Triangles ABC and DEF share circumcircle Ω and incircle ω so that points A,F,B,D,C, and E occur in this order along Ω. Let ΔA be the triangle formed by lines AB,AC, and EF, and define triangles ΔB,ΔC,…,ΔF similarly. Furthermore, let ΩA and ωA be the circumcircle and incircle of triangle ΔA, respectively, and define circles ΩB,ωB,…,ΩF,ωF similarly. (a) Prove that the two common external tangents to circles ΩA and ΩD and the two common external tangents to ωA and ωD are either concurrent or pairwise parallel. (b) Suppose that these four lines meet at point TA, and define points TB and TC similarly. Prove that points TA,TB, and TC are collinear. Nikolai Beluhov
Problem
Source: TSTST 2021/6
Tags: USA TSTST, geometry
13.12.2021 20:02
Really good geometry problem, appropriate for its placement. Or maybe I just like the techniques used. 6a. Note that the intersection of the two common external tangent to two circles is the center of the unique positive homothety mapping one circle to the other. Let fA be the positive homothety mapping ΩA to Ω, and define fD similarly. It suffices to show that fA and fD both map ωA and ωD, respectively, to the same circle. Let △A′B′C′ be the image of ΔA under fA and define △D′E′F′ similarly. Let their incenters be JA and JD. Let BC intersect EF at X. Let l=XI be the interior angle bisector of these two lines. Also let X′ be the circumecenter of △AID. Our first claim is that X′ lies on l. This follows since if we let l′ be the interior angle bisector of BF and CE, then angle chasing immediately shows that l⊥l′ but also that l′ is parallel to A1D1 where A1 and D1 are the midpoints of "minor" arcs ⌢BC and ⌢EF respectively. This shows A1D1⊥l. But we also know AA1 intersects DD1 at I and that from isogonal conjugates (read: angle chasing) in similar △A1ID1 and △AID that A1D1⊥l implies that X′ lies on l as claimed. Define B1 and its variants in a similar fashion. We have from CA||C′A′ that B1 is still the "minor" arc midpoint of ⌢C′A′. As a combined corollary, B1C1 is the perpendicular bisector of A′JA. We also already know B1C1 is the perpendicular bisector of AI and hence passes through X′. From AB||A′B′ and CA||C′A′ we have that BC′||B′C and moreover that both latter lines are parallel to the "interior" angle bisector of BC and B′C′. But B′C′||EF, so B′C||XI. Now create a dummy point N at the intersection of AI and BC. We get: ∠IAA′=180∘−∠ACA1−∠ACA′=180∘−∠ACA1−∠B′CB=180∘−∠ACA1−∠B′CB=180∘−∠INX−∠IXN=∠AIX′=∠IAX′So A′ also lies on AX′ which also gives that JA lies on l. Similarly JD lies on l. Recall that within our angle chase we showed ⌢AA′=∠BXE. Similarly we also have ⌢DD′=∠BXE. Thus, IJA=AA′=DD′=IJD, which proves JA=JD. From Poncelet's Porism, since the incircles of △A′B′C′ and △D′E′F′ have the same center, they are necessarily identical, which is exactly what we wanted to show. ◼ 6b. Suppose that the common tangents to ωA and ωD meet at P. Let IA be the incenter of ΔA and define ID similarly. In the spirit of 6a, let f′A be the positive homothety mapping ωA to ω, and define f′D similarly. They have centers at A and D respectively. Let the image of ΔA under f′A be △AB∗C∗ and define △DE∗F∗ similarly. From 6a, we are already given that these share a circumcircle, Γ. Let A∗1 be the intersection of AI with Γ and define D∗1 similarly. Also let I∗A be the image of the A-excenter of Δ under f′A and define I∗D similarly. We know that A∗1 is the midpoint of II∗A. Define f(X)=Pow(X,ω)−Pow(X,Ω). Linearity of Power of a Point states this function is linear. We have IAPIDP=rArD=rAr:rDr=AIAAI:DIDDI so after applying Menelaus twice we have that P lies on AD and APDP=AIAIAI:DIDIDI. So now we compute: F(P)=1IAIAIA−IDIDID(IAIAIA⋅F(A)−IDIDID⋅F(D))=1IAIAIA−IDIDID(IAIAIA⋅Pow(A,ω)−IDIDID⋅Pow(D,ω))=1IAIAIA−IDIDID(IAIAIA⋅(AI2−r2)−IDIDID⋅(DI2−r2))=1IAIAIA−IDIDID(IAIAIA⋅AI2−IDIDID⋅DI2)−r2=1IAIAIA−IDIDID(II∗AAI⋅AI2−II∗DDI⋅DI2)−r2=2IAIAIA−IDIDID(IA∗1⋅AI−ID∗1⋅DI)−r2=2IAIAIA−IDIDID(Pow(I,Γ)−Pow(I,Γ))−r2=−r2The locus of points where F is constant is necessarily a line, so we are done. ◼
13.12.2021 20:25
Part (a) is implied by the following, which I haven't been able to prove: Quote: Let B′,C′ be the points on AB,AC such that ¯B′C′ is parallel to ¯EF and tangent to ω (i.e. △AB′C′ is the image of ΔA under the homothety at A sending ωA to ω). Similarly define E′,F′. Then AB′C′DE′F′ are concyclic. can someone help? (amusingly, this statement is its own converse, if we consider AB′C′DE′F′ as the original hexagon)
13.12.2021 20:46
@above maybe something like the following: Construct B’, C’ instead such that B’ lies on AB, C’ lies on AC, ∠EDC=∠FDB′, and ∠FDB=∠EDC′. By Pascal B’C’ is parallel to EF. The goal is to show this is the same definition as yours, i.e. B’C’ is tangent to ω. Let X = B’C’ cap BC. We use DDIT on the point D and the quadrilateral ABXC’ and an inconic tangent to its four sides plus the side DE. (This inconic exists except when three of the five sides concur; if that happens, replace DE with DF and DF with DE in the argument below.) Then the involution is just “reflection across internal angle bisector of EDF”, so DF is tangent to the said inconic as well, so this inconic is just the incircle ω, which shows that B’C’ is tangent to it. The rest is just angle chasing to prove AB’C’D concyclic (D is the Miquel point of ABXC'). A similar argument shows AE’F’D concyclic; finally, because ∠F′AB′=∠EAC=∠EDC=∠F′DB′, the six points lie on the same circle.
14.12.2021 04:52
There was a lot of work put into the official solutions for this problem. I hope you enjoy some of them. Let I and r be the center and radius of ω, and let O and R be the center and radius of Ω. Let OA and IA be the circumcenter and incenter of triangle ΔA, and define OB, IB,…,IF similarly. Let ω touch EF at A1, and define B1, C1,…,F1 similarly.
14.12.2021 09:27
the vertices of ΔA are A,XA,YA in clockwise order, OA and IA are the circumcenter and incenter of ΔA resp. denote the center of mixtilinear incircle associated with A and △ABC with MA. Part a:
Let P=AD′∩A′D. It is obvious that P is fixed (since in (1) we already proved that (A′,D′) and (A,D) are anti-homologous pairs) so TA is on the polar of P wrt Ω. ◼
[1] https://artofproblemsolving.com/community/q3h186117p1022749
24.04.2022 09:26
27.08.2022 21:21
Great problem! It took me a lot of time but it worthwhile! First notice that by Poncelet's Porism, we can take D as any point. Now make some construction to caracterize the extimilicenter of (ΩA,ΩD). Let E′∈DE,F′∈DF such that E′F′∥BC and E′F′ is tangent to ω (i.e. the intersections of the reflection of BC over I. Similarly, define points B′ and C′. Let I be the center of ω. Claim 1: Points A,E′,F′ and D′ are concyclic. Proof: Notice that since EE′ and F′F concurr at D and A∈(DEF), our claim is equivalent to show that A is the Miquel Point of lines ¯DE′E,¯DFF′,¯EF and ¯E′F′. Therefore, if we let H:=EF∩E′F′, it suffies to show that points AHEF′ are concyclic. In other words, we have to show that ∡(AH,AE)=∡(F′H,F′F)=∡(E′F′,FD)=∡(BC,FD)==∡CAF+∡BADOn the other hand ∡(AH,AF)=∡HAC+∡CAFSo we need to show that ∡BAD=∡HAC, i.e., that AH and AD are isogonal conjugate with respect to ∠BAC. Now we use unheterred moving points. Fix△ABC and let SA be the tangency point of the A-mixtilinear incircle with Ω and let XA be the tangency point of ω with ¯BC. Animate point D over Ω, and let Q be the second intersection of DXA with Ω and let P:=¯EF∩¯BC. First notice that by DDIT P,Q and A are colinear. Then D has degree 2, then P also has degree 2 (since the map D→P is projective); then Q has degree 2 and therefore line EF (the reflection of ¯BC over ¯DI has degree 2. Therefore H has degree at most 2. On the other hand the isogonal of AD has degree 1. Then we only have to check that H lies on the isogonal conjugate of AD for 4 choices of D. The degenerate cases D=A,B,C are trivial; to finish, consider the case D=SA, then ¯EF∥¯BC (since ¯SA