Let ABC be a scalene triangle with circle Γ. Let P,Q,R,S distinct points on the BC side, in that order, such that ∠BAP=∠CAS and ∠BAQ=∠CAR. Let U,V,W,Z be the intersections, distinct from A, of the AP,AQ,AR and AS with Γ, respectively. Let X=UQ∩SW, Y=PV∩ZR, T=UR∩VS and K=PW∩ZQ. Suppose that the points M and N are well determined, such that M=KX∩TY and N=TX∩KY. Show that M,N,A are collinear.
Problem
Source: Cono Sur 2021 #6
Tags: geometry
01.12.2021 03:31
Applying Multiple converse of pascal and 4 conics theorem and 3 conics theorem should give that X,Y,T,K lie on (ABC) i believe. @below I wish i wasnt confused about that theorem during the test, at least i would've got 2 points for my proof using conics...
01.12.2021 03:46
@above you are right
01.12.2021 08:27
√bc inversion + involution + pascals
01.12.2021 09:36
a solution by cadaeibf posted yesterday , in a deleted afterwards post Quote: We have ∠UAQ=∠WAS. By √bc-inversion+reflection, we have AUAQ=AQ∗AU∗=AWAS, implying AUQ∼AWS and so ∠AUQ=∠AWS, which implies that UQ and WS concur on Γ (they form the same arc). Similarly, X,Y,K,T lie on Γ. This problem was also 2020 Cono Sur Shortlist G5
01.12.2021 18:46
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01.12.2021 19:23
Solved with Mathslion. The converse of Pascal makes a lot of sense here, so apply it on AVWXUZ (since UZ||PQ||VW so they intersect at infinity) and we get X lies on the circumcircle. By another application on AVWKZU we get K lies on the circle as well. Similarly we get X,Y,K,T lie on Γ. Now by Reim's (↔ just angles) we get the following cyclic quadrilaterals: XYQR, XYPS, KPST,KTRQ. Complete the quadrilateral XYKT by adding XY∩KT=D. It's enough to have that A lies on the polar of D (the line MN), this would mean that DA is tangent to Γ. This follows by looking at the radical center of the circles: APS,KPST,XYPS,ABC (notice that the radical of APS and ABC is the tangent at A)
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01.12.2021 20:46
By the angles, we have that (B,C;Q,S)=(B,C;P,R)A=(B,C;U,W)⟹X∈Γ, since X=UQ∩SW. Similarly, Y, K, Y∈Γ. Let D be the intersection point of the tangent to Γ by A and BC. We will prove that D, X, Y are collinear (Similarly, D, T, K are collinear): By inversion √bc + reflection, we have that X′ and Y′ lies on BC, and D′ lies on Γ such that BC∥AD′, so, D, X, Y are collinear ⟺ D′, X′, Y′, A are cyclic ⟺ X′C′=Y′B′, by the isosceles trapezoids, ⟺XCXA⋅c=YBYA⋅b. But, (B,A;Y,C)V=(B,Q;P,C)U=(B,X;A,C)⟹YBBC÷YAAC=BABC÷XAXC⟹XCXA⋅c=YBYA⋅b. Now, by Brocard's Theorem in XYKT, we have that MN is the polar of D, but, D lies in the polar of A, so, A lies in the polar of D, so, M, N, A are collinear.
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19.02.2022 03:26
MathLuis wrote: Applying Multiple converse of pascal and 4 conics theorem and 3 conics theorem should give that X,Y,T,K lie on (ABC) i believe. @below I wish i wasnt confused about that theorem during the test, at least i would've got 2 points for my proof using conics... I want revenge of this problem... After learning √bc inversion this became trivial (sadly i did not think abt it on the test :c) Continuing my progress we do a √bc inversion, since by Reim's we have AKPQ,ARST,AYPR,AXQS cyclic, we get that K is send to BC∩WZ=K′, T is send to BC∩UV=T′, Y is send to BC∩VZ=Y′ and X is send to BC∩UW=X′. Now let T the interesction of the tangent from A to Γ and let A′ be a point on Γ such that AA′∥BC. Then T is send to A′, now since by symetry w.r.t. the bisector line of BC we have that AA′K′T′,AA′X′Y′ isosceles trapezoids and this means that KT,XY,BC are concurrent at T and by duality we are done
07.03.2022 04:24
MathLuis wrote: Applying Multiple converse of pascal and 4 conics theorem and 3 conics theorem should give that X,Y,T,K lie on (ABC) i believe. @below I wish i wasnt confused about that theorem during the test, at least i would've got 2 points for my proof using conics... to prove this, it is enough to note a simple involution from which it turns out that {R,C;S,B}={U,C;V,B}⇒T∈Γ. Could you tell me what the theorem you mention is about (or failing that, provide a link to that topic)? i'm not familiar with it
08.06.2024 23:47
almost trivial with projective Note that (B,P;Q,C)=(B,R;S,C)=(B,W;Z,C)Γ implies that K∈Γ, and similarly T,X,Y∈Γ. By Brocard it suffices to show that the tangent ℓ to Γ through A passes through XY∩KT, but this is true because (B,Y;A,C)Γ=(B,P;Q,C)=(B,R;S,C)=(B,A;X,C)Γ=(C,X;A,B)Γso XY passes through ℓ∩BC, and similarly so does KT.
19.07.2024 17:52
YLG_123 wrote: By the angles, we have that (B,C;Q,S)=(B,C;P,R)A=(B,C;U,W)⟹X∈Γ, since X=UQ∩SW. Similarly, Y, K, Y∈Γ. Let D be the intersection point of the tangent to Γ by A and BC. We will prove that D, X, Y are collinear (Similarly, D, T, K are collinear): By inversion √bc + reflection, we have that X′ and Y′ lies on BC, and D′ lies on Γ such that BC∥AD′, so, D, X, Y are collinear ⟺ D′, X′, Y′, A are cyclic ⟺ X′C′=Y′B′, by the isosceles trapezoids, ⟺XCXA⋅c=YBYA⋅b. But, (B,A;Y,C)V=(B,Q;P,C)U=(B,X;A,C)⟹YBBC÷YAAC=BABC÷XAXC⟹XCXA⋅c=YBYA⋅b. Now, by Brocard's Theorem in XYKT, we have that MN is the polar of D, but, D lies in the polar of A, so, A lies in the polar of D, so, M, N, A are collinear. Really nice solution! I proved that X∈Γ with something of a spiral similarity and was trying to prove that X,Y and the point you named D are collinear, but I was struggling, thaks for the help.