Three circles ω1,ω2,ω3 are tangent to line l at points A,B,C (B lies between A,C) and ω2 is externally tangent to the other two. Let X,Y be the intersection points of ω2 with the other common external tangent of ω1,ω3. The perpendicular line through B to l meets ω2 again at Z. Prove that the circle with diameter AC touches ZX,ZY. Proposed by Iman Maghsoudi - Siamak Ahmadpour
Problem
Source: 4th Iranian Geometry Olympiad (Advanced) P4
Tags: IGO, Iran, geometry
15.09.2017 15:20
Excellent problem! bgn wrote: Three circles ω1,ω2,ω3 are tangent to line l at points A,B,C (B lies between A,C) and ω2 is externally tangent to the other two. Let X,Y be the intersection points of ω2 with the other common external tangent of ω1,ω3. The perpendicular line through B to l meets ω2 again at Z. Prove that the circle with diameter AC touches ZX,ZY. Proposed by Iman Maghsoudi - Siamak Ahmadpour Draw tangents ¯ZX′,¯ZY′ to ⊙(AC) with X′,Y′ on ω2. Let V,W be contact points of {ω1,ω2},{ω2,ω3} respectively. Let P,Q be contact points of ¯XY with ω1,ω3, respectively. Let M be the mid-point of ¯AC and T=¯PV∩¯QW. Lemma 1. ¯XY,¯VW,¯AC are concurrent at a point S. (Proof) Note that A,C lie on ¯ZV,¯ZW, for homothety reasons. Thus, ∠AVB=∠CWB=90∘, so we have ZV⋅ZA=ZB2=ZW⋅ZChence A,C,V,W lie on a circle. Consequently, ¯VW passes through the exsimilicenter of ω1,ω3, proving the lemma. ◼ Lemma 2. T lies on line ¯ZM. (Proof) Notice that T lies on ω2; in fact it is the mid-point of arc ¯XY not containing Z, due to homothety reasons. Consequently, T has equal powers in ω1,ω3. However, ¯ZM is the radical axis of ω2,ω3; so the lemma must hold. ◼ As an immediate corollary, we see that ¯ZM bisects the angle XZY, so ¯XY∥¯X′Y′. Now for the grand finale: Lemma 3. S lies on ¯X′Y′. (Proof) By Dual of Desrague's Involution Theorem applied on ⊙(AC) and point Z, we see that (¯ZA,¯ZC),(¯ZB,¯ZB),(¯ZX′,¯ZY′) form pairs of an involution. Project on ω2, so (V,W),(X′,Y′),(B,B) are pairs of an involution on ω2. Consequently, lines ¯AC,¯VW,¯X′Y′ concur. ◼ Now it is clear as day that ¯XY,¯X′Y′ coincide; just as we desired. ◼
16.09.2017 12:22
Here is the solution I found during the contest ( unfortunately i wasn't able to fully write it down due to a lack of time ) . Let J be the intersection point of AC and XY . Let O1 and O3 be the centers of ω1 and ω3 respectively.Then J,O1,O3 are collinear .Let D=ω1∩ω2 and G=ω3∩ω2 .Consider the inversion centered at J that sends ω1 to ω3 .This inversion also sends ω3 to ω1 and AC to AC. Then the inversion must send ω2 to a circle which is tangent to ω1,AC and ω3 at three different points . But this type of circle must be unique and ω2 itself is such a circle .So ω2 is fixed under this inversion .Then this inversion takes B to B , D to G. So J,D,G are collinear and JB2=JA⋅JC . Let the common tangent of ω1 and ω2 meet AB at N . Then NA=ND=NB . So D lies in the circle with diameter AB (from now we will write the circle with diameter AB as (AB) ). So ∠ADB=90∘ .So , ∠ADB+∠BDZ=90∘+90∘=180∘ . Thus A,D,Z are collinear , in the same way C,G,Z are collinear . X′Z and Y′Z be the tangent from Z to (AC) such that {X′,Y′}∈ω2 .X′Z and Y′Z touches (AC) at R,P respectively .Let K=RP∩AC. By la hires theorem the polar of K wrt (AC) goes though Z .So ZB is polar of K wrt (AC). So K is the inverse of B wrt (AC) .Now its well known that (K,B;A,C)=−1 . Let K′ be the reflection of B wrt J . Now as JB2=JA⋅JC so inversion wrt (K′B) takes A to C⟹(K′,B;A,C)=−1⟹K=K′ . So J is the midpoint of KB . Now let us prove a lemma , Lemma : Let A,B,C,D be four points in circle ω . Tangents to ω at A & D meet at M ,tangents to ω at B & C meet at N . Let J=AB∩CD . Then J,M,N are collinear . [asy][asy] import graph; size(7 cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.2) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -33., xmax = -6, ymin = -2., ymax = 18.5; /* image dimensions */ pen ffdxqq = rgb(1.,0.8431372549019608,0.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); /* draw figures */ draw(circle((-15.591406596940846,4.170602969551047), 5.838881922231399), linewidth(0.4) + blue); draw((-32.09037581295504,0.7047951396101702)--(-10.059940647473814,2.3009969812082036), linewidth(0.4) + red); draw((-15.464881863893629,10.008113876923124)--(-32.09037581295504,0.7047951396101702), linewidth(0.4) + red); draw((-20.346084727918672,16.96841178922098)--(-20.795591909873973,1.5231518688861532), linewidth(0.4) + ffdxqq); draw((-10.059940647473814,2.3009969812082036)--(-20.346084727918672,16.96841178922098), linewidth(0.4) + ffdxqq); draw((-20.632815068113167,7.116234095227272)--(-22.24083871628236,4.364123020944482), linewidth(0.4) + red); draw((-20.795591909873973,1.5231518688861532)--(-22.24083871628236,4.364123020944482), linewidth(0.4) + red); draw((-15.464881863893629,10.008113876923124)--(-7.513228800524669,9.835766313536771), linewidth(0.4) + red); draw((-10.059940647473814,2.3009969812082036)--(-7.513228800524669,9.835766313536771), linewidth(0.4) + red); draw((-32.09037581295504,0.7047951396101702)--(-7.513228800524669,9.835766313536771), linewidth(0.4) + linetype("4 4")); draw((-15.464881863893629,10.008113876923124)--(-20.795591909873973,1.5231518688861532), linewidth(0.4) + yellow); draw((-20.632815068113167,7.116234095227272)--(-10.059940647473814,2.3009969812082036), linewidth(0.4) + ffdxqq); draw((-20.632815068113167,7.116234095227272)--(-20.795591909873973,1.5231518688861532), linewidth(0.4) + ffxfqq); draw((-15.464881863893629,10.008113876923124)--(-10.059940647473814,2.3009969812082036), linewidth(0.4) + ffxfqq); /* dots and labels */ dot((-20.795591909873973,1.5231518688861532),linewidth(3.pt) + dotstyle); label("A", (-21.161718685573057,0.5293791919886705), NE * labelscalefactor); dot((-10.059940647473814,2.3009969812082036),linewidth(3.pt) + dotstyle); label("B", (-9.759484813801835,1.5754556939860207), NE * labelscalefactor); dot((-20.632815068113167,7.116234095227272),linewidth(3.pt) + dotstyle); label("D", (-21.632453111471868,7.381180280071315), NE * labelscalefactor); dot((-15.464881863893629,10.008113876923124),linewidth(3.pt) + dotstyle); label("C", (-15.251386449287974,10.310194485663896), NE * labelscalefactor); dot((-20.346084727918672,16.96841178922098),linewidth(3.pt) + dotstyle); label("E", (-20.115642183575698,17.266603223946277), NE * labelscalefactor); dot((-32.09037581295504,0.7047951396101702),linewidth(3.pt) + dotstyle); label("J", (-32.35473725694481,-0.3074820096092098), NE * labelscalefactor); dot((-22.24083871628236,4.364123020944482),linewidth(3.pt) + dotstyle); label("M", (-22.8877449138687,3.4583933975812515), NE * labelscalefactor); label("N", (-7.405812684307776,8.95029503306734), NE * labelscalefactor); dot((-18.027242665298672,5.929570143860911),linewidth(3.pt) + dotstyle); label("F", (-18.180400654880582,4.818292850177807), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Proof: Let E=AD∩BC , F=AC∩BD . Now by brocards theorem JF is the polar of E . By la hires theorem the polar of E goes though M,N , So J,M,F,N are collinear . [asy][asy]import graph; size(18.8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); /* default pen style */ real xmin = -33., xmax = -4., ymin = -11., ymax = 9.; /* image dimensions */ pen xfqqff = rgb(0.4980392156862745,0.,1.