Wednesday, May 28, 2008

Geometry Problem 111



See complete Problem 111
Orthogonal Circles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 10:05 PM
Labels: , , , ,

9 comments:

  1. Kasun SampathJuly 3, 2009 at 12:32 PM

    Angle COB = 180 - 2OCB, because ∆OCB is Isosceles. Angle BDC = 1\2COB = 90 - OCB.
    Angle EFC = 2BDC. Angle ECF = (180 - EFC)\2 = OCB. Angle ACB = 90 because ∆ACB is a right trangle. therefore OCB + ACO =90, ACF = OCB therefore ACO + ACF = OCF = 90
    Q.E.D
    Is there a better way to prove this?

    ReplyDelete
  2. 김은배December 15, 2011 at 6:30 PM

    각BAC=각BDC=a
    각ACO=a
    OC과 FC는 수직
    각OCF=90도

    ReplyDelete
  3. Nilton LapaMay 12, 2012 at 3:48 PM

    Solution to Problem 111.
    Let ang(BAC) = a.
    We have ang(BDC) = a and ang(ACO) = a (because triangle AOC is isosceles). In the circle of center F, ang(CFE) = 2*ang(BDC) = 2a.
    Then angle ACO is inscribed in arc EC, OC is tangent to the F circle and perpendicular to the radius CF. So ang(OCF) = 90º.

    ReplyDelete
  4. PravinMay 13, 2012 at 8:41 AM

    Join CB.
    CF is tangent to circle (O) at C and CA is a chord of circle (O)
    ∴ ∠FCA = angle in the alternate segment = ∠CBA
    ∆OAC is isosceles.
    ∴ ∠OCA = ∠OAC = ∠BAC
    Hence ∠OCF = ∠FCA + ∠OCA = ∠CBA + ∠BAC = 90° (∵∠ACB = 90°)

    ReplyDelete
  5. Nilton LapaMay 13, 2012 at 5:36 PM

    To Pravin: I have a doubt on your solution of problem 111. The fact that CF is tangent to circle (O) is not given in the enunciate. Besides, if CF is tangent to circle O, then it is perpendicular to the radius OC and ang(OCF) - 90º, this would be imediate.

    ReplyDelete
  6. PravinMay 14, 2012 at 6:46 AM

    To Nilton Lapa: Thank you!. Please see the following Revised Proof.
    ∠EDC = ∠BDC = ∠BAC (angles in the same segment)
    ∠BAC = ∠OAC = ∠OCA (since OA = OC)
    ∴ ∠EDC = ∠OCA
    Next in the circle (F) through D, E, C
    ∠EDC = (1/2)∠EFC (angle subtended by arc EC at circumference = half the angle subtended by arc EC at the centre F)
    Join EF. ∆ECF is isosceles (∵ FE = FC). So ∠CEF = ∠ECF
    ∠EFC = 180° - ∠CEF - ∠ECF = 180° - 2∠ECF
    ∴ ∠OCA = ∠EDC = (1/2)∠EFC = 90° - ∠ECF
    Hence ∠OCF = ∠OCA + ∠ECF = 90°

    ReplyDelete
  7. Sumith PeirisApril 29, 2016 at 11:39 PM

    If < EFC = 2@ then < EDC = @ = < BAC = OCA

    Similarly if < EFD = 2€ then < ACD = €

    So < OCD = @+€

    But in isoceles Tr. CDF, < DCF = 90 - @-€

    So < OCF = 90

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  8. Sailendra TNovember 30, 2020 at 5:52 PM

    Considering usual triangle notations in tr. ABC, m(BDC)=A
    => m(EFC)=2A
    => m(EDC)=90-A (Since EFC is isosceles)

    Since AOC isosceles,m(ACO)=A=>m(FCO)=m(FCE)+m(ECO)=90-A+A=90

    ReplyDelete
  9. MarcoDecember 28, 2022 at 12:37 AM

    Let <CAB=x & <CBA=y
    <BOC=2x & <AOD=2y (< at centre is twice < at circumference)
    <DOC=180-2x-2y (adj. <s on st. line)
    <DAC=<DOC/2=90-x-y (converse of < at centre is twice < at circumference)
    <ADB=90
    <DEC=<ADB+<DAC=180-x-y
    <DFC=2*(180-(180-x-y))=2x+2y
    FD=FC
    So <FCD=(180-2x-2y)/2=90-x-y=<DAC
    Since <FCD=<DAC, we can conclude that CF is a tangent to circle 1 and so <OCF=90

    ReplyDelete

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