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A circle has center at origin and radius 1. The points X, Y

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Joined: 11 Apr 2012
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A circle has center at origin and radius 1. The points X, Y  [#permalink]

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25 Aug 2012, 21:04
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A circle has center at origin and radius 1. The points X, Y, and Z lie on the circle such that the length of arc XYZ is (2/3)π.What is the length of line segment XZ?

A. √2
B. √3
C. 2
D. 2+√3
E. 2+√2

I figured we could eliminate c,d,e based on the rule that for a triangle,the 3rd side must be greater than difference of other 2 sides & lesser than the sum of other two sides.I am stuck after that. is there an neater method?
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Joined: 22 Mar 2011
Posts: 604
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Re: A circle has center at origin and radius 1.The points X,Y,  [#permalink]

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26 Aug 2012, 00:43
3
vinay911 wrote:
A circle has center at origin and radius 1.The points X,Y,and Z lie on the circle such that the length of arc XYZ is (2/3)π.What is the length of line segment XZ?

a)√2

b)√3

c)2

d)2+√3

e)2+√2

I figured we could eliminate c,d,e based on the rule that for a triangle,the 3rd side must be greater than difference of other 2 sides & lesser than the sum of other two sides.I am stuck after that. is there an neater method?

If the arc XYZ is $$2/3\pi$$, the central angle XOZ is $$120^o$$ (2/3*360=120).
Then you can calculate XZ from the isosceles triangle XOZ which has angles 30-30-120, and XO = OZ = 1 = radius.
The height to the base of the triangle is 0.5 (opposing the $$30^o$$ angle), so half of the base is $$0.5*\sqrt{3}{$$, and $$XZ=\sqrt{3}.$$

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Intern
Joined: 11 Apr 2012
Posts: 37
Re: A circle has center at origin and radius 1. The points X, Y  [#permalink]

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26 Aug 2012, 08:56
Thanks Eva.
I have a clarification here: I am not sure how you got the length of the perpendicular to side XZ as 0.5.Is it by trigonometry(sin 30 ?) ?
Also how do you say the perpendicular from the center divides XZ equally ?
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Re: A circle has center at origin and radius 1. The points X, Y  [#permalink]

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26 Aug 2012, 08:59
2
vinay911 wrote:
Thanks Eva.
I have a clarification here: I am not sure how you got the length of the perpendicular to side XZ as 0.5.Is it by trigonometry(sin 30 ?) ?
Also how do you say the perpendicular from the center divides XZ equally ?

Yes, is "hidden" trigonometry... In a right triangle of type 30-60-90 the leg opposed to the 30 degrees angle is half of the hypotenuse.

Good to remember: in a right triangle of type $$30-60-90$$ the sides are $$x-x\sqrt{3}-2x$$.
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Re: A circle has center at origin and radius 1. The points X, Y  [#permalink]

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26 Aug 2012, 09:14
Gotcha! i always knew this but din't strike me here

+1 EvaJ and you hit the half century kudos mark
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Re: A circle has center at origin and radius 1.The points X,Y,  [#permalink]

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07 Sep 2012, 12:55
1
EvaJager wrote:
vinay911 wrote:
A circle has center at origin and radius 1.The points X,Y,and Z lie on the circle such that the length of arc XYZ is (2/3)π.What is the length of line segment XZ?

a)√2

b)√3

c)2

d)2+√3

e)2+√2

I figured we could eliminate c,d,e based on the rule that for a triangle,the 3rd side must be greater than difference of other 2 sides & lesser than the sum of other two sides.I am stuck after that. is there an neater method?

If the arc XYZ is $$2/3\pi$$, the central angle XOZ is $$120^o$$ (2/3*360=120).
Then you can calculate XZ from the isosceles triangle XOZ which has angles 30-30-120, and XO = OZ = 1 = radius.
The height to the base of the triangle is 0.5 (opposing the $$30^o$$ angle), so half of the base is $$0.5*\sqrt{3}{$$, and $$XZ=\sqrt{3}.$$
Yes, is "hidden" trigonometry... In a right triangle of type 30-60-90 the leg opposed to the 30 degrees angle is half of the hypotenuse.

