $S$ be the second intersection point of $k_1$ with $XT$. On the arc $\overarc{TS}$ not containing $A$ and $B$ , a point $C$ is choosen. Let $CY$ be the tangent line to $k_2$ with $Y\in{k_2}$ , such that the segment $CY$ doesn't intersect the segment $ST$ .If $I=XY\cap{SC}$ , prove that : $(a)$ the points $C,T,Y,I$ are concyclic. $(b)$ $I$ is the $A-excenter$ of $\triangle ABC$
Problem
Source: Azerbaijan Math Olympiad Training
Tags: geometry, TST
15.12.2019 20:11
Claim Two circles $\omega_1$ and $\omega_2$ are tangent to each other at $T$. If a line intersects $\omega_1$ at $A$ and $B$ and tangent to $\omega_2$ at $X$, then second intersection of $TX$ and $\omega_1$ $S$ is the midpoint of one of the arcs $\overarc{AB}$. Proof If we can prove that the tangent from $S$ to $\omega_1$ is parallel to $AB$, then we are done. Also $AB$ is the tangent to $\omega_2$ at $X$ so if $O_1S\parallel O_2X$ then we are also done. But showing this easy because we know that $O_1, T, O_2$ and $S,T,X$ collinear. So we have $$\angle{O_1ST}=\angle{O_1TS}=\angle{O_2TX}=\angle{O_2XS}~~\text{and we are done}$$ So we have $\overarc{SCT}=\overarc{TX}$. That is equivalent to $\angle{TCI}=\angle{TYI}$, which implies $C,T,I,Y ~~\text{concyclic}.$ $(a)$ is done.
03.01.2022 23:19
All angle chases left to readers
04.12.2022 21:19
You can solve the problem using angle chase only, observing equal angles carefully or just trying random equations is enough to solve the problem Let $O_1,O_2$ be centers of circles $k_1,k_2$ respectively. We have homothety between triangles $O_1TS, O_2TX$ $(\clubsuit)$ which is obvious. Let $Z= O_1O_2 \cap XA$. Claim 1. $S$ is midpoint of arc $AB$ i.e $AS=BS$. By tangent property, we have $ \angle O_2XB=90^{\circ}$. Then $$ \angle O_1ZA = \angle XZO_2 = 90^{\circ}-\angle XO_1Z \stackrel{(\clubsuit)}{=} 90^{\circ} - \angle ZO_2S$$Hence $O_1S \perp AB$. Then it is well known that $AS=BS$. $\square$. Claim 2. $TIYC$ are cyclic. $$ \angle TCI= 180^{\circ} - \angle TCS \Rightarrow TO_1S=360^{\circ}-( 360^{\circ}- 2\angle TCS)= 2 \angle TCS$$But We have $$ 2 \angle TCS \stackrel{(\clubsuit)}{=} 2 \angle XO_2T = \angle IYT$$Hence we have proven our claim. $\square$ Let $D$ be intersection of circle $(TIYC)$ and line $AC$. Claim 3. $ \angle BCI= \angle ABS$. $$ \angle BCI = \angle BCT + \angle TCI = ( \angle BAS- \frac{\angle TO_1S}{2}) + \angle TYI= \angle BAS$$Since we know that $ \frac{ \angle TO_1S}{2} = \angle TYI$. $\square$ Claim 4. $ \angle BCA = 180^{\circ}- 2\angle ABS$ Direct consequence of Claim 1. $\square$ Claim 5. $CI$ is bisector of angle $ \angle ECD$. Direct consequence of Claim 4. and Claim 3. $\square$ Claim 6. $SI$ is tangent to circle $(TIX)$ By Claim 2 and basic angle chase, we can see that $$\angle TIC= \angle TXI$$But this is a property of tangent line, hence conclusion. $\square$ Claim 7. $SB$ is tangent to circle $(BTX)$. Similiar to Claim 6, we have $$ \angle SBT = \angle BXT.$$$\square$ Claim 8. $SB=SI$ Direct consequence of Claim 5,6 since $S$ lies on radical axis of the circles ang $S$ is power point of each circle, hence it is well known that $SB=SI$, consequently $SY=SA$ by claim 1.$\square$ Claim 9. $I$ is excenter of $ABC$. Well known after proving our latter claims, if $A,B,C,S$ are cyclic points such that $AS=BS$, and if we know that $IS=BS=AS$ and $IS$ is angle bisector of external angle of $C$, then $I$ is $A-excenter$ of $ABC$. Which we have shown by our claims, hence we are done! [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(17.36337394623623cm); real labelscalefactor = 0.1; /* changes label-to-point distance */ pen dps = linewidth(0.1) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -8.452434540931394, xmax = 8.910939405304834, ymin = -7.926393976978712, ymax = 0.41421048017120354; /* image