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GMATPrep DS Geometry

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myfish Rising GMAT Star Default Avatar
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GMATPrep DS Geometry Post Sun Apr 08, 2012 5:47 pm
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  • Lap #[LAPCOUNT] ([LAPTIME])
    Can someone please prove (1) sufficient. It took me only a few seconds to prove (2), but (1) seems to be more tricky. Anyone? Thanks.
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    Post Sun Apr 08, 2012 6:15 pm
    We know that angles BAO and BOA are equal; let's call them x. We also know that for any given triangle, an exterior supplementary angle is equal to the sums of the two other angles (the ones to which the exterior angle is not supplementary). Thus, we can say that CBO must be 2x. We can also say that BCO is 2x, since triangle BCO is isosceles (sides BO and OC are both radii, so they are equal in length).

    We know that COD is 60 degrees. We also know that COD + COA = 180, so COA must be 120 degrees. COA is made up of angles BOA and BOC. BOC is in a triangle with BCO (2x) and CBO (2x), so it is equal to 180 - 4x.

    AOB (x) + BOC (180 - 4x) + COD (60) are equal to 180 degrees.

    x + 180 - 4x + 60 = 180

    180 - 3x + 60 = 180

    -3x = -60

    x = 20 = angle BAO

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    myfish Rising GMAT Star Default Avatar
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    Post Sun Apr 08, 2012 6:47 pm
    Now that's rather a bit more tricky than (2). Thanks.

    [quote="Bill@VeritasPrep"]We know that angles BAO and BOA are equal; let's call them x. We also know that for any given triangle, an exterior supplementary angle is equal to the sums of the two other angles (the ones to which the exterior angle is not supplementary). Thus, we can say that CBO must be 2x. We can also say that BCO is 2x, since triangle BCO is isosceles (sides BO and OC are both radii, so they are equal in length).

    We know that COD is 60 degrees. We also know that COD + COA = 180, so COA must be 120 degrees. COA is made up of angles BOA and BOC. BOC is in a triangle with BCO (2x) and CBO (2x), so it is equal to 180 - 4x.

    AOB (x) + BOC (180 - 4x) + COD (60) are equal to 180 degrees.

    x + 180 - 4x + 60 = 180

    180 - 3x + 60 = 180

    -3x = -60

    [b]x = 20 = angle BAO[/b][/quote]

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    Post Sun Apr 08, 2012 7:00 pm
    Refer to the figure below,


    OC is the radius of the circle.
    Hence, AB = OC implies, AB = OC = OD = OB

    Hence, triangle ABO is isosceles with AB = OB.
    Hence, angle BAO = angle BOA = x (say)
    Hence, angle ABO = (180 - 2x)

    Now on straight line AC, angle ABO = (180 - 2x)
    Hence, angle CBO = 180 - (180 - 2x) = 2x

    Again triangle CBO is isosceles with OB = OC
    Hence, angle BCO = CBO = 2x
    Hence, angle BOC = (180 - 4x)

    Now on straight line AD, (angle AOB + angle BOC + angle COD)= 180
    Hence, (x + (180 - 4x) + angle COD) = 180
    => angle COD = 3x

    Statement 1: angle COD = 3x = 60
    Hence, angle BAO = x = 20

    Sufficient

    Statement 2: angle BCO = 2x = 40
    Hence, angle BAO = x = 20

    Sufficient

    The correct answer is D.

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