GMAT Prep - Geometry

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by taketotask » Thu Aug 20, 2009 10:20 am
hey,

just looking at some of the explanations for this question. I believe PQO is a 45-45-90 triangle since PO and PQ are both radii of the circle then PO=PQ.

Knowing this, draw a line down from Q to the x-axis, let's call it point Z.

This forms another 45-45-90 triangle. Since QZ is 1, OZ must also be 1. Hence S = 1.

Hope this helps.

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Do we assume (0,0) for point O?

by ILAKSHMIR » Mon Aug 24, 2009 8:14 am
The solution for this problem holds only when we assume that O is the origin with co-ordinates (0,0). I am surprised that GMAT's q is not clearly worded. (Usually they are)

Do others share this view?

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by cat er ina » Sun Apr 25, 2010 6:51 am
kadishmj wrote:
akram wrote:
tutonaranjo wrote:No luck solving this. Please help.
Image
Options are:
a) 1/2
b) 1
c) sqrt 2
d) sqrt 3
e) sqrt 2 / 2
B is correct answer.

(s,t) will be (1, sqrt 3), radius is 2 .
Angle O is 60 degress on left and 30 degrees on right since sin 360=sqrt 3/2
B is correct, but akram's reasoning is not. I have no idea where he comes up with sin 360 = sqrt 3/2

It is a 30-60-90 triangle, and Angle O is 30 degrees on the left and 60 on the right (not what akram said). This may clear things up:

https://www.themathpage.com/aTrig/30-60-90-triangle.htm
THANKS FOR POSTING THIS LINK!!

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by r2kins » Sun Apr 25, 2010 7:25 am
lunarpower wrote:here's an awesome way to approach this.

see that 90 degree angle there? ok. that means that this question is really asking where the point (-rad3, 1) would go if the paper were rotated clockwise by 90 degrees.

so...

draw that point on your paper, and then physically rotate the paper by 90 degrees.

originally the x coordinate was -√3 (to the left). when you rotate the paper, this is now upward, so it's positive √3 in the y direction.
originally the y coordinate was positive 1 (upward). when you rotate the paper, this is now to the right, so it becomes positive 1 in the x direction.

ergo, new coordinates = (1, √3)

sweet
real sweet! jus a lil imagination and whoa...no calculation at all!

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by venmic » Thu Aug 26, 2010 3:30 pm
This is a mirror image problem
Step 1: Check the signs in the cooridnates
Step 2: Mirror the point (s,t) to (-^3,1)
tutonaranjo wrote:No luck solving this. Please help.
Image
Options are:
a) 1/2
b) 1
c) sqrt 2
d) sqrt 3
e) sqrt 2 / 2

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by skalevar » Mon Oct 18, 2010 5:32 pm
This might be a stupid question, but aren't diagrams in problem solving supposed to be accurate?

Upon initial inspection, it looks as if point S is the same height (1) and the same horizontal distance away from 0, just on the other side of the axis (sqrt 3).

I can see from the discussion that is incorrect, but can someone please explain why? I'd like to fully understand so I know when and when not to use the diagrams provided in PS.

Thanks!

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by Geva@EconomistGMAT » Tue Oct 19, 2010 12:49 am
skalevar wrote:This might be a stupid question, but aren't diagrams in problem solving supposed to be accurate?

Upon initial inspection, it looks as if point S is the same height (1) and the same horizontal distance away from 0, just on the other side of the axis (sqrt 3).

I can see from the discussion that is incorrect, but can someone please explain why? I'd like to fully understand so I know when and when not to use the diagrams provided in PS.

Thanks!
This question is the single exception to the "trust the diagram" rule that I recall ever seeing on any official materials - this diagram is supposed to be drawn to scale, but is clearly misleading.

By the GMAC's own instructions, GMAT diagrams are drawn to scale, so you can trust the figure, except for the following:
1) the figure explicitly states that the diagram is not drawn to scale (with the sentence: note: figure not drawn to scale)
2) In DS questions, the diagram conforms to the question stem, but not necessarily to the statements.

In simple words: trust the diagram in Problem solving questions (unless noted otherwise), but do not trust the diagram in DS questions - a DS diagram is probably there to make you think you know something that you don't really know until the statements tell you so. If a DS diagram presents a right triangle but doesn't explicitly label it as such, then the triangle can be drawn in a different way - until a statement later on will probably "lock" it as a right angle. Ignoring this property of DS diagrams is dangerous, as it affects you judgment of a sufficiency - you may think a statement is sufficient on its own, but that's because you assume the angle is a 90 degree just from the diagram, not from any information supplied by the question.
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