GMATPrep: value of point s

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California4jx: Master | Next Rank: 500 Posts; Posts: 227; Joined: Thu Aug 14, 2008 10:43 amFollowed by: 1 member; Thanked: 7 times; GMAT Score:650

by California4jx » Sat Jul 04, 2009 11:09 am

Seems like my assumption about the right angle at the center 0 was wrong. If not, how else are we going to find the value of point s - given the seemingly obvious answer -- D -- is wrong.

Question:

In the figure ahove, points P and Q lie on the circle with center 0. What is the value of s ?

A- 1/2

B- 1

C- sqrt(2)

D- sqrt(3)

E- sqrt(2)/2

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Source: Beat The GMAT — Problem Solving | Sat Jul 04, 2009 11:09 am

abhinav85: Master | Next Rank: 500 Posts; Posts: 179; Joined: Mon Dec 29, 2008 4:41 am; Thanked: 2 times

by abhinav85 » Sat Jul 04, 2009 11:19 am

IMO B

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LevelOne: Senior | Next Rank: 100 Posts; Posts: 69; Joined: Fri Jun 19, 2009 11:46 am; Location: London; Thanked: 1 times

by LevelOne » Sat Jul 04, 2009 11:32 am

I've recently been given this solution:

t/s*(-1/root(3)) = -1 ---> since lines OP and OQ are perpendicular

also let
s^2+t^2=r^2=3+1=4 ---> (since P lies on the same circle and O is origin)

solve for s

s^2+3s^2=4 => s=+/-1

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California4jx: Master | Next Rank: 500 Posts; Posts: 227; Joined: Thu Aug 14, 2008 10:43 amFollowed by: 1 member; Thanked: 7 times; GMAT Score:650

by California4jx » Sat Jul 04, 2009 11:56 am

LevelOne wrote:I've recently been given this solution:

t/s*(-1/root(3)) = -1 ---> since lines OP and OQ are perpendicular

also let
s^2+t^2=r^2=3+1=4 ---> (since P lies on the same circle and O is origin)

solve for s

s^2+3s^2=4 => s=+/-1

how did u replace t^2 with 3s^2 ?

Can u expand more on your first solution, if u can please.

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California4jx: Master | Next Rank: 500 Posts; Posts: 227; Joined: Thu Aug 14, 2008 10:43 amFollowed by: 1 member; Thanked: 7 times; GMAT Score:650

by California4jx » Sat Jul 04, 2009 11:56 am

abhinav85 wrote:IMO B

ok, how ?

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ssmiles08: Master | Next Rank: 500 Posts; Posts: 472; Joined: Sun Mar 29, 2009 6:54 pm; Thanked: 56 times

by ssmiles08 » Sat Jul 04, 2009 12:28 pm

IMO you don't need any math here to solve the problem.

you can notice in the diagram, to be a 90 degree angle, the two points must be in reverse order with respect to the quadrant they are in. think about (1,0 and 0,1) the angle b/w them is 90 degrees. The same with (-1,0) and (0,1).

here the semi-circle is in two quads. the point (s) is in the first quad which means it is positive. since the point given is [-sqrt(3),1],

(s,t) would be the reverse with respect to the first quadrant.

[1, sqrt(3)] would be (s,t) in this case, which would make s = 1.

hope that made sense.

if you are looking for a more mathematical approach, you can find it by making two 30-60-90 triangles with the x axis. Check out the link below.

https://www.beatthegmat.com/what-is-the- ... 27181.html

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California4jx: Master | Next Rank: 500 Posts; Posts: 227; Joined: Thu Aug 14, 2008 10:43 amFollowed by: 1 member; Thanked: 7 times; GMAT Score:650

by California4jx » Sun Jul 05, 2009 5:09 am

ssmiles08 wrote:IMO you don't need any math here to solve the problem.

you can notice in the diagram, to be a 90 degree angle, the two points must be in reverse order with respect to the quadrant they are in. think about (1,0 and 0,1) the angle b/w them is 90 degrees. The same with (-1,0) and (0,1).

here the semi-circle is in two quads. the point (s) is in the first quad which means it is positive. since the point given is [-sqrt(3),1],

(s,t) would be the reverse with respect to the first quadrant.

[1, sqrt(3)] would be (s,t) in this case, which would make s = 1.

hope that made sense.

if you are looking for a more mathematical approach, you can find it by making two 30-60-90 triangles with the x axis. Check out the link below.

https://www.beatthegmat.com/what-is-the- ... 27181.html

Thanks for pointing this link !

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