dimensions */ pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pair O_1 = (-2.931964724884746,-3.8024524499625474), T = (0.8742639629400576,-3.409431768769399), O_2 = (3.2746912527656806,-3.161570264383272), X = (2.989000375524323,-0.7653509042751026), A = (-6.231485873291579,-1.8646696264546856), B = (-0.18052350870167239,-1.1432395375098683), Y = (3.2621406720134565,-5.574727735834093), C = (0.46688818028571666,-5.560189918664959), I = (3.155480164343957,-3.6966797838417103), D = (1.8990123066977722,-6.350298641735329); /* draw figures */ draw(circle(O_1, 3.8264660040128478), linewidth(1.) + blue); draw(circle(O_2, 2.4131901083619876), linewidth(1.) + blue); draw(circle((1.8689646290206325,-4.711798000093499), 1.638776133458893), linewidth(1.) + red); draw(Y--T, linewidth(1.)); draw(C--Y, linewidth(1.) + linetype("4 4")); draw(T--I, linewidth(1.)); draw(circle((2.174924803378208,-2.2819768347386704), 1.7212998722243344), linewidth(1.) + linetype("4 4") + eqeqeq); draw(C--B, linewidth(1.) + linetype("4 4")); draw(I--Y, linewidth(1.)); draw(O_2--Y, linewidth(1.)); draw(O_2--T, linewidth(1.)); draw(X--T, linewidth(1.)); draw(X--B, linewidth(1.)); draw(B--T, linewidth(1.)); draw(B--(-2.478960038004651,-7.602008849534833), linewidth(1.) + qqwuqq); draw(B--A, linewidth(1.)); draw(A--(-2.478960038004651,-7.602008849534833), linewidth(1.) + qqwuqq); draw(O_1--T, linewidth(1.)); draw(O_1--(-2.478960038004651,-7.602008849534833), linewidth(1.)); draw(X--I, linewidth(1.) + linetype("4 4")); draw(Y--(xmax, -0.005200895880482592*xmax-5.5577616818514635), linewidth(1.)); /* ray */ draw(O_2--X, linewidth(1.)); draw(B--I, linewidth(1.)); draw(circle((1.4979810354523042,-1.740557071626131), 1.7816188541066467), linewidth(1.) + linetype("4 4") + eqeqeq); draw(C--D, linewidth(1.)); draw(T--(-2.478960038004651,-7.602008849534833), linewidth(1.)); draw(circle((-2.478960038004651,-7.602008849534833), 6.855546039900086), linewidth(1.) + linetype("4 4") + qqwuqq); draw((-2.478960038004651,-7.602008849534833)--I, linewidth(1.) + qqwuqq); draw((2.989000375524323,-0.7653509042751024)--(xmax, 0.11922567774782582*xmax-1.1217164998354958), linewidth(1.)); /* ray */ draw(A--I, linewidth(1.) + linetype("4 4")); draw(A--C, linewidth(1.) + linetype("4 4")); /* dots and labels */ dot(O_1,linewidth(4.pt)); label("$O_1$", (-2.8950634140847016,-3.715161846487533), NE * labelscalefactor); dot(T,linewidth(4.pt)); label("$T$", (0.9068696972441321,-3.314958361084484), NE * labelscalefactor); dot(O_2,linewidth(4.pt)); label("$O_2$", (3.308090609662343,-3.069378949587158), NE * labelscalefactor); dot(X,linewidth(4.pt)); label("$X$", (3.026129063128386,-0.6772535709280221), NE * labelscalefactor); dot(A,linewidth(4.pt)); label("$A$", (-6.196742168659741,-1.7960042233047282), NE * labelscalefactor); dot(B,linewidth(4.pt)); label("$B$", (-0.14821221881841495,-1.0683615225719112), NE * labelscalefactor); dot((-2.478960038004651,-7.602008849534833),linewidth(4.pt)); label("$S$", (-2.422095658608387,-7.83543863938711), NE * labelscalefactor); dot(Y,linewidth(4.pt)); label("$Y$", (3.298995075903183,-5.479695395764614), NE * labelscalefactor); dot(C,linewidth(4.pt)); label("$C$", (0.5066662118410971,-5.488790929523775), NE * labelscalefactor); dot(I,linewidth(4.pt)); label("$I$", (3.189848670793264,-3.624206508895931), NE * labelscalefactor); dot(D,linewidth(4.pt)); label("$D$", (1.9346650120291995,-6.280102366570714), NE * labelscalefactor); dot((-2.4789600380046535,-7.6020088495348315),linewidth(4.pt)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Remark: There is a case where Circles touch themselves such that one of them inside of it, still, one may use directed angles to avoid configuration issues and I have a weird mistake in Claim 1 with definition of Z, will fix it later