AB

This topic has expert replies
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

AB

by goyalsau » Thu Jan 06, 2011 4:50 am
Tough one for me,
Attachments
Untitled.jpg
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Thu Jan 06, 2011 5:35 am
Image

Each of the answer choices contains √3. This means that we should look for a 30:60:90 triangle.

In a 30:60:90 triangle, the sides are proportioned x : x√3 : 2x. Since the side opposite the 60 degree angle is x√3, and the answer choices represent CB, the answer choices indicate the following: that angle CDO = 60 and angle COD = 30. See the drawing above.

This means that triangle CDO and triangle BOE are each 30:60:90 triangles.
Looking at triangle CDO, since CD = 6, CO = 6√3.
Looking at triangle BOE, since BE=2, BO = 2√3.
Thus, CB = 6√3 - 2√3 = 4√3.

The correct answer is B.

To confirm that angle COD=30, let's examine triangle BAO. Since OD bisects angle COD = 30, angle BOA = 60 and angle BAO = 30. So triangle BAO also is a 30:60:90 triangle, as indicated in the drawing above.
The shorter leg in this triangle is BO = 2√3.
The longer leg is AB = AE+EB = 4+2 = 6.
Does BO*√3 = AB? Yes, because 2√3*√3 = 6.
Last edited by GMATGuruNY on Fri Jan 07, 2011 9:56 am, edited 6 times in total.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Thu Jan 06, 2011 6:40 am
Here is another approach without assuming anything.

Image

Refer to the image above.

A perpendicular EM on AO is drawn from E.
Now triangle OEB is congruent to triangle OEM.
Thus, EM = EB = 2 cm
and, OM = OB = x xm (say)

In the right-angled triangle AEM, say AM = y cm, thus
  • .... (EM)² + (AM)² = (AE)²
    => (2)² + (y)² = (4)²
    => y² = 4² - 2² = 12
    => y = √12 = 2√3
Thus, AM = 2√3 cm

In the right-angled triangle ABO,
  • .... (AB)² + (OB)² = (OA)²
    => (AE + EB)² + (OB)² = (OM + AM)²
    => (4 + 2)² + (x)² = (x + 2√3)²
    => 36 + x² = x² + (4√3)x + 12
    => (4√3)x = 36 - 12 = 24
    => x = 2√3
Thus, OB = 2√3 cm

Now triangle ODC and triangle OEB are similar. Thus,
  • .... OB/OC = BE/CD = 2/6 = 1/3
    => OC = 3*(OB)
    => (OB + BC) = 3*(OB)
    => BC = 2*(OB) = 2*(2√3) = 4√3
The correct answer is B.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/

User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Thu Jan 06, 2011 8:21 am
Hats off to your approach Sir, But i would like to ask one thing.
Anurag@Gurome wrote:Here is another approach without assuming anything.

Image

Refer to the image above.

A perpendicular EM on AO is drawn from E.
Now triangle OEB is congruent to triangle OEM.
Thus, EM = EB = 2 cm
and, OM = OB = x xm (say)
I am not able to understand why EM will be equal to EB, why it will be equal I can draw a different triangle with some different dimensions. But i don't think its compulsory that every time EM will be equal to EB, I am sure there must be some reasoning behind this.
Please Clear my doubt sir...
Anurag@Gurome wrote: In the right-angled triangle AEM, say AM = y cm, thus
  • .... (EM)² + (AM)² = (AE)²
    => (2)² + (y)² = (4)²
    => y² = 4² - 2² = 12
    => y = √12 = 2√3
Thus, AM = 2√3 cm

In the right-angled triangle ABO,
  • .... (AB)² + (OB)² = (OA)²
    => (AE + EB)² + (OB)² = (OM + AM)²
    => (4 + 2)² + (x)² = (x + 2√3)²
    => 36 + x² = x² + (4√3)x + 12
    => (4√3)x = 36 - 12 = 24
    => x = 2√3
Thus, OB = 2√3 cm

Now triangle ODC and triangle OEB are similar. Thus,
  • .... OB/OC = BE/CD = 2/6 = 1/3
    => OC = 3*(OB)
    => (OB + BC) = 3*(OB)
    => BC = 2*(OB) = 2*(2√3) = 4√3
The correct answer is B.
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.

User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Thu Jan 06, 2011 8:27 am
I think i got it,

It like drawing a Circle with Center E, and Radius of the Circle is EM or EB , & OA and OB are tangents on the circle from the same point. In that case there length will also be equal...............
When we join a tangent and the radius of the circle, Angle is always and Always 90 degree.

Thanks for the thoughtful explanation.................
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Thu Jan 06, 2011 9:01 am
goyalsau wrote:Hats off to your approach Sir, But i would like to ask one thing.
Anurag@Gurome wrote:Image
I am not able to understand why EM will be equal to EB, why it will be equal I can draw a different triangle with some different dimensions. But i don't think its compulsory that every time EM will be equal to EB, I am sure there must be some reasoning behind this.
Please Clear my doubt sir...
For the triangles OEB and OEM,, all the corresponding angles are equal in measure. Thus they are similar. Which means (EB/EM) = (OB/OM) = (OE/OE) = 1. Hence not only EB = EM, but all the corresponding sides are equal in length. For two triangle if their corresponding sides are equal in length and their corresponding angles are equal in size, we say the triangles are congruent in nature.

