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Friday, December 9, 2016
Geometry Problem 1294 Triangle, Altitudes, Four Squares, Center, Midpoint, Concyclic Points
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Problem 1294
ReplyDeleteIs <AO_1B+<AKB=90+90=180,then A,K,B and O_1 are concyclic.So <O_1KA=O_1BA=45,but AO_3=O_3J_1, AM=MC so MO_3//CJ_1 ,(CJ is perpendicular in AB ) then MO_3 is perpendicular in AB.Is <MO_3A=<CJ_1A=45=<O_1KA.Therefore the O_1,O_3,M and K are concyclic.Νow <CJB=90=<CO_2B then the point C,J,B and O_2 are concyclic.So
<CJO_2=<CBO_2=45,<AJC=90 and <O_3JA=45 then <O_3JA+<AJC+<CJO_2=180 so the
Point O_3,J and O_2 are collinear. Τherefore <O_2O_3O_1=90 . Similar <O_2O_4O_1=90.
So the point O_1,O_2, O_4 and O_3 are concyclic.Now <O_1KA=45=<O_2KC so <O_1KO_2=90=<O_1O_3O_2=O_1O_4O_2 then the point O_1,O_2, K,O_4 and O_2 are
Concyclic.Therefore the points O_1, O_2,O_3,O_4, M and K are concyclic.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
AO1BK is concyclic so
ReplyDelete< AKO1 = < ABO1 = 45
Now MO3 // CJ1 by applying the midpoint theorem to Tr. ACJ1.
Hence < AO3M = < AJ1C = 45
So < AKO1 = 45 = < AO3M.
So O1O3MK is concyclic
Similarly we can show that O2O4KM is concyclic and so the result follows.
Sumith Peiris
Moratuwa
Sri Lanka
Sumith
DeleteRefer to your solution.
In my opinion, concyclic of O1,O3,M,K and concyclic of O2,O4,K,M are not enough to conclude that O1,O2,O3,O4,K and M are concyclic. Please explain.
Peter
Peter
DeleteLet me add...
Since BJCO2 is concyclic < CJO2 = 45 and O2JO3 are collinear and so O1O2 is the diameter of circle O1O2O3.
Similarly we can show that O1O2 is the diameter of circle O1O2O4.
Hence O1O2O3O4 is concyclic and so together with my proof above the result is complete
Sumith