Romanttics of Geometry, 8374

 

Peru Geometrico, Yusuf Zemzem, https://www.facebook.com/photo?fbid=10222549589361542&set=gm.4029198667192996

 

aops, 'collinear midpoints'

 (*) At https://stanfulger.blogspot.com/2021/07/own-tetrahedron.html I have proved the statement marked (*) in my proof here attached

Peru Geometrico, S.R. https://www.facebook.com/photo/?fbid=10222534642747886&set=gm.4022044794575050

 

Stanley Rabinowitz, Peru Geometrico, equilateral

 My non-trigo solution to this nice problem

aops, https://artofproblemsolving.com/community/c4t48f4h2628074_ltabc_ltacd105o_ltdcb30o_addb__2019_ecuador_tst_11

 

aops, https://artofproblemsolving.com/community/c6t48f6h2628662_od_____ab_wanted_2_circumrcircles

 

aops, https://artofproblemsolving.com/community/c4t48f4h2627419_formula_in_right_triangle

 

own tetrahedron

 General 3d problem, I hope you like it and post your solutions

Pr. 8616 Romantics of Geometry, S.R. https://www.facebook.com/photo?fbid=10223742689747225&set=gm.4185693751544299

 

Kyiv, https://artofproblemsolving.com/community/c6h2620399p22645030

 Solution with my son (I made ... the drawing, he the proof, so it's even!)

Angles puzzle, https://www.facebook.com/photo?fbid=10219330441082995&set=gm.791695651498757

 My solution to this old and nice problem

Trapezoid, https://artofproblemsolving.com/community/c4t48f4h2619384_geometry

 

Problem, Geometria Top, Jaihind Singh, https://www.facebook.com/photo?fbid=4183912448395804&set=gm.1008177553285947

 

A good geometry problem, aops https://artofproblemsolving.com/community/u36948h2620907p22664372

 

aops https://artofproblemsolving.com/community/c4t48f4h2619461_quadrilaterals_and_angles

 My solution

Cono sur 2011 https://artofproblemsolving.com/community/c6t48f6h603538_cono_sur_olympiad_2011_problem_5

 

aops, Kyiv https://artofproblemsolving.com/community/q1h2619778p22638964

 Corrected text and solution!

Kedlaya, https://artofproblemsolving.com/community/c6t48f6h24483_abcd_cyclic_gt_incenters_of_bcd_etc_form_rectangle

 

aops, Equilateral, https://artofproblemsolving.com/community/c4h2615750p22599678

 Let D, E points onto the sides BC, AB of the equilateral triangle ABC, and F onto (BA so that CD=BE=AF, and K midpoint of DE. Find <CKF.

Proof:
Construct the equilateral triangle BEG, outside ABC, DCEG is a parallelogram, see that triangles FEG and CAF are congruent, s.a.s, thus FG=FC, and the median FK of the isosceles triangle CFG is perpendicular bisector as well.

Problem, Romantics of Geometry, S. Rabinowitz, https://www.facebook.com/photo?fbid=10223251625550927&set=gm.3977192989061044

 My solution

aops, equilateral triangle, https://artofproblemsolving.com/community/c4t48f4h2615750_a_geometry_problem_involving_triangles

 D, E are points onto the sides BC, BA of the equilateral triangle ABC so that BE=CD, F is a point onto (BA so that AF=BE and K is the midpoint of DE. Find <CKF.

Proof
Take G reflection of C about K; triangle BEG is equilateral. See that triangles CAF and FEG are congruent, s.a.s., thus GF=CF and FK is perpendicular bisector of CG, i.e. <CKF=90.

Geometria Top, Eslam Alenani, https://www.facebook.com/photo?fbid=3047188738897926&set=gm.1003076040462765

 In the interior of the square ABCD there is a point P so that <PAD=x, <CDP=45+x. Find ratio DP/AP.

Proof
Clearly <ADP=45-x, so <ADP+<PAD=45, that is, DP passes midpoint E of the outer arc AB of the circle (AB). If F is midpoint of CD, tan(<DEF)=DF/EF=1/3, i.e tan(<DAP)=1/2, that is, A-P-F are collinear. Notice that triangles DAF and PBA are similar, i.e. PB=2AP. Finally notice that triangles DAP and BDP are similar, i.e. AP/PD=PD/PB, hence PD.PD=AP.PB=AP.2AP wherefrom PD=APsqrt(2).

aops, 'A great perpendicular problem', https://artofproblemsolving.com/community/c6t48f6h2611905_a_great_perpendicular_problem

 Reworded text:
Let CBVE be a cyclic kite, CV being a diameter of its circumcircle. Take F, G onto VE, BE so that FG||BV, call K the midpoint of BG. Prove that KF_|_CK.

