coordinate geometry

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mjjking: Master | Next Rank: 500 Posts; Posts: 353; Joined: Sat Jan 20, 2007 1:29 am; Location: Italy; Thanked: 7 times; GMAT Score:720

by mjjking » Mon Mar 09, 2009 5:16 am

thanks

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Source: Beat The GMAT — Problem Solving | Mon Mar 09, 2009 5:16 am

Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member; Thanked: 32 times

Re: coordinate geometry

by Vemuri » Mon Mar 09, 2009 5:27 am

I chose sqrt(3) & was surprised to see it is wrong. I will now ponder on how the value of s can be 1 until I see a good explanation for it

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Eye-on-the-Prize: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Fri Feb 27, 2009 9:29 am; Location: New York, NY

by Eye-on-the-Prize » Mon Mar 09, 2009 9:59 am

This may be a typo, as I also got sqrt(3) as well...

Envision three triangles, one on the left, one in the middle, and one on the right.

Left Triangle
Start by solving for the hypotenuse of the triangle on the left...
c^2 = (1^2)*(sqrt(3)^2)
c=2
Since our sides = 1, 2, sqrt(3) we know that this is a 30, 60, 90 triangle.

Middle Triangle
Since the top angle of the left triangle is 30 degrees, the angle to the right (as part of the triangle in the middle) must be 60 degrees. The hypotenuse of the left triangle as we found out in step one is = 2, which is the short side in the middle triangle. Set this equal to xsqrt(3) and solve for x to get the length of the adjacent side. x = 2sqrt(3)

Right Triangle
2sqrt(3) is the hypotenus of the triangle on the right. This must also be a 30, 60, 90 triangle, with the top angle measuring 30 degrees. Set 2sqrt(3) = 2x in order to find the length of the short side. x = sqrt(3), which is equal to s.
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Expect to win

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maihuna: Legendary Member; Posts: 1578; Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members; Thanked: 82 times; GMAT Score:720

by maihuna » Mon Mar 09, 2009 10:24 am

The two lines connecting the centre are perpendicular, try to find out the slope of the same and so multiply them to be -1 and get the value of unknown...

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mjjking: Master | Next Rank: 500 Posts; Posts: 353; Joined: Sat Jan 20, 2007 1:29 am; Location: Italy; Thanked: 7 times; GMAT Score:720

by mjjking » Mon Mar 09, 2009 11:25 am

can somebody enlighten us?

Beat The GMAT - 1st priority
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Loving my wife: MOST IMPORTANT OF ALL!

REAL THING 1 (AUG 2007): 680 (Q43, V40)
REAL THING 2 (APR 2009): 720 (Q47, V41)

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lilu: Master | Next Rank: 500 Posts; Posts: 148; Joined: Wed Dec 10, 2008 5:13 pm; Location: SF, CA; Thanked: 12 times

by lilu » Mon Mar 09, 2009 11:31 am

This question was posted a couple of days ago.
This thread has a great explanation from Brent:
https://www.beatthegmat.com/test-question-t32381.html

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hoodibaba: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Sun Feb 22, 2009 3:55 am; Thanked: 1 times

by hoodibaba » Mon Mar 09, 2009 11:37 am

OK I cant seem to draw anything here but I will try my best to explain.

The OA (shown in the diagram) is 1 which seems to be right.

Draw a line dropping from P towards the X-axis to meet the X-axis perpendicularly at P'. Similarly draw a line dropping from Q towards the X-axis to meet the X-axis perpendicularly at Q'.

Now we have two triangles to work with.
1. Triangle PP'O
2. QQ'O

These triangles are similar. Why?
because
a) Angle PP'O = angle QQ'O = 90
Say Angle POP' = x. => QOQ' would be 90-x (sum of angles in a straight line is 180).
Hence angle P'PO = 90-x and angle Q'QO = x (sum of angles of a triangle is 180)

So we have 2 triangles that are similar. So thier sides are in equal proportion.

However if you notice, the length of hypotenuese of each of the triangles is nothing but the radius of the circle. So we are in luck. i.e not only are the two triangles similar, they are also equivalent.

So QQ' = P'O as they are sides opposite equal angles.

Since QQ' = x-cordinate of point Q and P'O = y-cordinate of point P, s=1

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