Saturday, October 4, 2014

Geometry Problem 1049: Hexagon inscribed, Circle, Circumcircle, Congruence, Angles

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view the complete problem 1049.

Online Math: Geometry Problem 1049: Hexagon inscribed, Circle, Circumcircle, Congruence, Angles, Polya's Mind Map.

8 comments:

  1. http://s21.postimg.org/4qgvbi7bb/pro_1049.png

    Connect OGt
    Since ODEG and OAFG are cyclic quadrilaterals
    We have x= ∠ (EGt)+ ∠ (FGt)= ∠ (ODE)+ ∠ (OAF)
    = 90- ½∠ (DOE)+90-1/2∠ (OAF)=180-1/2(∠DOE+∠ OAF)
    But ∠ (DOE)+ ∠ (AOF)= 360-60-60-26-60= 152
    So x=180-.5*152= 104

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  2. Ang.AOB=ang.COD=ang.EOF=60
    ang.DOE+angFOA=152(=180-28)

    x=360-ang.EGO-ang.FGO
    ang.EGO=180-ang.EDO, ang.EDO=90-ang.DOE/2
    ang.FGO=180-ang.FOA, ang.FOA=90-ang.FOA/2

    Result: x=104degree

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  3. Ang.ODG=a ---> ang.EGO=180-a, and ang.DOE=180-2a
    Ang.AOB=ang.COD=ang.EOF=60 ---> ang.FOA=360-60*3-28-(180-2a)=2a-28
    ---> ang OAF=(180-(2a-28))/2=104-a ---> ang.FGO=76+a
    x=360-ang.EGO-ang.FGO=360-(180-a)-(76+a)=360-180-76=104

    x=104deg.

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  4. Denote <DOE + <AOF by y
    AB, CD, EF each subtends 60 deg at O. They sum upto 180 deg.
    BC, DE, AF subtend angles whose sum is y + 28deg.
    Thus y + 28 + 180 = 360, so y = 152
    The ray OG divides angle x into 2 angles which are equal to ODE and OAF (Property of cyclic quadrilaterals).
    ΔODE being isosceles, <ODE = <OED = 90 deg - ½ <DOE
    Similarly <OAF = <OFA = 90 deg - ½ <AOF
    So x = 180 deg - ½ y = 180 deg - 76 deg = 104 deg

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  5. Problem 1049:
    Solution sent by Fukuchan Thanks.
    Hi, Antonio,

    See
    attached picture here.

    Ans. x=104 degree.

    From Fukuchan

    ReplyDelete
  6. Denote the sum <DOE + <AOF by y
    AB, CD, EF each subtends 60 deg at O. They sum to 180 deg.
    BC, DE, AF subtend at O, angles whose sum is y + 28.
    Thus y + 28 deg+ 180 deg = 360 deg, so y = 152 deg
    The ray O through G divides angle x into 2 angles which are equal to <ODE and <OAF (Property of cyclic quadrilaterals).
    ΔODE being isosceles, <ODE = <OED = 90 deg- ½ <DOE
    Similarly <OAF = <OFA = 90 deg - ½ <AOF
    So x = 180 deg - ½ y = 180 deg - 76 deg= 104 deg

    ReplyDelete
  7. Problem 1049:
    More generally ∠ EGF = 90 deg + (½) ∠BOC

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  8. The 6 angles about O add upto 360, the unknown 2 add upto 152 since the known 4 add upto 208.

    These 2 angles are the 2 vertex angles of the 2 isoceles Tr.s whose balance 4 angles must thus add upto 360 - 152 = 208.

    Now x is obviously 1/2 of that and is thus 104.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete