Geometry Problem

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vivekjaiswal: Master | Next Rank: 500 Posts; Posts: 107; Joined: Tue Sep 15, 2009 7:18 pm; Location: Pune, India; Thanked: 6 times; GMAT Score:630

by vivekjaiswal » Wed Sep 30, 2009 3:23 am

What is the circumference of the circle above? (see attachment)
(1) The length of arc XYZ is 18.
(2) r = s

OA is [spoiler](C)[/spoiler]

I took this question from set 5 of 1000 DS series.
It seems there is something wrong in the diagram. Even after assuming the two angles in the triangle to be r and s, I couldnt do it.

Any help out there!

Cheers,
Vivek.

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Source: Beat The GMAT — Data Sufficiency | Wed Sep 30, 2009 3:23 am

vaibhav.iit2002: Master | Next Rank: 500 Posts; Posts: 295; Joined: Tue Jul 15, 2008 10:07 am; Thanked: 4 times; GMAT Score:690

by vaibhav.iit2002 » Wed Sep 30, 2009 7:00 am

If XZ is parallel to base of triangle then only I can think of C to be the answer.

1 and 2 alone are insufficient

1 and 2:
as r=s, all angles are 60deg
say center of circle is C. [symbol L : angle]
LOYZ = LOZY = 30deg => LYOZ=120deg
similarly LYOX=120deg, now LXOZ=360-120-120=120deg
hence all arcs are of same length. Now as arc XYZ=18, circumference = 18/2x3 = 27

Hence C.

So ques. misses that assumption.

Any inputs??

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vivekjaiswal: Master | Next Rank: 500 Posts; Posts: 107; Joined: Tue Sep 15, 2009 7:18 pm; Location: Pune, India; Thanked: 6 times; GMAT Score:630

by vivekjaiswal » Wed Sep 30, 2009 7:36 pm

vaibhav.iit2002 wrote:If XZ is parallel to base of triangle then only I can think of C to be the answer.

1 and 2 alone are insufficient

1 and 2:
as r=s, all angles are 60deg
say center of circle is C. [symbol L : angle]
LOYZ = LOZY = 30deg => LYOZ=120deg
similarly LYOX=120deg, now LXOZ=360-120-120=120deg
hence all arcs are of same length. Now as arc XYZ=18, circumference = 18/2x3 = 27

Hence C.

So ques. misses that assumption.

Any inputs??

hey vaibhav.iit2002,

thanks for the reply, but am confused as to how you came to the conclusion that all the angles of the triangle are 60deg?
may be i am unable to see the obvious; could you please help me with this deduction.

Thanks,
Vivek

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vaibhav.iit2002: Master | Next Rank: 500 Posts; Posts: 295; Joined: Tue Jul 15, 2008 10:07 am; Thanked: 4 times; GMAT Score:690

by vaibhav.iit2002 » Wed Sep 30, 2009 7:43 pm

vivekjaiswal wrote:
hey vaibhav.iit2002,

thanks for the reply, but am confused as to how you came to the conclusion that all the angles of the triangle are 60deg?
may be i am unable to see the obvious; could you please help me with this deduction.

Thanks,
Vivek

As angles are r,s,s and r=s so this makes r+s+s=180 => 3s=180 => s=60 => hence all angles 60deg

Did I answer ur doubt or i missed what exactly u wanna know??

But as I said my assumption of parallel lines makes this question solvable else I don't know how to solve.

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vivekjaiswal: Master | Next Rank: 500 Posts; Posts: 107; Joined: Tue Sep 15, 2009 7:18 pm; Location: Pune, India; Thanked: 6 times; GMAT Score:630

by vivekjaiswal » Wed Sep 30, 2009 7:54 pm

vaibhav.iit2002 wrote:
vivekjaiswal wrote:
hey vaibhav.iit2002,

thanks for the reply, but am confused as to how you came to the conclusion that all the angles of the triangle are 60deg?
may be i am unable to see the obvious; could you please help me with this deduction.

Thanks,
Vivek
As angles are r,s,s and r=s so this makes r+s+s=180 => 3s=180 => s=60 => hence all angles 60deg

Did I answer ur doubt or i missed what exactly u wanna know??

But as I said my assumption of parallel lines makes this question solvable else I don't know how to solve.

OOOkkkk

I had been assuming the top angle of the triangle to be labeled as 'y' 'cos its not really clear over there. You have cleared my doubt bang on man!

Now that makes our life simpler, even without the assumption that XZ is parallel to the base of the triangle we now know that the arc XZ will cast an angle of 120deg on the center of the circle. That means it is 120/360 = 1/3rd of the perimeter of the circle. Now if XYZ is given, we know that it is 2/3rd of the perimeter and hence the perimeter = (3/2)*XYZ.

Hope that helps.
Cheers,
Vivek.

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vaibhav.iit2002: Master | Next Rank: 500 Posts; Posts: 295; Joined: Tue Jul 15, 2008 10:07 am; Thanked: 4 times; GMAT Score:690

by vaibhav.iit2002 » Wed Sep 30, 2009 8:06 pm

OOPs !!

i missed something ..u r right ..we don't need any assumption here.

Thanks

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glorydefined: Senior | Next Rank: 100 Posts; Posts: 43; Joined: Tue Mar 31, 2009 9:54 am; Thanked: 2 times

by glorydefined » Wed Sep 30, 2009 8:45 pm

I do not agree with this, two angles equal means that the triangle is isoceles not equilateral....so it cannot be assumed that the angle y is 120...answer is E...

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vivekjaiswal: Master | Next Rank: 500 Posts; Posts: 107; Joined: Tue Sep 15, 2009 7:18 pm; Location: Pune, India; Thanked: 6 times; GMAT Score:630

by vivekjaiswal » Wed Sep 30, 2009 8:49 pm

glorydefined wrote:I do not agree with this, two angles equal means that the triangle is isoceles not equilateral....so it cannot be assumed that the angle y is 120...answer is E...

glorydefined,

This is exactly what had confused me. The top angle of the triangle which appears to be a 'y' is actually an 'r'
Its the picture which is not clear

Hope that clears your doubts...
Cheers,
Vivek.

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