In the triangle ABC, variable points D,E,F are on the sides[lines] BC,CA,AB respectively so the triangle DFE is similar to the triangle ABC in this order. Circumcircles of BDF and CDE intersect respectively the circumcircle of ABC at P and Q for the second time. Prove that the circumcircle of DPQ passes through a fixed point.
Problem
Source: Iran MO Third Round 2022 Mid-Terms P2
Tags: geometry, circumcircle, Iran, Miquel point, geometry proposed
16.08.2022 22:12
Very interesting problem...Any ideas?
17.08.2022 11:00
Let the tangent of ⊙(ABC) at A intersect BC at G.Let FE∩AG=H,FD∩AC=I,ED∩AB=J.We are going to prove that ⊙(DPQ) pass though the fixed point G. Claim:A,H,J,Q,E are concyclic;A,F,P,H,I are concyclic. By angle chasing:∠FED=∠ACB=∠HAB.So A,H,J,E are concyclic. ∠AQE=∠AQC−∠EQC=∠ABC−∠EDC=∠AJE.So A,J,Q,E are concyclic. So A,H,J,Q,E are concyclic.Likewise,A,F,P,H,I are concyclic. Notice that:∠QDC=∠QEC=∠QHA.So G,H,D,Q are concyclic. Likewise,G,H,D,P are concyclic. So we can get:G,P,D,Q,H are concyclic,which implies that G∈⊙(DPQ).Done.◻
17.08.2022 20:14
Lemma 1. The secend intersection of (BDF),(CED) is a fix point. Proof.call that point X. Note that ∠FXD=180−∠B and ∠EXD=180−∠B So ∠EXF=180−∠A . So AEXF is cyclic. We have : ∠BXC=∠A+∠ABX+∠ACX=∠A+∠FDX+∠EDX=2∠ASimilarly ∠AXB=∠B+∠C=∠AXC So X is a fix point. Also , its well-known that such X , is in fact the midpoint of A-symmedian chord in (ABC). Now we shall prove that the desired point is T , the intersection of the tangent through A in (ABC) with BC. Let O be the center of (ABC). consider an inversion through T withbradious TA. since X was the midpoint of symmedian chord and the other tangent through T to (O) is the intersection of A-symmedian with (O) , X goes to O . B,C swaps and D goes to another point D′ on BC. So P,Q go to (D′OC)∩(O) and (BD′O)∩(O) respectively and we shall prove that in this case D′,P′,Q′ are collinear. Now apply another inversion , this time with center O and radius OA. So D′ goes to some point on the circle.(BOC) and P′,Q′ would be D″ and D''C \cap (O) and we should prove that P'',Q'',D'',O are concyclic. This is obvious since OD" is the angle bisector of BD"C and so PD"Q where we also have OP''=OQ'' and P''D''=Q''D'' only occures when BD''=CD'' which is a trivial case and we're done
24.08.2022 22:08
Let O be the circumcenter of ABC, T = AA \cap BC, the projection of A onto OT be X, the midpoints of AB and AC be M and N respectively, and the projection of A onto BC be L. Since X is the A-Dumpty point, it's well-known that BXOC is cyclic and (AMN), (BLM), (CNL) concur at X. Now, observe LMN \overset{-}{\sim} AMN \overset{+}{\sim} ABC \overset{-}{\sim} DFEso LMN \overset{+}{\sim} DFE. Hence, the Spiral Center Lemma implies (AMN), (BLM), (CNL), (AEF), (BDF), and (CDE) concur. It follows that X is the second intersection point between (BDF) and (CDE). By Radical Axes, we know BP, CQ, DX concur at some point R. Now, let D^* denote the inverse of D wrt (ABC). It's easy to see D^* \in (BXOC). Moreover, because X and T are clearly inverses wrt (ABC) as well, we know (BPTD^*) and (CQD^*T) are cyclic. Thus, applying Radical Axes again yields R \in D^*T. Now, consider the inversion about R with radius \sqrt{RB \cdot RP}. By POP, we have B \leftrightarrow P, C \leftrightarrow Q, D \leftrightarrow X, and D^* \leftrightarrow T. Since BXCD^* is cyclic, it follows that PDQT is cyclic, which suffices. \blacksquare Remark: I determined that T is the fixed point by considering the case where AD \perp BC and performing \sqrt{bc}-inversion.
31.12.2022 16:01
Take foot of perpendicular from A to BC be D', midpoint of AB and BC be F' and E',respectively. Let T be the A-Dumpty point. Take O as the circumcenter of (ABC) Notice that BF'D'T, CE'D'T and AE'F'T are cyclic and D'E'F' is similar to DEF. Note that the spiral homothety center that sends DEF to D'E'F' must be on circle BF'D'T and CE'D'T so it must be T. Also converse of this is also true: if you take any D on BC, E and F are defined as the intersections of (TDC) and (TDB) with AC and AB Rewrite the problem as follow: T is the A-Dumpty point of ABC. D is taken on BC and (TDB) and (TDC) intersect (ABC) at P and Q. Show that (DPQ) passes through a fixed point. If you take inversion with center T problem turns to this: D is taken on (BOC), BD and CD intersect circumcircle at P and Q. Show that (DPQ) passes through a fixed point. Indeed, that fixed point is O in the inverted picture. In the original picture, it is OT\cap BC or it can be defined as the intersection of tangent from A with BC.
04.01.2023 19:49
06.01.2023 10:53
Suppose that S is Miquel point of D, E, F with respect to \triangle ABC. Since \triangle DFE \sim \triangle ABC, we have \angle{DFE} = \angle{ABC}. But \angle{DBS} = \angle{DFS} then \angle{SBF} = \angle{SFE} = \angle{SAE}, which means AC touches (ABS). Similarly, AB touches (ACS). So S is A - Dumpty point of \triangle ABC. Let S' be inversion of S through (ABC), O be circumcenter of \triangle ABC. We have \angle{SFE} = \angle{SAE} = \angle{SBA} = \angle{SDF}Similarly, \angle{SEF} = \angle{SDE}. Hence EF is common tangent of (DSF), (DSE). So we have \angle{PS'Q} = \angle{PS'O} + \angle{QS'O} = \angle{OPS} + \angle{OQS} = \angle{POQ} - \angle{PSQ}= \angle{POQ} - \angle{PSD} - \angle{DSQ} = \angle{POQ} - \angle{PFE} - \angle{FEQ} + \angle{EFD} + \angle{DEF}= \angle{POQ} + \angle{PDF} + \angle{QDE} - 180^{\circ} - \angle{EDF} = \angle{POQ} + \angle{ACP} + \angle{ACQ} - 180^{\circ} - \angle{EDF}= 360^{\circ} - \angle{ACP} - \angle{ACQ} - 180^{\circ} - \angle{EDF} = 180^{\circ} - \angle{PDF} - \angle{PDE} - \angle{EDF} = 180^{\circ} - \angle{PDQ}or P, Q, D, S' lie on a circle