Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 27 Aug 2016, 22:46

### 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 June
Open Detailed Calendar

# What is the least possible distance between a point on the

Author Message
Manager
Joined: 03 Feb 2010
Posts: 68
Followers: 2

Kudos [?]: 91 [0], given: 4

### Show Tags

20 Apr 2010, 09:58
I think a lot of people are making this question way harder than it has to be.

Start by graphing the circle at (0,-1), (1,0), (0,1), (,-1,0)

Then graph the line at (0,-3) and (4,0)

What is the area? Simple...6

Now figure out hypotenuse of the 3, 4, 5 triangle. Are should still be 6 because it is the same triangle. Use 5 as your base and figure out h= 2.4. Now minus the 1 from the circle and it equals 1.4

20 sec problem easy.
Director
Joined: 21 Dec 2009
Posts: 591
Concentration: Entrepreneurship, Finance
Followers: 18

Kudos [?]: 611 [0], given: 20

### Show Tags

20 Apr 2010, 10:04
mmphf wrote:
gmatbull wrote:
ritula wrote:
I tried calculating OD from Tri OBD:
Using 30:60:60 degrees
OD = 2, since OB (side facing 90 deg) = 4
Least distance = 2 – 1 = 1

Again 30:60:90 (x=3/2)
OD = 3/2(\sqrt{3})
OD = 2.5
Least distance = 2.5 – 1 = 1.5

It's not a 30:60:90 triangle. The original triangle (your OAB) is a pythagorean triple 3:4:5. You can find OD by getting the height of the original triangle, using AB as a base.

Thanks, i got it now.
_________________

KUDOS me if you feel my contribution has helped you.

Intern
Joined: 20 Apr 2010
Posts: 8
Location: Europe/Netherlands
Schools: HBS, Booth, Tuck, Fuqua, LBS
WE 1: International 4 years Industry
Followers: 0

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

### Show Tags

20 Apr 2010, 13:26
fully agree with ksharma.

Just plot the points on x and y-axis for the two equations (1,1,-1,-1...4,-3 ) and you'll easily find the solution from the height of the 3,4,5 triangle.

easy under 1min question.
Manager
Joined: 27 Feb 2010
Posts: 105
Location: Denver
Followers: 1

Kudos [?]: 307 [0], given: 14

### Show Tags

20 Apr 2010, 14:53
Only 10% of Gmat takers will be able to get to it in less than 1.5 mins. I think in real GMAT, educated guess is the best way to approach this.
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 220

Kudos [?]: 1510 [0], given: 235

### Show Tags

20 Apr 2010, 15:58
Least distance between the circle and the given line will be the perpendicular distance between the given line and the tangent at a point on the circle where it is parallel to the given line.

Thus the required distance = Distance of given line from origin - radius

= 2.4-1 = 1.4

Reason : Line drawn form the center to tangent will be perpendicular and now if we extend it to the given line it will be perpendicular to the given line as well.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Manager
Joined: 04 Dec 2009
Posts: 71
Location: INDIA
Followers: 2

Kudos [?]: 9 [0], given: 4

### Show Tags

20 Apr 2010, 21:11
basik21 wrote:
Common guys, it's not as tough as seems to be...
just easy, X^2+Y^2=1 is circle with center (0,0) and radius 1, just simply x=0, y= +/-1, and y=0, x=+/-1, and y=3/4-3 is just a line, which forms right triangle(3,4,5) with origin.
Assume that Z is perpendicular to hypothenuse, r is radius of circle, which is 1, d is Distance from our circle to line.
so Area of this triangle is 1/2*3*4=1/2*5*Z, so Z is equal to 12/5.
And, Z=r+d=>12/5=1+d=> d=7/5=1.4

Good one , really simple method , thanks
_________________

MBA (Mind , Body and Attitude )

Manager
Affiliations: CPA, CFA
Joined: 18 Apr 2010
Posts: 76
WE 1: credit analyst 4yrs
WE 2: equity analyst 1yr
WE 3: tax associate 7months
Followers: 3

