GMAT Prep - Geometry (2)

[RonPurewal]

Confusion regarding this problem...

we know that each triangle is 30 60 90 ....

also that radius is 2... so if we draw a line joining both the points...we form a triangle... with sides equal(radius)...

each angle is 45 ..

now - is this correct.... that means my total sum of angels is not 90 for the angel p.. between a line joining pint p to q and a perpendicular to x axis...

am i missing something ??

no, you're not missing anything.
the problem is that the diagram is not drawn to scale. (if it were, it would be easy to infer the coordinates, because you could just equate the known distances with any unknown distances: i.e., because of the symmetry of the problem, anything looking like the known "1" could be assumed to be 1, and anything looking like the known "√3" could be assumed to be √3.

in reality, the diagram is nowhere near being symmetrical with respect to the y-axis. therefore, the segment connecting p and q is not parallel to the x-axis, and so you are correct that the designated angle would NOT be 90 degrees.

[guy.b]

Hi,

I understand the solution BUT the figure (posted in the beginning of this thread) is confusing. When I read the question and saw the image I immediately answered S=sqrt(3).
Why the authors of the question didn't add the comment "figure not drawn to scale"? How should I use the figures in Geometry problems in the GMAT? Can I make assumption based on figures?

[RonPurewal]

Hi,

I understand the solution BUT the figure (posted in the beginning of this thread) is confusing. When I read the question and saw the image I immediately answered S=sqrt(3).
Why the authors of the question didn't add the comment "figure not drawn to scale"? How should I use the figures in Geometry problems in the GMAT? Can I make assumption based on figures?

yeah -- they should have included that warning. i'm not sure why they didn't.

however, if there's one thing i could tell you about guessing answers on this test, it would be this:
if you think that you can solve a gmat problem in less than ten seconds, your solution is probably wrong.

there are exceptions to this rule -- for instance, if you know some really obscure fact or shortcut -- but, if you just think "hey, symmetry, i'm done!", then it's a pretty safe bet that you're picking a "sucker answer".

[chaouky]

hi,

there is something I don't get here :

we have OPQ a rectangle triangle isocele with OP=OQ=2
and PQ = sqrt(8) = 2*sqrt2)

Why don't we have s = sqrt(8) - sqrt (3) ??

Thanks !!

[tim]

PQ is not a horizontal line..

[Buix0065]

Hello,

What got me on this problem--and I think will get me on any similar problem is that both lines start from (0,0) which is the center point of the circle.

If this is the case, from center point to the edge is the radius right? So I would assume OP = OQ

Then, since <POQ is a 90 degree angle, OP and OQ are same length and same angle, so P and Q must be same distance from 0.

Please let me know what I'm missing so I don't make this mistake again!

[RonPurewal]

Hello,

What got me on this problem--and I think will get me on any similar problem is that both lines start from (0,0) which is the center point of the circle.

If this is the case, from center point to the edge is the radius right? So I would assume OP = OQ

Then, since <POQ is a 90 degree angle, OP and OQ are same length and same angle, so P and Q must be same distance from 0.

Please let me know what I'm missing so I don't make this mistake again!

actually, nothing that you wrote there is wrong -- the two points really are the same distance away from the origin.

the problem, though, is that the problem isn't asking for a distance from the origin; the problem is only asking for one coordinate (namely, the x-coordinate).
therefore, the problem has little to do with distance from the origin.
for instance, (5, 0) and (4, 3) are both 5 units away from the origin, but none of their coordinates are the same.

[hsgross]

Isn't the easiest way to complete this problem simply by recognizing that there are two perpendicular lines (angle =90) so then to find point (s,t) we can just use the reciprocal slope, which would leave us with 1 for s?

[RonPurewal]

Isn't the easiest way to complete this problem simply by recognizing that there are two perpendicular lines (angle =90) so then to find point (s,t) we can just use the reciprocal slope, which would leave us with 1 for s?

it makes no difference what is the "easiest" way to solve the problem; your only goal should be to learn as many ways as possible to solve it.
("easiest" is a meaningless judgment on a forum like this one anyway, because judgments of what is easy and what is hard will vary considerably from individual to individual.)

[steven.shears]

in the spirit of trying another method, i used similar triangles. since we know that when we drop a perpendicular we form a 30-60-90 right triangle with point P on the left hand side of the page, you can draw a line connecting point P to Q. you therefore have a rectangle and a new 30-60-90 triangle PQO. Since this new triangle shares a side with the triangle on the left, we can use similar triangles.

2/PQ = sqrt (3)/2. Length of PQ = 4/sqrt(3). Since we need to find the horizontal distance, s, we take PQ - sqrt (3) [the sqrt of 3 being the horizontal distance left of point 0) and we are left with distance of point O to horizontal point S is 1.

[RonPurewal]

in the spirit of trying another method, i used similar triangles. since we know that when we drop a perpendicular we form a 30-60-90 right triangle with point P on the left hand side of the page, you can draw a line connecting point P to Q. you therefore have a rectangle and a new 30-60-90 triangle PQO.

nope. again, you're assuming that the diagram is drawn to scale. read the earlier posts in the thread again -- it isn't. (this is actually an issue; PS problems with not-to-scale diagrams should be explicitly marked as such.)

* if you connect P and Q, you don't get a rectangle, because segment PQ is not horizontal. try drawing the diagram to scale, and you'll see what i mean.
if you don't understand what the correct diagram should look like, let me know, and i'll upload a drawn-to-scale version when i get the chance (can't right now, don't have access to the proper software).

* it should be clear that PQO is not a 30º-60º-90º triangle, because it's isosceles -- two of its sides are radii of the same circle.
in fact, this observation guarantees that PQO is a 45º-45º-90º triangle.

[steven.shears]

ahh I see that now, PQ is not horizontal - thanks

[RonPurewal]

ahh I see that now, PQ is not horizontal - thanks

yep.

this problem is troubling, because, as far as i know, it's the only instance of an official multiple-choice problem on which a not-to-scale diagram has not been labeled as such. (by contrast, NO diagram on data sufficiency is "to scale"; in fact, it's not even really possible for data sufficiency diagrams to be "to scale", because there will generally be lots of unknown quantities in them.)

to anyone reading this thread -- if you know of any other official problems in which gmac has included a not-to-scale diagram but failed to label it as such, please let us know. thanks!

[Khush]

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 negative rad 3 (to the left). when you rotate the paper, this is now upward, so it's positive rad 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, rad3)

sweet

wow!! that is a fantastic approach, Ron.
Thanks !

[RonPurewal]

You're welcome.

Just keep in mind that this problem doesn't represent the official problems in general. In EVERY other instance, a not-to-scale diagram has been marked as such ... on multiple-choice problems.

(On DS, NO diagram is "to scale", because, by definition, the diagrams without the statements contain undetermined lengths/angles.)