passes through the arc midpoint of BAC) and H is the reflection of XA over I; thus indeed ASA and AH are isogonal due to the well known facts that AH passes through YA, the tangency point of the A-excircle with ¯BC, and the fact that ATA and AYA are isogonal. ◻. Similarly we show that A,B′,C′,D are concyclic. Further, let K:=BC∩B′C′ and let D′ and A′ be the second intersections of lines AH and DK, respectively. Then, we have that D′ is the reflection of D over the perpendicualr bisector of BC and similarly A′ is the reflection of A over the perpendicular bisector of EF. Now let OA and OD be the centers of ΩA and ΩD, respectively. We then claim the following Claim 2: Lines AOA,DOD and the perpendicuar bisector of AD concurr. Proof: Our claim is equivalent to show that ∡(AD,AOA)=∡(DOD,AD)Fist note that since AD and AH are isogonal conjugate with respect to ∠BAC and AOA is isogonal to the perpendicular from A to EF with respect to ∠BAC, we have that ∡(AD,AOA)=90+∡(EF,AH)=90+∡=90+∡BAF+∡D′AB+∡ABESimilarly, ∡(DOD,AD)=90+∡CBD+∡CDA′+∡A′DATherefore it suffies to show ∡BAF+∡D′AB+∡ABE=∡CBD+∡BDF+∡FDA′which is obvious by the isogonal definition of A′ and D′. ◻ Now let R be the intersection of AOA,DOD and the perpendicular bisector of AD. Considering the homothety with center A sending △A to △AB′C′, we must have that the center of (AB′C′) must lie on AOA and at the perpendicular bisector of AD, (by the homothety and by Claim 1) therefore, R is the center of (AB′C′D). Analogously, we have that R is the center of (AE′F′D) and then the six points A,D,B′,C′,E′,F′ must lie on a circle centered at R. Considering the homothetys again we can state the following: Lemma wrote: Let IA and ID are the centers of ωA and ωD. We have that (i) The points A,D,B′,C′,E′,F′ lie on a circle ΓA which is tangent to ωA at A and to ωD at D. (ii) The lines OAIA,RI and ODID are pairwise parallel (equivalently, they concurr at the line at infinity). Now we are able to establish part (a); apply Monge's Theorem to (ΩA,ΩD,ΓA) and (ωA,ωD,ω) to conclude that the extimilicenters of (ΩA,ΩD) and (ωA,ωD) both lies on AD. Then, it suffies to show that lines OAOD,IAID and AD concurr; indeed, this follows from part (ii) of the Lemma applying Desargue's Theorem to △AOAIA and △DODID. Now we have that TA lies on AD; we need to identify a second line. Let SD be the tangency point of the D-mixtilinear incircle of △DEF. We then have the following: Claim 3: TA lies on SASD. Proof: The key idea is to notice that the line E′F′ is the line passing through the intersections of Ω with the tangents from SA to ω; then we can find nice relations by swaping A↔SA and D↔SD and apply the Lemma. In which follows, dentoe by △(P,Q) the corresponding of ΩA by swapping A↔P and D↔Q for any points P,Q∈Ω and further let Ω(P,Q) be the circumcircle of △P,Q (note that this is posible by Poncelet's Porism; for intance we have △(D,A)=△A.) Now consider the configuration generated by D↔SD; then it follows by applying the part (i) of the Lemma that Ω(SD,SA)=Ω(A,D); similarly, we have that Ω(SA,SD)=Ω(D,A). Then from applying the part (a) of the problem to the configuration (A,D)↔(SD,SA) we have that the extimilincenter of (Ω(SA,SD),Ω(SD,SA)) lies on SASD; but by we have just shown that (Ω(SA,SD),Ω(SD,SA))=(ΩD,ΩA).◻. Then we finally have that TA=AD∩SASD (the case D=SA is done by continuity). To finish, let P:=ATA∩DTD; therefore TA lies on the polar of P with respect to Ω. Now notice that P is a point not depending at A or D (i.e. it can be determined uniquely by ω and Ω). Indeed, by applying Monge's Theorem to the incircle, circumcircle and mixtilinear incircle, P is the extimilincenter of ω and Ω. Another way to see that P is fixed, is to notice that the map A⇒SA is projective and furthermore an involution (since SA can be defined as the point such that the second intersection of AI and SAI cut Ω in diametrically oppositive points), and therefore ASA must pass through a fixed point.
