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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