The explanation you've provided is more complicated than required. In fact the reasoning behind your explanation (i.e. why the lengths of tangents from a given point to a particular circle will be equal) is also what I've explained above.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/

User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Thu Jan 06, 2011 9:43 am
Anurag@Gurome wrote:
Anurag@Gurome wrote:Image

For the triangles OEB and OEM,, all the corresponding angles are equal in measure. Thus they are similar. Which means (EB/EM) = (OB/OM) = (OE/OE) = 1.


Sir, As far as i know when two triangle are similar ratio of there sides is equal , It does not mean that there ratio is equal to 1, When its ratio is equal to one then we can say that they are congruent.

Take for example A right Angle Triangle with sides 3 , 4 , 5 & 6, 8 , 10

Both these triangles are similar but there ratio of sides are 1/2.
Anurag@Gurome wrote: Hence not only EB = EM, but all the corresponding sides are equal in length. For two triangle if their corresponding sides are equal in length and their corresponding angles are equal in size, we say the triangles are congruent in nature.

The explanation you've provided is more complicated than required. In fact the reasoning behind your explanation (i.e. why the lengths of tangents from a given point to a particular circle will be equal) is also what I've explained above.
I hope i am making my point this time....
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Thu Jan 06, 2011 9:46 am
goyalsau wrote:Sir, As far as i know when two triangle are similar ratio of there sides is equal , It does not mean that there ratio is equal to 1, When its ratio is equal to one then we can say that they are congruent.

Take for example A right Angle Triangle with sides 3 , 4 , 5 & 6, 8 , 10

Both these triangles are similar but there ratio of sides are 1/2

I hope i am making my point this time....
Obviously you can't take the ratio as 1.
But in this case note that the side OE is common two both the triangles. Thus the ratio of the lengths of their corresponding sides will be equal to 1 as (OE/OE) = 1.

Hope it is clear now.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/

User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Thu Jan 06, 2011 9:59 am
Anurag@Gurome wrote:
goyalsau wrote:Sir, As far as i know when two triangle are similar ratio of there sides is equal , It does not mean that there ratio is equal to 1, When its ratio is equal to one then we can say that they are congruent.

Take for example A right Angle Triangle with sides 3 , 4 , 5 & 6, 8 , 10

Both these triangles are similar but there ratio of sides are 1/2

I hope i am making my point this time....
Obviously you can't take the ratio as 1.
But in this case note that the side OE is common two both the triangles. Thus the ratio of the lengths of their corresponding sides will be equal to 1 as (OE/OE) = 1.

Hope it is clear now.
thanks sir, If i will be given the opportunity to choose the person for the next fields Medal , With no second thought You will be my fields medal winner....
:)
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.

Senior | Next Rank: 100 Posts
Posts: 95
Joined: Wed Sep 22, 2010 8:26 am
Thanked: 1 times
Followed by:1 members

by RACHVIK » Fri Jan 07, 2011 7:22 am
Is there a simpler way??
Rachvik

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Fri Jan 07, 2011 9:46 am
RACHVIK wrote:Is there a simpler way??
I think that using the answer choices is the quickest way to solve, as in my post above. I've added a drawing to make the explanation easier to follow. The time to solve was less than 1 minute.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

User avatar
Legendary Member
Posts: 543
Joined: Tue Jun 15, 2010 7:01 pm
Thanked: 147 times
Followed by:3 members

by anshumishra » Sun Jan 09, 2011 8:09 pm
goyalsau wrote:Tough one for me,
Please see the attached diagram.

tan z = 6/(x+y) = 2/y => 6y = 2x+2y => x= 2y

tan 2z = AB/BO = 6/Y

Since; tan 2z = 2tan z/(1 - tan^2 z)
=> 6/y = 2*(2/y)/(1-4/y^2)
=> 6y^2 -24 = 4y^2
=> y^2 = 12
=> y = 2*sqrt(3)
BC = x = 2y = 4*sqrt(3)
Attachments
AB_diagram.png
Thanks
Anshu

(Every mistake is a lesson learned )

User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Sun Jan 09, 2011 10:09 pm
anshumishra wrote:
Since; tan 2z = 2tan z/(1 - tan^2 z)
Is it a formula.....
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.

User avatar
Legendary Member
Posts: 543
Joined: Tue Jun 15, 2010 7:01 pm
Thanked: 147 times
Followed by:3 members

by anshumishra » Mon Jan 10, 2011 4:37 am
goyalsau wrote:
anshumishra wrote:
Since; tan 2z = 2tan z/(1 - tan^2 z)
Is it a formula.....
Yes. Sometimes having a math background, gives you multiple options.
Thanks
Anshu

(Every mistake is a lesson learned )

User avatar
GMAT Instructor
Posts: 1449
Joined: Sat Oct 09, 2010 2:16 pm
Thanked: 59 times
Followed by:33 members

by fskilnik@GMATH » Mon Jan 10, 2011 11:59 am
Hi there!

Please follow the steps in the figure.

Focus: CB = 2x = ?

01. OB = CB/2 = 2x/2 = x because triangles EAD and EBO are similar with ratio of similarity 2:1 in that order.

01. Triangles OBE and OEM are congruent (therefore "arrows for x and 2" are justified) ;

02. Triangle AEM is 30-60-90 (hypothenuse is twice one of the other sides) therefore AM = 2*sqrt(3)

03. Pythagoras on triangle OBA (remember we are looking for 2x):

(2*sqrt(3) + x)^2 = 6^2 + x^2
12+4x*sqrt(3) = 36
6+2x*sqrt(3) = 18
2x*sqrt(3) = 12 then multiplying by sqrt(3) we get 2x = 12*sqrt(3)/3 = 4*sqrt(3)

Regards,
Fabio.

Image
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br