Kudos [?]: 26 [0], given: 25

### Show Tags

21 Apr 2010, 05:48
Area of triangle = 1/2 base * height
=> 1/2 * 3 * 4 = 1/2 * 5 * perpendicular on hypotenuse
=> perpendicular = 2.4

=> distance from circle = 1.4 (i.e. 2.4 - radius)
_________________

http://gmatclub.com/forum/my-gmat-experience-750-q-51-v-40-awa-98031.html

Manager
Joined: 30 Jun 2004
Posts: 177
Location: Singapore
Followers: 1

Kudos [?]: 23 [0], given: 5

### Show Tags

15 Jun 2010, 20:30
Thanks to all the guys who provided the solution.
Intern
Joined: 10 Aug 2009
Posts: 2
Followers: 0

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

### Show Tags

06 Sep 2010, 18:39
Prasun, how r u calculatin this?

the length of the perpendicular from the centre(0,0)= 3/ sq root?
Manager
Joined: 17 Mar 2010
Posts: 186
Followers: 3

Kudos [?]: 146 [0], given: 9

### Show Tags

07 Sep 2010, 00:13
absolutely amazing explaination
SVP
Joined: 16 Nov 2010
Posts: 1673
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 34

Kudos [?]: 471 [4] , given: 36

### Show Tags

22 Apr 2011, 04:34
4
KUDOS

x intercept (OA) = 4

y intercept (OB) = 3

5 - Hypotenuse

1/2 * OD * 5 = 1/2 * 12

=> OD = 12/5

So distance = 12/5 - 1 = 7/5

= 1.4

Attachments

Circle.png [ 8.06 KiB | Viewed 1079 times ]

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2021
Followers: 157

Kudos [?]: 1560 [0], given: 376

### Show Tags

22 Apr 2011, 05:03
ritula wrote:
What is the least possible distance between a point on the circle $$x^2 + y^2 = 1$$ and a point on the line $$y = \frac{3}{4}x - 3$$ ?

(A) $$1.4$$
(B) $$\sqrt{2}$$
(C) $$1.7$$
(D) $$\sqrt{3}$$
(E) $$2.0$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

Similar post:
tough-tricky-set-of-problms-85211.html#p638361
_________________
Manager
Joined: 20 Jan 2011
Posts: 87
Followers: 3

Kudos [?]: 40 [0], given: 23

### Show Tags

22 Apr 2011, 14:05
How do we know the perpendicular is passing through origin?
I could not figure our that. I missed something.

Anyway I used derivative(have some blurred memories of this:) to minimize distance.

Say point (a,b) on circle which is nearest to the line
Now distance between a line and a point is
d = |b-ma-c|/sqrt(m^2+1) where m is the slope of line and c is intercept

= |b-(3/4)a +3|/ sqrt( (3/4)^2 +1)

(a,b) on circle, hence a^2 + b^2 = 1 => b = sqrt(1-a^2)

Use this in distance eqn and minimize the numerator using derivative.
I wont go into that. Just do df/da = 0 and remember a should be + and b should be - as this point is in 4th quad.

So I got a = 3/5 and b = -4/5, this is the point of tangent to the circle parallel to given line

d = 1.4
_________________

Conquer the Hell and make it Haven. Brain is your hell and Success is your haven!

"Kudos" is significant part of GMAT prep. If you like it, you just click it

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6830
Location: Pune, India
Followers: 1926

Kudos [?]: 11961 [0], given: 221

### Show Tags

22 Apr 2011, 14:54
hellishbrain wrote:
How do we know the perpendicular is passing through origin?
I could not figure our that. I missed something.

The point is not whether the perpendicular on the line will pass through the center. The point is that the perpendicular from the center will give the minimum distance. Look at the diagram below and forget the circle for the time being. What is the minimum distance from the center (0,0) to the line? It will the perpendicular from the center to the line, right? Now think, what will be the shortest distance from the circle to the line? It will be 1 unit less than the distance from the center to the line, isn't it?