17.05.2023 23:04
Cant believe this geo was once something i called impossible and now here i'm :O, i'm pretty sure this is one of the best geos i've ever seen in my whole life so far (May 17th of 2023), congrats to Nikolai Beluhov for making such a good problem . Let B′,C′ points in AB,AC respectivily such that B′C′∥EF and B′C′ tangent to ω, let E′F′ points in DE,DF respectivily such that E′F′∥BC and E′F′ tangent to ω, now let DE hit BC,AC at D2,E2 respectivily and let DF hit AB,BC at F2,D1 respectivily and finally let EF hit AB,AC at F1,E1 respectivily. Now let ΩA∩Ω=A′ and ΩD∩Ω=D′. Claim 1: (ΩA∩ωA)∼(ΩD∩ωD) (this implies part a) Proof: Let D″ the miquel point of complete quad \{B'C',BC, AB, AC \}, now clearly \angle BD''B'=\angle (BC, B'C')=\angle CD''C' so D''B, D''C' is isogonal w.r.t. \angle B'D''C, now by DDIT over that complete quad and inconic \omega we get that D''I bisects \angle B'D''C, \angle BD''C', let \ell_{D''} the tangent from D'' to AB'D''C', we have \angle (\ell_{D''}, D''B')=\angle B'C'D''=\angle BCD''= \angle CD''\infty_{E'F'} so D''I bisects \angle E'DF' as well this means that D''=D so AB'DC' is cyclic, now clearly (AB'DC') is tangent to \Omega_A but we can also show it is tangent to \Omega_D, notice that this is equivalent to \angle D_1D_2D-\angle B'DC'=\angle D_1DB' but this holds as \angle D_1D_2D-\angle B'DC'=\angle BD_2D-\angle BDC=\angle EDC=\angle D_1DB' thus (AB'DC') is tangent to \Omega_D as well, now notice that by the same process u get that AE'DF' is cyclic and (AE'DF') tangent to both \Omega_A and \Omega_D so they are basically the same circle, hence AF'B'DC'E' is cyclic and (AF'B'DC'E') tangent to \Omega_A and \Omega_D so now to finish we have (\Omega_A \cap \omega_A) \sim ((AF'B'DC'E') \cap \omega) \sim (\Omega_D \cap \omega_D) thus we are done. Claim 2: T_A, T_B, T_C all lie in the polar of X_{57} w.r.t. \Omega (implies part b) Proof: Notice that here \triangle ABC and \triangle DEF share same X_{57}, now by Taiwan TST 2014 R3 P3 we have that D' is the A-mixtilinear intouch point of \triangle ABC and A' is the D-mixtilinear intouch point of \triangle DEF, by monge over \Omega_A, \Omega_D, (AF'B'DC'E') we have that AD passes through T_A so make a \sqrt{T_AA \cdot T_AD} inversion with center T_A, this sends \Omega_A \to \Omega_D but since ADD'A' is cyclic we have that A' \to D' so A'D' \cap AD=T_A, now by monge its known that AD' \cap DA'=X_{57} so by Brokard over \Omega we have that T_A lies in the polar of X_{57}, but now u can do the same for all T_B, T_C so T_A,T_B,T_C make a line which is the polar of X_{57} w.r.t. \Omega thus we are done . Wow, what a journey, waiting for more problems like this coming soon .
18.05.2023 15:11
Denote with O_\bullet and I_\bullet the centers of \Omega_\bullet and \omega_\bullet, respectively. We first prove the following problem: Lemma wrote: Let \triangle ABC have incircle \omega and let P be some point on the circumcircle \Omega. If \triangle PQR is the triangle which has incircle \omega and QR\parallel BC prove that APQR is cyclic.
Now we begin with some Claims. Claim 1: A and D are antihomologous points w.r.t. \Omega_A and \Omega_D. Proof. Let AD\cap\Omega_A=\left\{A,D'\right\}, it then suffices \angle AD'O_A=\angle ADO_D \Longleftrightarrow \angle EDA-\angle EDO_D=\angle CAD-\angle CAO_A\Longleftrightarrow \angle EDA-\angle CAD=\angle ABC-\angle DFEwhich is seen to be correct. Claim 2: \left\{\Omega_A\cup\omega_A\right\}\sim\left\{\Omega_D\cup\omega_D\right\} which implies a). Proof. Taking the homothety centered at A, \chi:\omega_A\rightarrow\omega we get from the lemma that D\in\chi(\Omega_A) and hence \chi(D')=D\Longrightarrow D'I_A\parallel DI_D and similarly A'I_D\parallel AI_A where A' is analogously defined. Let T_A be the homothetic center of \triangle AD'I_A\sim\triangle A'DI_D and we see that T_A is the exsimilicenter of \Omega_A and \Omega_D and by Monge, the exsimilicenter of \omega_A and \omega_D is on AD hence is T_A. From the Taiwan TST 2014 problem we know that S_A, the A-mixtilinear intouch point of \triangle ABC, is on \Omega_D and similarly S_D\in\Omega_A. Claim 3: S_A and S_D are antihomologous w.r.t. T_A. Proof. This is true because of Claim 1 and the fact that pairs of antihomologous points form cyclic quadrilaterals (here AS_DS_AD is cyclic). And for the final blow: T_A is on the polar of the exsimilicenter X of \Omega and \omega w.r.t. \Omega. This is due Monge X=AS_A\cap DS_D and by Claims 1 and 3, T_A=S_AS_D\cap AD hence Brokard's theorem on AS_DS_AD finishes.