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AB is perpendicular to CO. Is A or B closer to C?

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AB is perpendicular to CO. Is A or B closer to C? [#permalink]

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New post 17 Oct 2017, 22:40
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AB is perpendicular to CO. Is A or B closer to C?

(1) OA is less than OB.
(2) ACBD is not a parallelogram.


[Reveal] Spoiler:
Attachment:
2017-10-18_1039.png
2017-10-18_1039.png [ 3.66 KiB | Viewed 415 times ]
[Reveal] Spoiler: OA

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AB is perpendicular to CO. Is A or B closer to C? [#permalink]

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New post 18 Oct 2017, 05:42
Bunuel wrote:
Image

AB is perpendicular to CO. Is A or B closer to C?

(1) OA is less than OB.
(2) ACBD is not a parallelogram.


[Reveal] Spoiler:
Attachment:
2017-10-18_1039.png


We need to find which is less \(AC\) or \(BC\)

\(AC^2= OC^2+OA^2\) and \(BC^2=OC^2+OB^2\). subtract the two to get

\(AC^2-BC^2=OA^2-OB^2\)-------------------(1)

Statement 1: Statement itself is sufficient to answer the question, because position of C is fixed.

Mathematically it can be deduced as - given \(OA<OB\), square both sides to get \(OA^2<OB^2\), or \(OA^2-OB^2<0\)

Hence equation 1 becomes \(AC^2-BC^2<0 = > AC<BC\). Sufficient

Statement 2: Nothing mentioned about distances of any point, hence insufficient

Option A

Last edited by niks18 on 18 Oct 2017, 08:32, edited 1 time in total.

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Re: AB is perpendicular to CO. Is A or B closer to C? [#permalink]

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New post 18 Oct 2017, 06:13
S1- OA>OB
OC is the same.
Therefore A is closer to C than B.
Sufficient
S2- No useful information.
Insufficient.
Answer A.
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Re: AB is perpendicular to CO. Is A or B closer to C? [#permalink]

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New post 23 Oct 2017, 17:07
Answer A.

St1, OA<OB
I viewed this problem as two triangles ACO and BCO. If leg OA is smaller than OB and they share a side in CO, a smaller OA will produce a smaller hypotenuse. Therefore point A is closer to C. Sufficient.

St2, ABCD is a parallelogram
Honestly, I have no idea what to do wit this information and assumed it was useless.
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Re: AB is perpendicular to CO. Is A or B closer to C?   [#permalink] 23 Oct 2017, 17:07
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