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pair C_2 = (-23.001762946342133,-3.026280956652035), A_2 = (-9.038448471151908,-3.035942661766724), D_2 = (-18.963717497650077,-3.0290750213731505), Z = (-18.956311477197165,7.674273028079407), R_2 = (-21.25877248998085,1.584073070684142), T = (-15.48936252473688,3.9303444172584294), K = (-32.579197490828996,-3.0196539950886834), J = (-25.771457494239606,-3.024364508230916), A = (-23.00176294634213,-3.026280956652034), B = (-18.96371749765007,-3.0290750213731497), C = (-9.038448471151908,-3.035942661766723), O_3 = (-9.0352642722373,1.5659335359059465), O_1 = (-23.00123588881407,-2.2645655116171963), D = (-22.497708218366977,-1.693014479697557), O = (-18.962422022444255,-1.1568249910872843), F = (-17.72478176926395,0.24801252900858514), G = (-13.623825214074284,1.9157665830543051), I = (-19.918700912776877,0.4527875517520047), H = (-18.961126547238454,0.7154250391985535), P = (-11.946932705322933,2.639234106568399), R = (-22.41617265779375,-0.23220332486901085), M = (-16.02010570874702,-3.0311118092093796); 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draw(G--(-9.035264272237299,1.565933535905946), linewidth(0.4)); draw((-9.035264272237299,1.565933535905946)--A_2, linewidth(0.4)); draw((-22.49770821836698,-1.693014479697558)--(-23.00123588881407,-2.264565511617198), linewidth(0.4)); draw((-23.00123588881407,-2.264565511617198)--C_2, linewidth(0.4)); draw((-25.77145749423961,-3.0243645082309167)--(-9.035264272237299,1.565933535905946), linewidth(0.4) + blue); draw((-25.77145749423961,-3.0243645082309167)--G, linewidth(0.4) + xfqqff); draw((-20.82925480831267,-1.0144971871690105)--O, linewidth(0.4)); draw((-17.724781769263945,0.2480125290085855)--O, linewidth(0.4)); draw(Z--(-11.946932705322935,2.639234106568399), linewidth(0.4)); draw(Z--(-22.888511240035744,-1.3115916691255134), linewidth(0.4)); draw(D_2--(-22.888511240035744,-1.3115916691255134), linewidth(0.4) + ffxfqq); draw(D_2--(-13.889598943492828,4.03470729171564), linewidth(0.4) + ffxfqq); draw(D_2--(-19.918700912776877,0.45278755175200514), linewidth(0.4) + ffxfqq); draw(Z--C_2, linewidth(0.4) + blue); draw(Z--A_2, linewidth(0.4) + blue); draw(T--C_2, linewidth(0.4) + blue); draw(R_2--A_2, linewidth(0.4) + blue); draw((-13.889598943492828,4.03470729171564)--J, linewidth(0.4) + green); draw(Z--M, linewidth(0.4) + yellow); draw(Z--D_2, linewidth(0.4)); /* dots and labels */ dot(C_2,linewidth(3.pt) + red); dot(A_2,linewidth(3.pt) + red); dot(D_2,linewidth(3.pt) + red); dot(Z,linewidth(3.pt) + red); label("Z", (-18.72764855770153,7.827165226927332), NE * labelscalefactor,red); label("w2", (-22.318067521242696,7.350885160335141), NE * labelscalefactor); label("w1", (-24.406372428608478,-1.6984361049165062), NE * labelscalefactor); label("w3", (-9.75160114884862,6.654783524546552), NE * labelscalefactor); dot(R_2,linewidth(3.pt) + blue); dot(T,linewidth(3.pt) + blue); dot(K,linewidth(2.pt) + blue); label("K", (-32.7229551298722,-3.640193299484673), NE * labelscalefactor,blue); dot(J,linewidth(2.pt) + blue); label("J", (-25.90848648478386,-3.640193299484673), NE * labelscalefactor,blue); dot(A,linewidth(2.pt) + blue); label("A", (-23.453812295424086,-3.530282514886475), NE * labelscalefactor,blue); dot(B,linewidth(2.pt) + blue); label("B", (-18.83755934229973,-3.5669194430858746), NE * labelscalefactor,blue); dot(C,linewidth(2.pt) + blue); label("C", (-9.531779579652222,-3.6768302276840727), NE * labelscalefactor,blue); dot(O_3,linewidth(2.pt) + blue); label("O3", (-8.79904101566423,1.0859704382378466), NE * labelscalefactor,blue); dot(O_1,linewidth(2.pt) + blue); label("O1", (-23.673633864620484,-2.541085453502692), NE * labelscalefactor,blue); dot(D,linewidth(2.pt) + blue); label("D", (-23.01416915703129,-1.881620745913503), NE * labelscalefactor,blue); dot((-20.82925480831267,-1.01449718716901),linewidth(2.pt) + blue); label("E", (-20.99913810606431,-1.6617991767171068), NE * labelscalefactor,blue); dot(O,linewidth(2.pt) + blue); label("O", (-19.350476337091326,-1.7350730331159057), NE * labelscalefactor,blue); dot(F,linewidth(2.pt) + blue); label("F", (-17.665177639918944,-0.41614361793752797), NE * labelscalefactor,blue); dot(G,linewidth(2.pt) + blue); label("G", (-14.257943317374776,1.3057920074342428), NE * labelscalefactor,blue); dot((-22.88851124003574,-1.311591669125514),linewidth(2.pt) + blue); label("X′", (-23.527086151822886,-1.3687037511219118), NE * labelscalefactor,blue); dot(I,linewidth(2.pt) + blue); label("I", (-20.04657797287992,0.7562380844432521), NE * labelscalefactor,blue); dot(H,linewidth(2.pt) + blue); label("H", (-19,.35), NE * labelscalefactor,blue); dot(P,linewidth(2.pt) + blue); label("P", (-11.949816840812598,2.881179920008416), NE * labelscalefactor,blue); dot(R,linewidth(2.pt) + blue); label("R", (-22.97753222883189,-0.08641126414293357), NE * labelscalefactor,blue); dot((-13.889598943492828,4.03470729171564),linewidth(2.pt) + blue); label("Y′", (-13.74502632258318,4.053561622389196), NE * labelscalefactor,blue); dot(M,linewidth(2.pt) + blue); label("M", (-16.163063583743558,-3.6768302276840727), NE * labelscalefactor,blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let DG meet (AB) & (BC) again at E & F respectively .Let ZB∩O1O2=O . Now O1A,O1D,OB are tangent to (AB) , so by our lemma OE is tangent to (AB) . In the same way OF is tangent to (BC)⟹OB=OE=OF⟹O is the circumcenter of △EBF. Let M be the midpoint of AC .{R,B,P}∈(ZM) . So ∠EAB=∠EDB=∠BZC. BZ⊥AC⟹AE⊥ZC. In the same way CF⊥ZA So CF & AZ meet at (AC) . AE,CF,ZB concur at the orthocenter H of △ZAC .The inversion with center Z and and radius ZR takes H to B . ZRBMP is cyclic ⟹H∈RP .Let I∈RP such that BI⊥RP .By simson's theorem I∈X′Y′ . ∠X′IB=∠X′RB=∠ZMB=∠IBK . So X′IY′ goes though the midpoint J of KB . Now {E,I,F}∈(HB)⟹OI=OB . Now as JI=JB so the line JX′IY′ is the reflection of AC wrt the line O1O3. So X′Y′ is the other common external tangent of ω1 and ω3 . ◼
01.03.2021 13:48
Here is a simpler solution. Let O be the midpoint of AC, and U,V the intersection of ZX,ZY with l respectively. We apply an inversion with center at Z and radius ZB. The inversion maps ω2,X,Y to l,U,V respectively, and ω1,ω3 to themselves. Also the other common external tangent of ω1,ω3 maps onto a cirle that passes Z,U,V and tangent to both ω1 and ω3. Now, by casey’s theorem for U,V,Z,ω1, we have UV⋅BZ+UA⋅VZ=UZ⋅VA. Similarly, we have UV⋅BZ+UZ⋅VC=UC⋅VZ for U,V,ω3,Z. Then from these two equations we got UZVZ=UOVO, which means O lies on the bisector of ∠UZV, hence the distances from O to UZ, VZ are equal. Furthermore, we also got UV⋅BZ=AO⋅(UZ+VZ). Finally, by the area relation we can easily deduce the distance from O to UZ is equal to AO, from which the claim will be obtained.
26.09.2022 13:25
anantmudgal09 wrote: Excellent problem! bgn wrote: Three circles ω1,ω2,ω3 are tangent to line l at points A,B,C (B lies between A,C) and ω2 is externally tangent to the other two. Let X,Y be the intersection points of ω2 with the other common external tangent of ω1,ω3. The perpendicular line through B to l meets ω2 again at Z. Prove that the circle with diameter AC touches ZX,ZY. Proposed by Iman Maghsoudi - Siamak Ahmadpour Draw tangents ¯ZX′,¯ZY′ to ⊙(AC) with X′,Y′ on ω2. Let V,W be contact points of {ω1,ω2},{ω2,ω3} respectively. Let P,Q be contact points of ¯XY with ω1,ω3, respectively. Let M be the mid-point of ¯AC and T=¯PV∩¯QW. Lemma 1. ¯XY,¯VW,¯AC are concurrent at a point S. (Proof) Note that A,C lie on ¯ZV,¯ZW, for homothety reasons. Thus, ∠AVB=∠CWB=90∘, so we have ZV⋅ZA=ZB2=ZW⋅ZChence A,C,V,W lie on a circle. Consequently, ¯VW passes through the exsimilicenter of ω1,ω3, proving the lemma. ◼ Lemma 2. T lies on line ¯ZM. (Proof) Notice that T lies on ω2; in fact it is the mid-point of arc ¯XY not containing Z, due to homothety reasons. Consequently, T has equal powers in ω1,ω3. However, ¯ZM is the radical axis of ω2,ω3; so the lemma must hold. ◼ As an immediate corollary, we see that ¯ZM bisects the angle XZY, so ¯XY∥¯X′Y′. Now for the grand finale: Lemma 3. S lies on ¯X′Y′. (Proof) By Dual of Desrague's Involution Theorem applied on ⊙(AC) and point Z, we see that (¯ZA,¯ZC),(¯ZB,¯ZB),(¯ZX′,¯ZY′) form pairs of an involution. Project on ω2, so (V,W),(X′,Y′),(B,B) are pairs of an involution on ω2. Consequently, lines ¯AC,¯VW,¯X′Y′ concur. ◼ Now it is clear as day that ¯XY,¯X′Y′ coincide; just as we desired. ◼ I think we also have to consider the following two things in this proof: Prove that Z does not lie inside ⊙(AC) (otherwise X′,Y′ don't exist). Consider the special case when S lies at infinity separately (otherwise Lemma 3 is not enough to imply {X,Y}={X′,Y′}). There might be ways to avoid considering these. Can someone please tell if there are ways to modify the main proof so that the above two things are also handled? Either way, I will put below a proof to handle these (in a different way though): To prove Z doesn't lie inside ⊙(AC) : In fact Z must strictly lie outside ⊙(AC). Let ℓ2 denote line XY. We will be invoking the fact that ℓ2 intersects ω2 at two distinct points. Let Oi be center of ωi. Note that dist(O1,ℓ)=dist(O1,ℓ2) , dist(O3,ℓ)=dist(O3,ℓ2) but dist(O2,ℓ)>dist(O2,ℓ2)where in the end we use ℓ2 intersects ω2 at two distinct points. It follows that O2 lies on the opposite side of line O1O3 as B. Since U,V lie on segments O2O1,O2O3 respectively, so points O2,B lie on opposite of line UV. This means ∠UBV>90∘Hence it follows ∠AZC=∠UZV=180∘−∠UBV<90∘thus Z lies outside ⊙(AC), as desired. ◻ Remark: Using a similar proof as above, one may not that if ℓ2 is tangent to ω2 (i.e. when X≡Y), then Z∈⊙(AC). To resolve the case when S is at infinity : Then ℓ∥ℓ2. Since tangent to ω2 at B is parallel to ℓ2, so there exists a (positive) inversionΦ at B swapping ω2 and ℓ2. Note Φ(ω1),Φ(ω3) are circles tangent to all three of Φ(ℓ)=ℓ , Φ(ω2)=ℓ2 , Φ(ℓ2)=ω2since Φ was positive, it follows ω1,ω3 must be fixed under Φ. This means A,C are fixed under Φ, and thus radius of Φ must equal BA,BC. Next note that X,Y are also fixed under Φ (since X,Y are intersection points of ω2,ℓ2 which swap under Φ). So radius of Φ must also equal BX,BY. We obtain BA=BC=BX=BY=radius of ΦIt follows X,Y∈⊙(AC) and B is center of ⊙(AC). Since ∠ZXB=∠ZYB=90∘, it follows ZX,ZY are tangents to ⊙(AC), as desired. ◻