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What is the least possible distance between a point on the [#permalink]
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01 Sep 2008, 04:25
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What is the least possible distance between a point on the circle \(x^2 + y^2 = 1\) and a point on the line \(y = \frac{3}{4}x  3\) ? * \(1.4\) * \(\sqrt{2}\) * \(1.7\) * \(\sqrt{3}\) * \(2.0\) == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



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Re: interesting distance [#permalink]
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01 Sep 2008, 13:24
Hi,
I believe it is = 2 (E).
Reasons: 1. The circle is having its center at origin with a radius of 1. 2. The Line crosses Y axis at 3 and X axis at 4. Put X=0 and Y=0 to get this. 3. If you can do a quick diagram you will know how it will look like. 4. The Minimum distance from any point on the circle to the line should be between a point on the circle's circumference and a point on the line. Circle's point at (1, 0) and line at (3,0) are the ones. So difference is 3  (1) = 2.
You might question, that why not any other point on the circle. For this the diagram would be best to visualize. Still to add if x<0 then from that point if a line is drawn to the circle, the length of this new line(distance) should be > 2. Similarly if x>0 then the distance will also be >2. The line moves away from the circle and hits X axis at (4,0).
Let me know your views.
Regards, Max.



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Re: interesting distance [#permalink]
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01 Sep 2008, 14:08
Good Q.
Equation of the circle with the given line tells us that it is a unit circle (circle with radius 1) with center (0,0)
Some points on the circle would (1,0) (1,0) (0,1) (0,1) and solving for x intercept and y intercept will give us (4,0) and (0,3)
Clearly the distance between (0,1) and (0,3) stands out as the least, which is 2 units.
My answer is E.



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Re: interesting distance [#permalink]
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01 Sep 2008, 14:13
Fasttrack wrote: Hi,
I believe it is = 2 (E).
Reasons: 1. The circle is having its center at origin with a radius of 1. 2. The Line crosses Y axis at 3 and X axis at 4. Put X=0 and Y=0 to get this. 3. If you can do a quick diagram you will know how it will look like. 4. The Minimum distance from any point on the circle to the line should be between a point on the circle's circumference and a point on the line. Circle's point at (1, 0) and line at (3,0) are the ones. So difference is 3  (1) = 2.
You might question, that why not any other point on the circle. For this the diagram would be best to visualize. Still to add if x<0 then from that point if a line is drawn to the circle, the length of this new line(distance) should be > 2. Similarly if x>0 then the distance will also be >2. The line moves away from the circle and hits X axis at (4,0).
Let me know your views.
Regards, Max. I don't agree with you. Shortest distance will be perpendicular line drawn from the given line to the surface of the circle. y=3/4x3 Line perpendicular to above line and passes throw the (0,0). y=4/3x find the intersection point of these two lilnes 4/3*x=3/4*x3 x(25/12)=3 >x= 36/25 y=4/3* 36/25 =48/25 distance between (36/25,48/25) and (0,0) is 60/25=2.4 Shortest distance = 2.41=1.4 A
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Last edited by x2suresh on 01 Sep 2008, 19:35, edited 1 time in total.



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Re: interesting distance [#permalink]
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01 Sep 2008, 18:32
x2suresh wrote: Fasttrack wrote: Hi,
I believe it is = 2 (E).
Reasons: 1. The circle is having its center at origin with a radius of 1. 2. The Line crosses Y axis at 3 and X axis at 4. Put X=0 and Y=0 to get this. 3. If you can do a quick diagram you will know how it will look like. 4. The Minimum distance from any point on the circle to the line should be between a point on the circle's circumference and a point on the line. Circle's point at (1, 0) and line at (3,0) are the ones. So difference is 3  (1) = 2.
You might question, that why not any other point on the circle. For this the diagram would be best to visualize. Still to add if x<0 then from that point if a line is drawn to the circle, the length of this new line(distance) should be > 2. Similarly if x>0 then the distance will also be >2. The line moves away from the circle and hits X axis at (4,0).
Let me know your views.
Regards, Max. I don't agree with you. Shortest distance will be perpendicular line drawn from the given line to the surface of the circle. y=3/4x3 Line perpendicular to above line and passes throw the (0,0). y=4/3x find the intersection point of these two lilnes 4/3*x=3/4*x3 x(25/12)=3 >x= 36/25 y=4/3* 36/25 =48/25 distance between (36/25,48/25) is 60/25=2.4 Shortest distance = 2.41=1.4 A good one ! good solution !



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Re: interesting distance [#permalink]
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01 Sep 2008, 19:09
very well done x2suresh!
I agree with what you have done.
You mentioned an important point about the shortest distance. Shortest distance is the perpendicular distance, which I missed
On a side note, I am looking at the answers and it has sqrt 2 and 1.4
Jeez that's kinda scary as sqrt 2 is 1.414 and I doubt GMAT wants to get to that level of hundreths to test us



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Re: interesting distance [#permalink]
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02 Sep 2008, 02:06
x2suresh wrote: distance between (36/25,48/25) and (0,0) is 60/25=2.4 So, you need to be able to calculate 48^2 + 36^2 without the help of Excel or a calculator, or am I missing something?
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Re: interesting distance [#permalink]
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02 Sep 2008, 03:45
Really tough question.
x2suresh, my question is, why should the perpendicular line pass through (0,0)?



