Sunday, February 21, 2010

Problem 428: Quadrant of a circle, Square, Angle bisector, Measurement

Proposed Problem
Click the figure below to see the complete problem 428 about Quadrant of a circle, Square, Angle bisector, Measurement.

Problem 428: Quadrant of a circle, Square, Angle bisector, Measurement.
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Complete Problem 428

Level: High School, SAT Prep, College geometry

7 comments:

  1. ang CFB = 90° ( draw symetrical on the left )

    draw at B tang to OB
    extend CF to G ( G on tangent )
    =>
    BCG is isoceles ( CD bisector )

    at right tr CFB and FBG, FB altitude, is common for two tr

    a² - x² = ( 2b )² - ( a - x )² ( BG = 2DE = 2b )
    a² - x² = 4b² - a² + 2ax - x²
    2ax = 2a² - 4b²
    ax = a² - 2b²

    x = (a² - 2b²)/a
    ---------------------

    b² + b² = R² ( from tr ODE , R = OB = OD )
    2b² = R²

    b² = R²/2 (1)
    ---------------------
    a² = R² + b² ( from tr COB )
    a² = R² + R²/2 (from (1) )

    a² = (3R²)/2 (2)
    ----------------------
    substitute (1) & (2) at x =>

    x = R√6/6

    a/x = (R√6/2)/(R√6/6) ( a from (2) )

    a/x = 3
    ----------------------

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  2. Let the side of the square be y. Then OC =√2 y and BC =√3 y. Draw FG perpendicular to OB. OG =x*OB/BC=x√(2/3) and FG = y+ x*OC/BC =y+x/√3. Now OF^2=OB^2=2y^2=(x√(2/3))^2+( y+x/√3)^2 or x^2 +2xy/√3 –y^2=0 which is a quadratic yielding x = y/√3 as the only admissible value. Hence x/BC = (y/√3)/√3 y = 1/3.
    Ajit

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  3. Erratum: Read OD=√2y (OC=y) in the proof above.
    Ajit

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  4. CF cuts OB at P, such that <CPB=<FCD=<DCB=<CBP making triangle PCB isosceles and P on circle. <FOB=2<FPB=<FCD+<DCB=<FCB so FCOB is cyclic. Using power of point P, we can say that
    a(a+x)=(PO=asqrt(2/3))*(PB=2asqrt(2/3))=2a^2*2/3. Solving for x gets desired answer.

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  5. Complete rectangle COBT and parellelogram COTU.

    T is the mid point of BU and so OT which is parellel to CU bisects FB perpendicularly since Tr. OFB is isoceles and so BF is an altitude of isoceles Tr. CUB.

    Writing the area if this Tr. in 2 ways ;

    FB. a = OB. OC. Squaring.

    FB^2 = 2a^2 . a^2/ 3 / a^2
    So FB ^2 = 8a^2/3 and so applying Pythagoras to Tr. BCF we see that this is = to a^2 - x^2 from which x^2 = a^2 / 9

    So x = a/3

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  6. Dear Antonio - do u have a proof for this which does not use Pythagoras?

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  7. Problem 428
    The power of point C of the circle (O, OB) is CF’.CB=CD’.CD or CF.CB=CD^2.Is CD^2=R^2/2
    α^2=OC^2+R^2 (F,F’ and D,D’ are symmetrical to the ΟΑ) ,then a^2=3R^2/2.But a.x=R^2/2
    so a.x=a^2/3 (a^2/3=R^2/2 ).Therefore x=a/3.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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