Attachment:

Ques2.jpg [ 17.76 KiB | Viewed 857 times ]

Since hypotenuse is 5,
(1/2)*3*4 = (1/2)*5*x
x = 2.4
So minimum distance is 2.4 - 1 = 1.4
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 20 Jan 2011 Posts: 87 Followers: 3 Kudos [?]: 40 [0], given: 23 Re: least distance [#permalink] ### Show Tags 22 Apr 2011, 15:13 Thanks. So take away is for any circle(centered anywhere) and any line, the perpendicular from the centre of circle to line would give the minimum distance. Right? _________________ Conquer the Hell and make it Haven. Brain is your hell and Success is your haven! "Kudos" is significant part of GMAT prep. If you like it, you just click it Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6830 Location: Pune, India Followers: 1926 Kudos [?]: 11961 [0], given: 221 Re: least distance [#permalink] ### Show Tags 22 Apr 2011, 15:20 hellishbrain wrote: Thanks. So take away is for any circle(centered anywhere) and any line, the perpendicular from the centre of circle to line would give the minimum distance. Right? Certainly! Just forget the circle itself. Think about the point at the center of the circle. If I had to find the minimum distance from that point to a line, it would be the perpendicular distance. So minimum distance of any point on the circle to the line would be that distance minus the radius. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Manager
Joined: 25 Aug 2008
Posts: 230
Location: India
WE 1: 3.75 IT
WE 2: 1.0 IT
Followers: 2

Kudos [?]: 50 [0], given: 5

### Show Tags

22 Apr 2011, 20:05
Thanks for the explanation guys and various methods to solve it..
_________________

Cheers,
Varun

If you like my post, give me KUDOS!!

Intern
Joined: 16 Apr 2010
Posts: 2
Followers: 0

Kudos [?]: 5 [0], given: 0

### Show Tags

22 Apr 2011, 20:27
Alternately, being a right angle triangle, with base can be taken as either of sides and height will be the perpendicular from vertex left aside.
Area = 0.5 * base * height = 0.5 * 3 * 4 = 6 (here we compute area)
Area = 0.5 * base * height = 0.5 * 5 * h, thus h = 2.4. However, h should be reduced by 1 as it's radius of circle. Hence, distance between circle and line is 1.4.
Intern
Joined: 11 Apr 2011
Posts: 32
Location: London
WE 1: 3 yrs f/t manager in aerospace (IT - Commercial - Int'l Mktg)
WE 2: 5 yrs freelance consultant to Gov't
Followers: 0

Kudos [?]: 3 [0], given: 3

### Show Tags

23 Apr 2011, 13:10
atletikos wrote:
yes, A
jallenmorris, nice explanation, but I'm not sure this way won't take more than 2 min.
I think the shortest way is as following:

We have right triangle with sides 3,4,5. Area is 1/2 (3x4) = 6. The shortest distance from the origin to the line y=3/4x-3 has to be perpendicular to this line and is 6/(5/2)=2,4. Thus, subtracting radius we have the answer: 2,4-1=1,4

great work!! it goes to show that there's usually a quick way hiding somewhere in a triangles question
_________________

Blogging my way through the applications/study process!!
http://mbahorizons.blogspot.com/

Director
Joined: 01 Feb 2011
Posts: 757
Followers: 14

Kudos [?]: 100 [0], given: 42

### Show Tags

23 Apr 2011, 14:04
we can see that this line forms a right angled triangle with origin.

let perpendicular distance from center of the circle/origin to line be h.

by comparing area of the triangle we have

1/2 * 4 * 3 = 1/2 * h * 5

h = 12/5 =2.4

minimum distance from the circle to the line = 2.4 -1 = 1.4
Re: least distance   [#permalink] 23 Apr 2011, 14:04

Go to page   Previous    1   2   3    Next  [ 42 posts ]

Similar topics Replies Last post
Similar
Topics:
3 If 25% of the company's employees contribute at least 4% of 7 30 Jan 2011, 13:12
M01 #20 - Contradiction between between the two statement? 5 20 Aug 2009, 20:37
22 least value of N (m09q33) 16 18 Mar 2009, 03:41
8 distance (m03q28) 26 27 Jan 2008, 23:26
Area of a rectangle is 80 What is the angle between the 10 23 Oct 2007, 12:42
Display posts from previous: Sort by

# What is the least possible distance between a point on the

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.