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# Geometry PS problem

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Intern
Joined: 24 Jun 2009
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Geometry PS problem [#permalink]

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25 Jun 2009, 12:38
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what is the greatest possible area of a triangular region with one vertex on center and two points on the circle?
1) PI/2
2) 1/2
3) SQRT3/4
4) 1
5) SQRT 2

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Manager
Joined: 30 May 2009
Posts: 215

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Re: Geometry PS problem [#permalink]

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25 Jun 2009, 13:13
Definitely some piece of information is missing here...Is the radius of the circle given...????

But if the radius of the circle is 1. Then the greatest area would be possible when angle at the center is right angle, with legs as the radius of the circle. Hence area = 1/2 *B * H = 1/2 * 1*1= 1/2.

What is the OA?

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Senior Manager
Joined: 23 Jun 2009
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Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
Re: Geometry PS problem [#permalink]

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26 Jun 2009, 00:30
radius is needed. I will solve as r as radius.
Area of that triangle is $$r*cos x * r*sin x /2$$ (x is half of the center angle).
This makes $$[m]r^2$$/2 * (sinx*cosx)[/m] taking derivative of that. we get.
$$r^2/2*$$ ($$cos^2$$x - $$sin^2$$x) it must be zero when the area is greatest.
simplfying that we get
$$cos^2$$x - $$sin^2$$x = 0 so sinx = cosx. That is x is 45 degrees. And this triangle is a right triangle.
So the area is $$r*cos x * r*sin x /2$$ = r * \sqrt{2} * r * \sqrt{2}/2 = $$r^2$$
If r = 1 than area is 1.

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Retired Moderator
Status: The last round
Joined: 18 Jun 2009
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Concentration: Strategy, General Management
GMAT 1: 680 Q48 V34
Re: Geometry PS problem [#permalink]

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26 Jun 2009, 01:45
@maliyeci:

The way you have answered it, I think the whole question is messed up. In my personal opinion, such solutions are not required at GMAT. Kindly correct me if some one thinks I am wrong!!

sdrandom seems to have correct answer!!
_________________

Kudos [?]: 1218 [0], given: 157

Senior Manager
Joined: 23 Jun 2009
Posts: 360

Kudos [?]: 134 [0], given: 80

Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
Re: Geometry PS problem [#permalink]

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26 Jun 2009, 02:13
I made a mistake in solution (the sides are not r*root 2 but r * root 2 / 2 so the area must be 2* r^2*root 2 ^2 * 1/4). Yes the answer will be 1/2.

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Intern
Joined: 24 Jun 2009
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Re: Geometry PS problem [#permalink]

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26 Jun 2009, 02:46
sdrandom1 wrote:
Definitely some piece of information is missing here...Is the radius of the circle given...????

But if the radius of the circle is 1. Then the greatest area would be possible when angle at the center is right angle, with legs as the radius of the circle. Hence area = 1/2 *B * H = 1/2 * 1*1= 1/2.

What is the OA?

Yes, the answer is 1/2. could you please help me understand why the greatest area is possible when the angle at the center is right angle?
thanks

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Current Student
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Re: Geometry PS problem [#permalink]

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26 Jun 2009, 11:42
See the following thread which discusses the same problem

triangular-circle-area-problem-79634.html

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Intern
Joined: 16 Jun 2009
Posts: 10

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Re: Geometry PS problem [#permalink]

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26 Jun 2009, 12:18
Nice problem/solution! I got really scared when I saw the cos and sin solution. yikes!

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Re: Geometry PS problem   [#permalink] 26 Jun 2009, 12:18
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# Geometry PS problem

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