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GMAT Prep : PS Geometry

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Senior | Next Rank: 100 Posts
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by Skywalker » Sun Oct 03, 2010 10:40 am
Hi.

There are 3 ways to solve for the sum. Please let me know how to proceed in each case..

Case 1
--------
The triangle is a 90 45 45 and thus the lengths of the side are in a certain ratio. Once we know that, using distance formulae we can cal the value of s & t. But how to ?

Case 2
---------
Use simple distance formulae & figure out the distance & seek value of s & t ?

Case 3
---------
Consider the OPY side to be mirror image of side OQY. This will give us the value of S easily .

Pls let me know which is the best approach here ?

OA = = option 2


Image
StAy HuNgry sTay fOOlisH
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by euro » Sun Oct 03, 2010 11:23 am
Drop a perpendicular from point P on the X-axis. Call it PT

Triangle OPT is right triangle having -
OT = -v3 (read as sq. root of 3)
PT = 1
Hence, by Pythagorean Theorem,
OP = 2 = Radius of Circle = OQ
Also, Triangle OPT has angles 30-60-90



Again drop a perpendicular from Q on X-axis. Call it QR.

Angle QOR = 180 - (90+30) = 60
Triangle QOR has angles 30-60-90
Since OQ (radius of circle) = 2, therefore OR = 1 and QR = v3

Answer: s = 1
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by Rahul@gurome » Sun Oct 03, 2010 9:29 pm
Solution:
Both PO and QO are radius and are equal.
Now P is (-sqrt 3, 1) and O is (0,0).
So length PO is sqrt(4) = 2.
Now length QO is sqrt (s^2 + t^2) = 2.
Or s^2 + t^2 = 4.
Slope of PO is 1/-(sqrt 3) = - 1/(sqrt 3).
Since PO is perpendicular to QO, product of their slopes is -1.
So slope of QO = t/s = sqrt (3)
So t = sqrt (3) * s.
Ot t^2 = 3 s^2.
So 4s^2 = 4.
Or s^2 = 1
So s = 1 or -1 but we omit the value -1 because point (s, t) is in the first quadrant.
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by blaster » Sun Oct 03, 2010 10:17 pm
radius of that circle is not changing . it must change only sign. so the answer is sqrt(3)
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by Ad_Astra_Per_Aspera » Mon Oct 04, 2010 9:27 pm
https://www.manhattangmat.com/forums/gma ... t2493.html

specifically look at the solution (Timestamp = Wed Mar 12, 2008 3:52 am) which talks about rotating the figure clockwise by 90 degrees - thats an awesome way of solving such questions.
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