Proof:
FG=FB; construct the parallelogram GCBL, we need to prove FL=FC. We have GL=BC=CE, FG||BV, GL||BC, so FG_|_GL. Triangles FGL and FEC are congruent, thus FL=FC, so the median FK is perpendicular bisector of CL as well, done.

Pr. 8537 Romantics of Geometry, Konstantin Knopp, https://www.facebook.com/photo?fbid=10222355364046501&set=gm.4147069368740071

 Let G be a point inside the cyclic hexagon ABCDEF so that <AGB=60, <EGC=90. If H is midpoint of BC, then G lies onto FH.

Proof:
If O is the center of the hexagon, G is intersection of the circles (AOB) and (CE). Noticing that AF is tangent to the circle (ABO), EF tangent to the circle (CE); as AF=EF, F belongs to the radical axis of the 2 circles. Noticing that BC is common tangent to the 2 circles, H belongs to their radical axis as well, and we are done. There are 2 internal points G, G', determining the radical axis FH of the 2 circles.

IMO 2014, P4 - Romantics of Geometry, 8511, https://www.facebook.com/photo?fbid=4201432579915126&set=gm.4136708653109476

 Let ABC be a triangle and P, Q points onto the side BC so that <BAP=<ACB, <CAQ=<ABC; on the lines AP, AQ take points M,N as symmetricals of A about P, Q respectively and call L interesection of BM with CN. Prove that ABLC is cyclic.

Proof. 
As constructed, both triangles ABP, CAQ are similar to CBA, and subsequently they are similar, AP=AQ and AP/CQ=BP/AQ ( 1 ). But AP=MP, QA=QN, so (1) becomes MP/CQ=BP/QN, which means triangles BPM and NQC are similar (<BPM=<CQN also), from their similarity getting <QCN=<BMP, i.e. BLNQ, CPML are cyclic, so <MLN=<BPM=<APQ=180-<BAC, done.

NOTE: As <AMN=<ANM=<MLN, AL is symmedian of triangle MLN, thus of triangle BLC as well and ABLC is harmonic quadrilateral.  

Pr. 8529, Romantics of Geometry, Konstantin Knopp, https://www.facebook.com/photo?fbid=10222350205397538&set=gm.4144235069023501

 Inside the regular hexagon ABCDEF there is a point G so that <DGE=60, <CGD=90. DG intersects AB at X. Find AX/XB.

Proof:
G is intersection of the circles (FOE) and (DE) of centers K, L respectively. Let M, N be midpoints of EF, OM; NL is midline of the trapezoid DEMO, that is, NL=3R/4 (R = AB). Easily KN = OM/6=R.sqrt (3)/12, consequently NL/KN=3.sqrt (3), but triangles KNL and XCE are similar, having perpendicular sides, so CE/CX=3.sqrt (3), but CE=R.sqrt (3), thus CX=R/3, hence AX/XB=2, done.

2018 Rusanovsky Lyc Olympiad, M. Plotnikov, https://artofproblemsolving.com/community/c4t48f4h2612208_midpoint_equidistant_from__lines__2018_rusanovsky_lyceum_olympiad_85

 Three parallel lines, l1, l2, l3 are intersected by another line l at points A, B, C respectively. Points E and D are taken onto l1, l3 so that BE=CB, BD=BA. Prove that midpoint K of the bisector BL of the angle DBE is equally apart of l1 and l3.

Proof:
Let l2 intersect DE at M, then EM/DM=AB/BC ( 1 ). Apply angle bisector theorem in tr. BDE, DL/EL=BD/BE=AB/BC ( 2 ), that is DL=EM, so N, midpoint of ML is midpoint of DE as well, so it lies at equal distance from l1 and l3. NK is midline in tr. BML, thus we are done.

Pr. 1027 Gogeometry, https://gogeometry.blogspot.com/2014/07/geometry-problem-1027-triangle-double.html

 Let D, E be points onto the sides BC, AB of triangle ABC so that <CAD=2<BAD=4x, AD=AC, DE=DC, find <ABC.

Proof:
If F is reflection of E about AD, then triangle CDF is equilateral, and easy angle calculation shows x=10, wherefrom <ABC=50. 

Greece test 2006, https://artofproblemsolving.com/community/c4t48f4h2610765_greece_tst_2006

 Let O be the circumcenter of triangle ABC, M, D midpoints of BC, CA, E projection of D onto AB. Prove that KE=KC.

