Let ABC be a triangle such that AB<AC. Let the excircle opposite to A be tangent to the lines AB,AC, and BC at points D,E, and F, respectively, and let J be its centre. Let P be a point on the side BC. The circumcircles of the triangles BDP and CEP intersect for the second time at Q. Let R be the foot of the perpendicular from A to the line FJ. Prove that the points P,Q, and R are collinear. (The excircle of a triangle ABC opposite to A is the circle that is tangent to the line segment BC, to the ray AB beyond B, and to the ray AC beyond C.) Proposed by Bozhidar Dimitrov, Bulgaria
Problem
Source: JBMO 2024
Tags: geometry, circumcircle, collinear, Balkan, JBMO
27.06.2024 14:35
27.06.2024 15:06
Lukaluce wrote: Let ABC be a triangle such that AB<AC. Let the excircle opposite to A be tangent to the lines AB,AC, and BC at points D,E, and F, respectively, and let J be its centre. Let P be a point on the side BC. The circumcircles of the triangles BDP and CEP intersect for the second time at Q. Let R be the foot of the perpendicular from A to the line FJ. Prove that the points P,Q, and R are collinear. (The excircle of a triangle ABC opposite to A is the circle that is tangent to the line segment BC, to the ray AB beyond B, and to the ray AC beyond C.) angleDQE=∠DQP+∠PQE=∠ABC+∠ACB=180−∠DAE so D,Q,E,A are concyclic. More over ∠ADIA=∠ARIA=∠AEIA=90 so D,A,R,E,IA,Q are concyclic. No w we have that: ∠DQP=∠ABC=∠DIAR=∠DQR So Q,P,R are collinear
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27.06.2024 15:11
Following last year, we have another high class JBMO Geometry problem by Bozhidar Dimitrov (Strudan_Borisov) from Bulgaria; or as several leaders said: bomba. The original submission also has a harder version, which hides the point R and requires to show that PQ passes through a fixed point as P varies - very beautiful, but of course not suitable for juniors. I only think it should have been P1, but oh well.
27.06.2024 15:17
From the cyclic quadrilaterals, we get that: ∠DQE=∠DQP+∠EQP=∠ABC+∠ACB=180−∠BAC=180−DAE, hence Q lies on (ADE). Also, ∠ARJ=∠AEJ=∠ADJ=90, so A,R,E,J,D are concyclic in a circle with diameter AJ and hence A,R,E,J,Q,D are all concyclic. Now we have that: AR⊥FJ=>AR∥BC. From the cyclic quadrilaterals and using that AR∥BC: ∠RQE=∠RAE=∠RAC=∠ACB=180−∠PCE=∠PQE and P,Q,R are collinear, as needed. Note that the same holds for any point J, not just the excenter, where D,E,F are the projections of J in the sides of the triangle.
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27.06.2024 15:29
Just using angle chasing.... ∠QDB=∠QPC=180−∠CEQ so, DAEQ is cyclic. Also,JDAE cyclic ⟹ JEADQ is cyclic.Also we know AREJ cyclic.So,by written some angles we deduce RQ∩BC=P.
27.06.2024 15:50
Kudos to Bozhidar Dimitrov for the cute problem! Here's a quick solution: Introduce Q′ as the second intersection of ¯RP and the cyclic ARDJE. Then ∠DQP=∠RAD=β=∠ABPhence BPQD is cyclic. Similarly, CPQE is cyclic as well, so Q=Q′, as desired.
27.06.2024 16:07
Notice that ARDJE is cyclic. Define Q′ as the second intersection of PR and (ARDJE). Then ∠PQ′D=180∘−∠BAR=∠ABC∠PQ′E=∠RAC=∠ACBThus BPQ′D and CPQ′E are concyclic, so Q=Q′, finishing the problem.
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27.06.2024 16:15
Since, ∠ARJ=∠JDA=∠AEJ=90A, R, E, J and D lie on a circle. Now, JF⊥ both BC and AR⟹BC∥AR ⟹∠RAB+∠ABC=180=∠RAB+∠PQD ⟹Q∈(AERJD) ⟹∠FPR=∠PRA=180−∠ADQ=∠BPQ ⟹P, Q and R lie on a single line.
27.06.2024 16:26
Surprisingly very easy. I think the original submission (showing PQ passes through a fixed point) wouldn't have been too hard for the Problem 2 spot. We first prove the following key claim. Claim : Points A , D , E , J , R and Q lie on the same circle. Proof : It is immediate that ADJE is cyclic, and that R lies on this circle as well, due to the right angles. Further, ∡DQE=∡DQP+∡PQE=∡ABC+∡BCA=∡BAC=∡DAEso Q also lies on this circle, finishing the proof of the claim. Now, let P′=¯QR∩¯BC. We simply note that since AR∥BC (since JF⊥BC quite clearly), we have ∡DQP′=∡DQR=∡DAR=∡CBA=∡P′BAso P′ lies on (DBQ). Thus, P′=P and it immediately follows that points P,Q, and R are collinear, as desired.
27.06.2024 18:41
Something more general holds: if E and F are fixed points on AB and AC respectively and D varies on BC, then the radical axis of (BDE) and (CDF) passes through a fixed point. For simplicity, assume D,E and F lie on the sides of △ABC (the same argument works in general). Let X be the point on (AEF) such that AX∥BC, which is clearly fixed. We claim that X is the desired point. If (BDE) and (CDF) meet a second time at K, then K lies on (AEF) by Miquel's theorem. The collinearity follows because ∠XKF=∠XAF=∠ACB=180∘−∠FKD. In fact, an even further generalization can be found in post #9 here.
27.06.2024 19:46
27.06.2024 20:35
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28.06.2024 00:04
Claim : A,R,E,J,Q,D are concyclic Proof : Clearly, the points R,E,D lie on the circle with AJ as diameter since they subtend 90∘ angle at AJ. Now, since BPQD and CPQE are cyclic, ∠DQP=∠B and ∠EQP=∠C ⟹∠DQE=180∘−∠A Let P′ be the intersection of QR with BC Claim : P=P′ Proof : ∠BPQ=180∘−∠BDQ=∠ARQ=∠ARP′=∠CP′R=∠BP′Q This means that P and P′ lie on the same line and subtend the same angle at BQ⟹∠PQP′=0⟹∠P=P′. And we are done.
06.07.2024 16:01
I solved it in 2 minutes it's very easy for jbmo
09.07.2024 23:43
By Miquel Theorem Q lies on (ADJE). Because it's diameter is AJ, R also lies on it. Now ∠DQP=∠ABC and ∠DQR=∠DJR=∠DJF=∠BAC, so P, Q and R are collinear
11.07.2024 10:00
trigadd123 wrote: Something more general holds: if E and F are fixed points on AB and AC respectively and D varies on BC, then the radical axis of (BDE) and (CDF) passes through a fixed point. Which, we just now realized, appears at the generalization of ISL 2018 G2 by Daniel Zhu (in post 8).
05.08.2024 00:53
JDAER is clearly cyclic and moreover AR∥BC as they are both perpendicular to RJ Claim: Q lies on (JDAER) Proof: ∠ADQ=∠BDQ=∠QPC=180∘−∠CEQ=180∘−∠AEQ⟹DAEQ cyclic ⟹JDAERQ cyclic. ◼ From here we can finish with 2 ways 1. ∠DQP=∠ABPAR∥BC=180∘−∠BAR=180∘−∠DAR=∠DQR.◼ 2. ∠EQP=∠ACB=∠CAR=∠EAR=∠EQR◼
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