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Re: interesting distance [#permalink]
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02 Sep 2008, 04:10
very nice solution x2suresh. question on one line x2suresh wrote: distance between (36/25,48/25) and (0,0) is 60/25=2.4 did u just recongnise that, or did u compute it very quikcly, or did u grind it the hard way (like i did)
A



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Re: interesting distance [#permalink]
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02 Sep 2008, 06:00
Hi All, Can it be replied by this way Please follow the file attached with the solution As per the circle equation x^2 + y^2 = 1 it shows that circle center is O (0,0) and radius is OC = 1 and as per the line equation y = 3/4x  3 we can see that it is a line passing through IVth Quadrant with intercepts as 4 and 3 on x and y axis respectively As in the figure point O is (0,0), OC = 1, OB = 4 and OA = 3 Our problem is to find CD CD = OD  OC OD = perpendicula distance from O to the line y = 3/4x  3 which is equal to 12/5 = 2.4 therefore CD = 2.4  1 = 1.4 Option is A
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Re: interesting distance [#permalink]
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02 Sep 2008, 06:34
IMO A.
The line has X and Y intercept as 4 and 3. So the line forms a rt triange with the origin with a base = v(4^2+3^2) = 5 Now area of the triange = (1/2)*5* height = (1/2)*4*3 ( rt triange ) => height = 12/5 = 2.4 So, distance from the circle = 2.41= 1.4



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Re: interesting distance [#permalink]
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02 Sep 2008, 06:37
sset009 wrote: very nice solution x2suresh. question on one line x2suresh wrote: distance between (36/25,48/25) and (0,0) is 60/25=2.4 did u just recongnise that, or did u compute it very quikcly, or did u grind it the hard way (like i did)
A
You know that 3^2+4^2=5^2 3:4 you will get 5 as the answer. (Similar to the right angle triangle with sides 3 and 4 then hyponteuous 5) 3:4 >5 36:48 >60 (without calculations) 36/25:48/25 > 60/25
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Re: interesting distance [#permalink]
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02 Sep 2008, 06:39
ssandeepan wrote: IMO A.
The line has X and Y intercept as 4 and 3. So the line forms a rt triange with the origin with a base = v(4^2+3^2) = 5 Now area of the triange = (1/2)*5* height = (1/2)*4*3 ( rt triange ) => height = 12/5 = 2.4 So, distance from the circle = 2.41= 1.4 This is the fastest way .. Good job sandeep.
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Re: interesting distance [#permalink]
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02 Sep 2008, 07:02
ssandeepan wrote: IMO A.
The line has X and Y intercept as 4 and 3. So the line forms a rt triange with the origin with a base = v(4^2+3^2) = 5 Now area of the triange = (1/2)*5* height = (1/2)*4*3 ( rt triange ) => height = 12/5 = 2.4 So, distance from the circle = 2.41= 1.4 Beautiful.
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Re: interesting distance [#permalink]
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02 Sep 2008, 09:34
ssandeepan wrote: IMO A.
The line has X and Y intercept as 4 and 3. So the line forms a rt triange with the origin with a base = v(4^2+3^2) = 5 Now area of the triange = (1/2)*5* height = (1/2)*4*3 ( rt triange ) => height = 12/5 = 2.4 So, distance from the circle = 2.41= 1.4 X2suresh's method made me look back at some line intersection basics (which I learnt in high school)....good application. Sandeep, very good approach.....easy and no big calculations involved.
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Re: interesting distance [#permalink]
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03 Sep 2008, 11:50
Hi Suresh,
I believe you are right. I missed that possibility when I was analyzing. I like Sandeep's method which makes this problem faster to solve.
Great going guys
Max.



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Re: interesting distance [#permalink]
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03 Sep 2008, 12:01
Fasttrack wrote: Hi Suresh,
I believe you are right. I missed that possibility when I was analyzing. I like Sandeep's method which makes this problem faster to solve.
Great going guys
Max. 4. The Minimum distance from any point on the circle to the line should be between a point on the circle's circumference and a point on the line.one more point to add: above situation possible when line is parallel to x axis i.e y=1
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Re: interesting distance [#permalink]
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04 Sep 2008, 07:51
A quick quesiton guys...
why can we use the following approach...
Required distance = Shortest distance from line to center of circle  radius of the circle
i.e distance between the line and (0,0)  radius



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Re: interesting distance [#permalink]
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04 Sep 2008, 08:32
Several people have pointed out an assumption that has been made in previous posts but has not yet been addressed, namely why is the shortest distance along the line that runs through the center of the circle.
I'll try to address that here but first I'll address another theorem that most people take for granted
Theorem: The shortest distance between a line and a point is along a line running through the point and perpendicular to the original line.
Proof: Compare distance along perpendicular line with distance from some other point on the line. Distance along perpendicular line will be one leg of a right triangle but distance from other point will be the hypotenuse of the same triangle so distance from other point will always be greater.
Theorem: Shortest distance from circle to point not on the circle is along the line running through the point and the center of the circle
Proof: Let X be a point not on the circle (remember the circle only is the points on the circumference). Let Y be a point on the circle and also on the line running through the center of the circle and the point X. Let Z be some other point on the circle. (Helpful to draw this). Designate the center of the circle as point C.
Form a triangle by drawing the line segments CZ, ZX, and CX.
Then CZ + ZX > CX since the sum of any two sides of the triangle is greater than the third
But CX = CY + YX
So CZ + ZX > CY +YX
And CZ = CY since both are radii
Therefore ZX > YX for any choice of Z
Therefore YX is always the shortest distance to the circle
These two theorems with a little bit more reasoning explain why the approach taken previous posters calculates the correct minimum distance. In my opinion this line of reasoning, though relying on very elementary properties, is a little more involved than your standard GMAT problem but at the same time is good practice for honing your reasoning skills.



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Re: interesting distance [#permalink]
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04 Sep 2008, 08:57
bhushangiri wrote: A quick quesiton guys...
why can we use the following approach...
Required distance = Shortest distance from line to center of circle  radius of the circle
i.e distance between the line and (0,0)  radius well,i did solve using the formula.. its just that we should be comfortable using the formula..




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