Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 837.
Wednesday, December 26, 2012
Problem 837: Triangle, Orthocenter, Circumcenter, Circle, Concyclic Points
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Since ∠ABO=90°−∠A, OB=OF, so ∠OFB=90°−∠A.
ReplyDeleteSince ∠FOG=2∠B, OF=OG, so ∠OFG=90°−∠B.
Therefore, ∠BFG=∠OFB+∠OFG=∠C.
Hence, A,F,G,C concyclic.
Since ∠BAD=90°−∠B, AB=AG, so ∠BAG=180°−2∠B.
So ∠AGC=∠BAG+∠B=180°−∠B.
Since O is the orthocenter of ΔABC, so ∠AOC=180°−∠B.
Therefore, ∠AGC=∠AOC.
Hence, A,O,G,C concyclic.
As a result, A,F,O,G,C concyclic.
solution by Michael Tsourakakis
ReplyDeleteq:tangent to the circle center Ο, at the point Β
q//AC ,so, angleIBC=α=angleC=β
angle α=angleγ(by chord and tangent).So β=γ, therefore, quadrilateral FGCA is,cyclic quadrilateral.
because O ,it is the center of the circumcircle of the triangle FGB, then ,angle δ=angle ε.
But, angle ζ=angleδ (because AHDB ,is, cyclic quadrilateral.), therefore, angle ζ= angle ε
So,OCGA , is, cyclic quadrilateral .
But, the circumcircle of the triangle AGC is unique. So A,F O G,C concyclic
see the image;http://img203.imageshack.us/img203/5730/26ggb.png
∠OGF = ∠OFG = (180 - ∠FOG)/2 = (180 - 2∠FBG)/2 = (180 - 2∠ABD)/2 = 90 - ∠ABD = ∠BAD = ∠FAO.
ReplyDeleteHence ,O,F,G,A concyclic. Similarly, O,G,C,F concyclic.
Q.E.D.
http://img209.imageshack.us/img209/5224/problem837.png
ReplyDeleteAdd lines per attached sketch
Let B=value of angle ABC
We have ∠FOG=2. ∠B
In right triangle CEB we have ∠BCE=90-∠B
In isosceles triangle FOG we have ∠OFG=90- ½. ∠FOG=90-∠B
So ∠BCE=∠OFG => F,O,G, C is cocyclic
Similarly we also have A,F,O, G cocyclic
So A,F,O,G,C is cocyclic
COE is perpendicular bisector of BF
ReplyDelete=> ΔOBE is congruent to ΔOFE
=> angle BCE = angle FCE
=> angle OCG = angle OCF
=> O,G,C,F concyclic
///ly so are O,F,A,C
Sorry, I don't get why ∠OCG=∠OCF would imply O,G,C,F concyclic.
DeleteCan you explain a bit more?
Prob 837 - Proof Corrected. Thank you Jacob!
DeleteCOE is perpendicular bisector of BF
=> ΔOBE is congruent to ΔOFE
=> ∠BCE = ∠FCE
=> ∠OCG = ∠OCF. Also ∠OCG = ∠OAF (each = 90°-B)
=> ∠OAF = ∠OCF
=> O,C,A,F concyclic
///ly so are O,G,C,A
Si me lo permites, Alejandro, voy a re-enunciar tu problema y lo demostraré con la menor cantidad de cálculos posibles:
ReplyDeleteDado tr ABC de ortocentro O, sea K el circuncírculo de tr ACO.
K corta a AB en F y a CB en G.
Demostrar que O es circuncentro de tr FGB.
Esta forma de enunciarlo te hace pensar directamente en "ortocentro--->circuncentro--->isogonalidad!" pues bien, usemos esa idea.
1- AFOGC es cíclico por tanto <OFG = <OCG = <BAD.
2- Trazamos la F-altura de tr FGB digamos FX y vemos que FX // AD, por tanto <BFX = <BAD.
De (1) y (2) tenemos <OFG = <OFX, por tanto FX y FO son isogonales conjugados.
Por simetría puedes demostrar que BO y GO son isogonales conjugados a las respectivas alturas, luego O es circuncentro de tr BFG.
http://www.youtube.com/watch?v=Jp0RmY0Hhx8
ReplyDelete< FOB = 2B and so < OGF = 90-B
ReplyDeleteBut < BAP = 90-B so AFOG is concyclic
Similarly CFOG is also concyclic
Hence AFOGC is concyclic
Sumith Peiris
Moratuwa
Sri Lanka