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M17-16

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M17-16 [#permalink] New post 26 May 2008, 02:55
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Which of the following can be a perimeter of a triangle inscribed in a circle of radius 1?

I. 0.001

II. 0.010

III. 0.100



I only
III only
II and III only
I, II, and III
not I, II, or III
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Re: M17-16 [#permalink] New post 26 May 2008, 11:48
E

2 < perimeter < 4
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Re: M17-16 [#permalink] New post 26 May 2008, 17:06
gmatnub wrote:
E

2 < perimeter < 4


can you explain why?
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Re: M17-16 [#permalink] New post 26 May 2008, 17:40
fresinha12 wrote:
gmatnub wrote:
E

2 < perimeter < 4


can you explain why?


My mistake, 4 is not the maximum
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Re: M17-16 [#permalink] New post 26 May 2008, 18:40
i think the answer should 1, 2 and 3 are all possible..
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Re: M17-16 [#permalink] New post 26 May 2008, 19:49
fresinha12 wrote:
i think the answer should 1, 2 and 3 are all possible..
can you leave some reasonings here? thanks
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Re: M17-16 [#permalink] New post 26 May 2008, 20:23
I think all are possible: Reason being, to construct a triangle all that we need is a set of 3 distinct points and in this case we can bring 3 points as close as possible and also make it lie the circle (your imagination should work here).
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Re: M17-16 [#permalink] New post 28 May 2008, 23:27
fresinha12 wrote:
i think the answer should 1, 2 and 3 are all possible..


hi fresinha,
the radius of the circle is 1 units
That means that the distance between the center of the circle and the vertices of the triangle is 1 i.e the radius .so when the internal distances are 1 how can the perimeter be <0

Please explain.
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Re: M17-16 [#permalink] New post 28 May 2008, 23:59
sondenso wrote:
Which of the following can be a perimeter of a triangle inscribed in a circle of radius 1?

I. 0.001

II. 0.010

III. 0.100



I only
III only
II and III only
I, II, and III
not I, II, or III


As we can see, a triangle with 3 sides: a, b, c which is inscribed in a circle with the radius is 1.
We assume a is the biggest side of the triangle.
We have the perimeter = a + b + c > a + a = 2a
Because the triangle is inscribled in a circle so a is always smaller than or equal to the diameter = 2.
So, we have the perimeter > 2a >= 4.
So E is the correct answer.
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Re: M17-16 [#permalink] New post 29 May 2008, 03:41
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The point of giving the radius is just confusing the test takers. It doesn't matter at all. We have to place all three vertices on a tiny arc of less than 0.001. The circle consists of infinite number of points. The circle isn't a straight line, so the three vertices will make a triangle.
Do you agree?
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Re: M17-16 [#permalink] New post 17 Jun 2008, 10:55
dzyubam wrote:
The point of giving the radius is just confusing the test takers. It doesn't matter at all. We have to place all three vertices on a tiny arc of less than 0.001. The circle consists of infinite number of points. The circle isn't a straight line, so the three vertices will make a triangle.
Do you agree?


boy this question did my head in today! nice tricky question and a great explanation. Can we update the explanation in the test with this one please. The OE in the test leaves a lot to be desired.
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Re: M17-16 [#permalink] New post 17 Jun 2008, 13:12
I updated the OE to this problem.
Thanks.
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Re: M17-16 [#permalink] New post 17 Jun 2008, 16:16
My reasoning for answer E

Even if it is not given the inscribed triangle is equilateral triangle , let's assume that the traingle is equilateral .

For an equilateral triangle inscribed in a circle side s = sqroot(3)*r ; with r =1 , side s = root(3) , so perimeter is 3*root(3) . Which is way off all the choices given in the question


intutively I think other triangle inscribed in the circle will have perimeter greater than equilateral triangle , but I don't have a good reasoning , I might be wrong . If some quant expert shed some light on this , would be great !
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Re: M17-16 [#permalink] New post 17 Feb 2012, 17:37
How did we come up with 0.001 ....why not 0.0001? What is the reasoning behind this lower bound? What is the logic behind 0.001 Does this have any connection with radius being =1?
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Re: M17-16 [#permalink] New post 17 Feb 2012, 17:49
Expert's post
teal wrote:
How did we come up with 0.001 ....why not 0.0001? What is the reasoning behind this lower bound? What is the logic behind 0.001 Does this have any connection with radius being =1?


The lower limit of the perimeter of an inscribed triangle in a circle of ANY radius is 0: P>0.

Answer is D.
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Re: M17-16 [#permalink] New post 22 Feb 2012, 20:10
For Bunuel -

Lower bound of P = 0 will mean that all the three points overlap ---> no triangle possible.

How do you justify a lower bound of 0? Can you please explain with a figure?
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Re: M17-16 [#permalink] New post 22 Feb 2012, 21:15
Expert's post
teal wrote:
For Bunuel -

Lower bound of P = 0 will mean that all the three points overlap ---> no triangle possible.

How do you justify a lower bound of 0? Can you please explain with a figure?


Look at my post above, it says P>0.

Hope it's clear.
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Re: M17-16 [#permalink] New post 27 Apr 2014, 12:31
I solved it like this!

Equilateral triangle has the least perimeter. So, any triangle will have a perimeter greater than that of an equilateral triangle.

Also, radius of the circle= 1 unit
This means, the distance from the center of the circle to the vertex of the triangle = 1 unit

we know that the value of the third side of a triangle lies between the sum and difference of the other two sides.
Therefore,
0<a<2, 0< b<2, 0<c<2 => 0<a+b+c<6
so, the perimeter is greater than 0.
Re: M17-16   [#permalink] 27 Apr 2014, 12:31
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