Consider two circles k1,k2 touching at point T. A line touches k2 at point X and intersects k1 at points A,B where B lies between A and X.Let S be the second intersection point of k1 with XT. On the arc \overarcTS not containing A and B , a point C is choosen. Let CY be the tangent line to k2 with Y∈k2 , such that the segment CY doesn't intersect the segment ST .If I=XY∩SC , prove that : (a) the points C,T,Y,I are concyclic. (b) I is the A−excenter of △ABC
Problem
Source: Azerbaijan Math Olympiad Training
Tags: geometry, TST
15.12.2019 20:11
Claim Two circles ω1 and ω2 are tangent to each other at T. If a line intersects ω1 at A and B and tangent to ω2 at X, then second intersection of TX and ω1 S is the midpoint of one of the arcs \overarcAB. Proof If we can prove that the tangent from S to ω1 is parallel to AB, then we are done. Also AB is the tangent to ω2 at X so if O1S∥O2X then we are also done. But showing this easy because we know that O1,T,O2 and S,T,X collinear. So we have ∠O1ST=∠O1TS=∠O2TX=∠O2XS and we are done So we have \overarcSCT=\overarcTX. That is equivalent to ∠TCI=∠TYI, which implies C,T,I,Y 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 O1,O2 be centers of circles k1,k2 respectively. We have homothety between triangles O1TS,O2TX (♣) which is obvious. Let Z=O1O2∩XA. Claim 1. S is midpoint of arc AB i.e AS=BS. By tangent property, we have ∠O2XB=90∘. Then ∠O1ZA=∠XZO2=90∘−∠XO1Z(♣)=90∘−∠ZO2SHence O1S⊥AB. Then it is well known that AS=BS. ◻. Claim 2. TIYC are cyclic. ∠TCI=180∘−∠TCS⇒TO1S=360∘−(360∘−2∠TCS)=2∠TCSBut We have 2∠TCS(♣)=2∠XO2T=∠IYTHence we have proven our claim. ◻ Let D be intersection of circle (TIYC) and line AC. Claim 3. ∠BCI=∠ABS. ∠BCI=∠BCT+∠TCI=(∠BAS−∠TO1S2)+∠TYI=∠BASSince we know that ∠TO1S2=∠TYI. ◻ Claim 4. ∠BCA=180∘−2∠ABS Direct consequence of Claim 1. ◻ Claim 5. CI is bisector of angle ∠ECD. Direct consequence of Claim 4. and Claim 3. ◻ Claim 6. SI is tangent to circle (TIX) By Claim 2 and basic angle chase, we can see that ∠TIC=∠TXIBut this is a property of tangent line, hence conclusion. ◻ Claim 7. SB is tangent to circle (BTX). Similiar to Claim 6, we have ∠SBT=∠BXT.◻ 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.◻ 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("O1", (-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("O2", (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