GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 14 Nov 2018, 07:37

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### Free GMAT Strategy Webinar

November 17, 2018

November 17, 2018

07:00 AM PST

09:00 AM PST

Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

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

Author Message
TAGS:

### Hide Tags

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

### Show Tags

25 Aug 2012, 20:04
1
6
00:00

Difficulty:

55% (hard)

Question Stats:

67% (02:32) correct 33% (02:46) wrong based on 185 sessions

### HideShow timer Statistics

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?
Director
Joined: 22 Mar 2011
Posts: 601
WE: Science (Education)
Re: A circle has center at origin and radius 1.The points X,Y,  [#permalink]

### Show Tags

25 Aug 2012, 23:43
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 $$XZ=\sqrt{3}.$$

_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

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

### Show Tags

26 Aug 2012, 07:56
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 ?
Director
Joined: 22 Mar 2011
Posts: 601
WE: Science (Education)
Re: A circle has center at origin and radius 1. The points X, Y  [#permalink]

### Show Tags

26 Aug 2012, 07:59
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 $$x-x\sqrt{3}-2x$$.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

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

### Show Tags

26 Aug 2012, 08:14
Gotcha! i always knew this but din't strike me here

+1 EvaJ and you hit the half century kudos mark
Senior Manager
Joined: 24 Aug 2009
Posts: 473
Schools: Harvard, Columbia, Stern, Booth, LSB,
Re: A circle has center at origin and radius 1.The points X,Y,  [#permalink]

### Show Tags

07 Sep 2012, 11:55
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.
_________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS.
Kudos always maximizes GMATCLUB worth
-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

Intern
Joined: 02 May 2013
Posts: 21
Re: A circle has center at origin and radius 1.The points X,Y,  [#permalink]

### Show Tags

24 Aug 2013, 21:05
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.
Director
Joined: 22 Mar 2011
Posts: 601
WE: Science (Education)
Re: A circle has center at origin and radius 1.The points X,Y,  [#permalink]

### Show Tags

24 Aug 2013, 21:57
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]

### Show Tags

25 Aug 2013, 00:20
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
Senior Manager
Joined: 08 Dec 2015
Posts: 294
GMAT 1: 600 Q44 V27
Re: A circle has center at origin and radius 1. The points X, Y  [#permalink]

### Show Tags

11 Jul 2016, 04:20
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.
Non-Human User
Joined: 09 Sep 2013
Posts: 8772
Re: A circle has center at origin and radius 1. The points X, Y  [#permalink]

### Show Tags

27 Jul 2018, 10:47
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: A circle has center at origin and radius 1. The points X, Y &nbs [#permalink] 27 Jul 2018, 10:47
Display posts from previous: Sort by