Hi Eva,

I would like to add few intermediate steps to the solution offered by you.
As Eva has correctly pointed that the central angle XOZ is $$120^o$$

Now drop a perpendicular from point X & extend the line OZ in the direction of that perpendicular & name that meeting point P
If you draw such triangle then you will that triangle XPO is 30-60-90 triangle.
We want to the height of the triangle XPO which XP & opposite to angle 60 degree.
Applying 30-60-90 theorem , we get XP = \sqrt{3}/2 and this height is the height of the original triangle XOZ

Property 1 - The Altitude to the base of an isosceles or equilateral triangle bisects the base & bisects the vertex angle.
Property 2 - The midpoint of the hypotenuse is equidistant from the three polygon vertices

Because the height (altitude) bisects the side XZ, thus is equidistant from X,O or Z
Therefore XZ = 2 x height = 2 \sqrt{3}/2 = \sqrt{3}

I hope this extra long explanation will help many & sorry that i didn't include the possible diagram.
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Re: A circle has center at origin and radius 1.The points X,Y,  [#permalink]

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24 Aug 2013, 22:05
EvaJager wrote:
vinay911 wrote:
A circle has center at origin and radius 1.The points X,Y,and Z lie on the circle such that the length of arc XYZ is (2/3)π.What is the length of line segment XZ?

a)√2

b)√3

c)2

d)2+√3

e)2+√2

I figured we could eliminate c,d,e based on the rule that for a triangle,the 3rd side must be greater than difference of other 2 sides & lesser than the sum of other two sides.I am stuck after that. is there an neater method?

If the arc XYZ is $$2/3\pi$$, the central angle XOZ is $$120^o$$ (2/3*360=120).
Then you can calculate XZ from the isosceles triangle XOZ which has angles 30-30-120, and XO = OZ = 1 = radius.
The height to the base of the triangle is 0.5 (opposing the $$30^o$$ angle), so half of the base is $$0.5*\sqrt{3}{$$, and $$XZ=\sqrt{3}.$$

Sorry EvaJager, in fact, I agree with the result that the central angle XOZ is 120 degree. However, I am afraid that the equation 2/3 * 360 = 120 is not correct (in fact 240). The result of 120 degree of central angle XOZ can be explained as follows:
The arc XYZ is 2/3pi . The circumference of circle (O,1) is 2pi. Therefore, arc XYZ makes up 1/3 of circle (O,1). And the result that the central angle XOZ is 1/3 * 360 = 120 degree can be inferred.
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Re: A circle has center at origin and radius 1.The points X,Y,  [#permalink]

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24 Aug 2013, 22:57
EvaJager wrote:
vinay911 wrote:
A circle has center at origin and radius 1.The points X,Y,and Z lie on the circle such that the length of arc XYZ is (2/3)π.What is the length of line segment XZ?

a)√2

b)√3

c)2

d)2+√3

e)2+√2

I figured we could eliminate c,d,e based on the rule that for a triangle,the 3rd side must be greater than difference of other 2 sides & lesser than the sum of other two sides.I am stuck after that. is there an neater method?

If the arc XYZ is $$2/3\pi$$, the central angle XOZ is $$120^o$$ (2/3*360=120).
Then you can calculate XZ from the isosceles triangle XOZ which has angles 30-30-120, and XO = OZ = 1 = radius.
The height to the base of the triangle is 0.5 (opposing the $$30^o$$ angle), so half of the base is $$0.5*\sqrt{3}{$$, and $$XZ=\sqrt{3}.$$

Above - 2/3*360=120 should be 2/3*180=120.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Intern
Joined: 22 Aug 2013
Posts: 3
Re: A circle has center at origin and radius 1.The points X,Y,  [#permalink]

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25 Aug 2013, 01:20
EvaJager wrote:
vinay911 wrote:
A circle has center at origin and radius 1.The points X,Y,and Z lie on the circle such that the length of arc XYZ is (2/3)π.What is the length of line segment XZ?

a)√2

b)√3

c)2

d)2+√3

e)2+√2

I figured we could eliminate c,d,e based on the rule that for a triangle,the 3rd side must be greater than difference of other 2 sides & lesser than the sum of other two sides.I am stuck after that. is there an neater method?

If the arc XYZ is $$2/3\pi$$, the central angle XOZ is $$120^o$$ (2/3*360=120).
Then you can calculate XZ from the isosceles triangle XOZ which has angles 30-30-120, and XO = OZ = 1 = radius.
The height to the base of the triangle is 0.5 (opposing the $$30^o$$ angle), so half of the base is $$0.5*\sqrt{3}{$$, and $$XZ=\sqrt{3}.$$

i know this is incorrect but please clarify, if the triangle was isoceles , then XZ is [square_root]2 *1 = [square_root]2
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Joined: 08 Dec 2015
Posts: 302
GMAT 1: 600 Q44 V27
Re: A circle has center at origin and radius 1. The points X, Y  [#permalink]

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11 Jul 2016, 05:20
Use the formula:

Radius of circumscribed circle = side * (sq. root 3 / 3 )

We have an equilateral triangle inscribed in a circle. (confirmed by the fact that the angle corresponding to 2pi/3 is 120º) So we can use the above formula.
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Re: A circle has center at origin and radius 1. The points X, Y  [#permalink]

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27 Jul 2018, 11:47
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