Let ABC be an inscribed triangle in circle (O). Let D be the intersection of the two tangent lines of (O) at B and C. The circle passing through A and tangent to BC at B intersects the median passing A of the triangle ABC at G. Lines BG,CG intersect CD,BD at E,F respectively. a) The line passing through the midpoint of BE and CF cuts BF,CE at M,N respectively. Prove that the points A,D,M,N belong to the same circle. b) Let AD,AG intersect the circumcircle of the triangles DBC,GBC at H,K respectively. The perpendicular bisectors of HK,HE, and HF cut BC,CA, and AB at R,P, and Q respectively. Prove that the points R,P, and Q are collinear.
Problem
Source: VMO 2021 P7 Vietnam National Olympiad
Tags: geometry, Concyclic, collinear, circumcircle
28.12.2020 21:16
I only had solved the first part of the problem: ∠BAC=∠CBD=∠BCD=∠BAG+∠GAC Let M, X and Y the midpoints of BC, BE and CF, respectively. BC is tangent to the circumcircle of ABG at B, and by power of a point: MB2=MG(MA)=MC2 BC is tangent to the circumcircle ACG at C. The reflection G′ of G relative to M lies on the circumcircle of ABC (because MB2=MG(MA)=G′M(MA)=BM(BC)). Aditionally, BGCG′ is a parallelogram (BC and GG′ intersect at M, the midpoint of these segments). All the properties implies: ∠CAG=∠BAC−∠BAG′=∠BCD−BCG′=∠GEC∠BAG=∠BAC−∠CAG′=∠CBD−CBG′=∠GFB ∠AEC=∠ACB,∠ACE=∠ABC,∠CAE=∠BAC ∠AFB=∠ABC,∠ABF=∠ACB,∠BAF=∠BAC So, we need to prove that ∠MAN=180∘−∠BDC=2∠BAC . If we prove that FMMB=CNNE, AFM∼ACN and AMB∼ANE, in other words: ∠MAB+∠CAN+∠ABC=2∠BAC. Finally, by Menelaus's theorem: FMMB⋅BXXG⋅GYYF=1⟹FMMB=XGBX⋅YFGYCNNE⋅EXXG⋅GYYC=1⟹CNNE=XGBX⋅YFGY I need help to solve the second part of the problem. I found that }H is the intersection of the reflections of GA, GB and GC to the bisectors of the angles BAC, ABC and ACB, respectively; ABKC is a parallelogram.
30.12.2020 09:30
a. Define T, X and Y as the midpoint of BC, BE and CF respectively. Define ΩB as a circle through A which tangent to BC at B. Define ΩC similarly. It's well known that ΩB and ΩC pass through G. Notice that ∠AGE=∠BAG+∠GBA=∠GBM+∠GBA=∠ABC=∠ECAHence, we must have E∈ΩC. Similarly, F∈ΩC. Therefore, by spiral similarity, we have △ABE∼△AFC. As X is midpoint of BE and Y is midpoint of CF, then we have △ABX∼△AFY. As △ABX∼△AFY, then we have AYGX and AXBM are both cyclic (by spiral similarity). Analogously, we can check that AYCN is cyclic. Therefore, ∡AMD=∡AXG=∡AYG=∡ANDSo, AMDN is cyclic. b. It is easy to check that H is the isogonal conjugate of G wrt △ABC. (G is the A-humpty point while H is the A-dumpty point). We will prove a generalization of this problem. Generalization: Let ABC be a triangle and P be a point on the plane. Define Q as the isogonal conjugate of P wrt △ABC. Let XA as the second intersection of AP and ⊙(BPC). We define XB and XC similarly. Let YA as the intersection of the perpendicular bisector of XBXC and BC. Define YB and YC similarly. Prove that YA,YB,YC are collinear. Proof: Define ZA,ZB,ZC as the circumcenter of △QXBXC,△QXCXA and △QXAXB respectively. By Desargues' it suffices to show that AZA, BZB and CZC are concurrent. Define U as the circumcenter of △XAXBXC. Claim: Isogonal conjugate of U wrt △ABC lies on AZA Proof: Do inversion with center A and radius √AB×AC continued by reflection across internal angle bisector of ∠BAC. It's easy to check that this process send XB to XC and Q to XA (Both converse are also true). Suppose this process send ZA to W. As ZA is the circumcenter of QXBXC, then W lies on the line through A and the circumcenter of XAXBXC which is U. So, U lies on AW. Furthermore, AW is isogonal to AZA wrt ∠ABC. So, the isogonal conjugate of U wrt △ABC lies on AZA. (Claim proven) By the claim, then we have AZA,BZB,CZC will concur at the isogonal conjugate of U which then finish the problem.
18.05.2021 15:40
We have: Lemma: Given points X,X′,Y,Y′ such that X and X′, Y and Y′ are images of one another by √bc inversion (inversion and reflection about angle bisector), we have that the centers of (XYY′) and (X′YY′) are isogonal wrt. ∠BAC. Proof: Obviously the two circles are images of one another by √bc inversion. Let ω be the image of (XYY′) by inversion around A with radius √AB⋅AC. We obviously have that (ABC),ω, and (XYY′) are coaxial, so their centers are collinear, and since ω reflects about the bisector of ∠BAC to (X′YY′), we are done. Using some of what @above said for isogonal conjugates and inversion in the configuration, this immediately implies part b) by Jacobi.