Proof:
Let BK intersect the perpendicular at C to BC at the point F, then OMCF is a rectangle, i.e. cyclic; also OMCD is cyclic, having 2 opposite angles of 90 degs each. Consequently O, M, C, F, D are concyclic points and <FDC=<FMC=<OCM=90-A=<ADE, so F-D-E are collinear, BCFE is cyclic, its circumcenter being K, midpopint of OM, done.

Ercole Suppa, Peru Geometrico, https://www.facebook.com/photo?fbid=10222396617697346&set=gm.3954117278034469

 Let E be midpoint of the side AD of the square ABCD, O the circumcenter of triangle DEB. The circle (DEB) intersects the circle (E, EA) at D and G. Prove that 3BG=2OA.

My proof:
If P is the center of the square, then obviously 3AP=2OA, so we need to prove BG=PA ( * ). As O lies on AC (perpendicular bisector of BD, the circle O will also pass through F, midpoit of AB. PEAF is a square, thus EF=AP ( 1 ). As EG=DE=BF, BEGF is an isosceles trapezoid, wherefrom BG=EF; with (1), our claim (*) has been proved.

Romantics of Geometry, Pr. 247 FJGC, https://www.facebook.com/103907057666827/photos/a.103973994326800/613689746688553

 Problem 247, FJGC
Let O be the circumcenter of triangle ABC and A' reflection of A about BO. AA' intersects BC at U. Prove that C and U are inverses of each other w.r.t. circle (B,BA)

Proof:

BC is angle bisector of <ACA', thus <A'CB=<ACB=<BAA', so AB is tangent to the circle (AUC), giving AB.AB=BU.BC, done.

aops, 2018 Kharkiv Lyceum, https://artofproblemsolving.com/community/c4t48f4h2599209_angles_wanted_ltbac2ltacb_ltbac2ltmnc_2018_kharkiv_lyceum_no_27__83

 In triangle ABC, with <BAC=2<ACB, let AD be inner bisector of <BAC. Let denote M, N the midpoints of AC and BD. It turned out that <BAC=2<MNC; find the angles of the triangle.

Proof 
Obviously DM_|_AC and <CAD=<CNM, thus ANDM is cyclic, AN_|_BC, so it is angle bisector of <BAD. So, if <ACD=<CAD=2x, <DAN=x and <ACD+<CAD+<DAN=90=5x, x=18, thus <A=36, <B=<C=72.

Geometria (solo trazos), reposted, https://www.facebook.com/photo?fbid=542149486954170&set=gm.1141154929709218

 In convex quadrilateral ABCD <BAC=48, <CAD=18, <ADB=30, <ACB=6, find <ACD.

Proof with Mustafa Yagci triangle
Take O the circumcenter of triangle ABC; as <ABD=90-<ACB, O lies onto BD and <AOD=<AOB=2<ACB=12 ( 1 ). Construct the equilateral triangle ACE, D inside it; as O lies onto the perpendicular bisector of AC, it follows that <AEO=30=<ADB and ADOE is cyclic, wherefrom <AED=<AOD=12. With <DAC=18 and (1) we get from Mustafa Yagci triangle <ACD=6, done.

Problem 1138, Gogeometry, https://gogeometry.blogspot.com/2015/07/geometry-problem-1138-square.html

 Let E, F be points onto extensions of the sides (AB, (AD of the square ABCD so that BE=DF and call G projection of A onto BF. Prove that GE_|_CG.

Proof, new
Construct the square AEKF; let AG cut KE at L. Well known, LE=AB=AD, BELC is a rectangle, BELG cyclic, thus BECG is cyclic, hence <CGE=<CBE, done.

2013 Rusanovsky Lyc. Olympiad, https://artofproblemsolving.com/community/c4t48f4h2607229_collinear_wanted_tangential_pentagon__2013_rusanovsky_lyceum_olympiad_78

 Let ABC be a triangle with <B=2<C, and I - its incenter. Prove that BI=AC-AB

Proof
Let BD be the angle bisector of B, D onto AC, E reflection of B about AI; it is onto AC and <AEI=<ABI=<ACB, so IE||BC, BCEI is an isosceles trapezoid, BI=CI=AC-AE=AC-AB, done.

aops. square Jean-Louis Ayme, https://artofproblemsolving.com/community/c6t48f6h2607184_a_circle_tangent_to_the_diagonal_of_a_square

 Let ABCD be a square, (A) the circle (A,AB), I midpoint of AB, segment CI intersects the circle (A) at Q. Prove that AC tangents the circle (BIQ).

Proof
Clearly CI passes through K, reflection of D about A, thus <IQB=45. Take O, midpoint of AC; as <BOI=45, QOIB is cyclic, OB is its circumdiameter, perpendicular to AC, done.