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In the x-y plane, point (p, q) is a lattice point if both p and q are integers. Circle C has a center at (–2, 1) and a radius of 6. Some points, such as the center (–2, 1), are inside the circle, but a point such as (4, 1) is on the circle but not in the circle. How many lattice points are in circle C? (A) 36 (B) 72 (C) 89 (D) 96 (E) 109

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07 May 2014, 02:10

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mikemcgarry wrote:

In the x-y plane, point (p, q) is a lattice point if both p and q are integers. Circle C has a center at (–2, 1) and a radius of 6. Some points, such as the center (–2, 1), are inside the circle, but a point such as (4, 1) is on the circle but not in the circle. How many lattice points are in circle C? (A) 36 (B) 72 (C) 89 (D) 96 (E) 109

Very basic question - Do such difficult questions really come on GMAT ? When I look at OG , I get a confidence boost but questions like these along with the time limit of ~ 2min / question make it difficult

Very basic question - Do such difficult questions really come on GMAT ? When I look at OG , I get a confidence boost but questions like these along with the time limit of ~ 2min / question make it difficult

Dear himanshujovi, I'm happy to respond.

I wrote this to be a super-challenging question. Could it appear on the GMAT? Maybe. If it did appear, it would about the hardest thing the GMAT would throw at you --- you would have to be acing everything else on the math section in order for the CAT to cough up a problem such as this in your direction. DO NOT interpret this as a typical GMAT math problem. DO NOT interpret this even as a problem necessary to break the 700 threshold. This would be more exemplary of the type of math problem you would have to solve to get the mythical 60 on the Quant sub-score. It's at the outer limit of what the GMAT even conceivably could ask. This can be solved in < 2 minutes, but that takes some serious intuition and reasoning. I show what I consider the most efficient solution in the blog at that link.

If I may say, your confidence should not be hinged to each new problem that comes along. In particular, if you have already solved hundreds of problems correctly, don't let a single new hard problem shake you. Confidence should be made of sterner stuff. If confidence is only based in external results, it will sway with the wind. Believe in yourself. Develop confidence from within. That's the confidence you need for excellence.

Does all this make sense? Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

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07 May 2014, 20:19

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Mike,

Here is how I solved the problem and used my intuition to guess the answer :

1. The range of x is 13 as it is for y 2. Therefore, a hypothetical square that inscribes the circle can have 169 (13 * 13) different co-ordinates including the ones that are on the square. 3. Since area of the circle constitutes most of the area inside the square, we can assume that number of lattice points in the circle should be close to this number 169.

Therefore I selected 109 as the plausible answer to this question.

Let me know if this is a right approach to solve such problems in less than 2 mins.

In the x-y plane, point (p, q) is a lattice point if both p and q are integers. Circle C has a center at (–2, 1) and a radius of 6. Some points, such as the center (–2, 1), are inside the circle, but a point such as (4, 1) is on the circle but not in the circle. How many lattice points are in circle C? (A) 36 (B) 72 (C) 89 (D) 96 (E) 109

The lattice points are all points with integer coordinates. They are evenly spread across the xy plane. Every time you move one step up/down/left or right from a lattice point, you reach another lattice point. So (-2, 1) which is a lattice point is no different from (0, 0) - an easier-to-work-with lattice point. So say the center of the circle is at (0, 0). Draw the circle. Find the coordinate of point P (which splits the quadrant into two equal halves) since that will tell us how many lattice points are inside the circle.

Attachment:

Ques3.jpg [ 25.21 KiB | Viewed 7866 times ]

P lies on the circle and the green line y = x \(x^2 + y^2 = 6^2\) \(2x^2 = 36\) \(x = 3\sqrt{2} = 4.2\) approx

So P is (4.2, 4.2). All points with equal x and y coordinates will lie on the green line i.e. (1, 1), (2, 2), (3, 3) and (4,4) and these 4 will be inside the circle. Now count the points inside the circle below the green line in the first quadrant. They are 5+4+3+2 (shown by black dots) = 14 Similarly there will be 14 points on the other side of the green line in the first quadrant. First quadrant will have a total of 4 + 14 + 14 = 32 points. All four quadrants will have 4*32 = 128 points but note that the 5 points on the x axis and y axis are double counted so subtract 4*5 = 20 out of the total 128 to get 108 points. But we did not count (0, 0) here so we have a total of 109 points.

Here is how I solved the problem and used my intuition to guess the answer : 1. The range of x is 13 as it is for y 2. Therefore, a hypothetical square that inscribes the circle can have 169 (13 * 13) different co-ordinates including the ones that are on the square. 3. Since area of the circle constitutes most of the area inside the square, we can assume that number of lattice points in the circle should be close to this number 169.

Therefore I selected 109 as the plausible answer to this question.

Let me know if this is a right approach to solve such problems in less than 2 mins.

Thanks

Dear kinghyts, Well, my friend, that's a guess answer. It's true that the number of points in the circle has to be less than 169, and some proportion of 169 roughly equivalent to the proportion that a circle takes up of a square, [pi]/4. The trouble is, [pi]/4*169 is very cumbersome to approximate, and if the numbers were a little closer together and the correct answer were not (E), then this method would have utterly failed.

See the brilliant solution of the great scholar VeritasPrepKarishma directly above.

Does all this make sense? Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

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04 Aug 2015, 02:54

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I have come up with a solution that may not be foolproof but that could help in solving this question quickly. This question can be interpreted as the number of unit squares that are present in the circle of area 36pi. Now the approximate value of 36pi is 113 which is close to 109.So E Please feel free to suggest flaws in this method(if any).

I have come up with a solution that may not be foolproof but that could help in solving this question quickly. This question can be interpreted as the number of unit squares that are present in the circle of area 36pi. Now the approximate value of 36pi is 113 which is close to 109.So E Please feel free to suggest flaws in this method(if any).

Dear tvrs09, I'm happy to respond! Bravo, my friend! That's a wonderful solution to this problem! Of course, the approximation method is successful, because the answers are spaced out relatively far. If the problem had answer choices such as {105, 109, 112, ...}, then approximation would not be successful. This is already a very challenging problem though, so I seriously doubt whether the GMAT would give a problem this hard and also make the answer choices so close: that is just not their style. Great solution, my friend! Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

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04 Aug 2015, 21:48

Not sure if this works but seems the easiest way would be to find area of the circle=36Pi and subtract the 4 lattice points on the circle. Maybe it only works in this example and not always.

Not sure if this works but seems the easiest way would be to find area of the circle=36Pi and subtract the 4 lattice points on the circle. Maybe it only works in this example and not always.

Dear jakeqs, As I expressed to tvrs09 above, what you are suggesting is an approximation method. Estimation is a perfectly fine strategy for the GMAT Quant. See: http://magoosh.com/gmat/2012/the-power- ... mat-quant/ If the answers are spread out, then estimation has a better chance of determining a unique answer. In what you suggested here, it was a totally lucky fluke that it wound up with the exact answer. Often, estimation does not produce an exact answer, but that's OK. Again, if the answer choices are widely space apart, then estimation is often enough to determine a single answer. Does all this make sense? Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

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05 Aug 2015, 19:41

mikemcgarry wrote:

jakeqs wrote:

Not sure if this works but seems the easiest way would be to find area of the circle=36Pi and subtract the 4 lattice points on the circle. Maybe it only works in this example and not always.

Dear jakeqs, As I expressed to tvrs09 above, what you are suggesting is an approximation method. Estimation is a perfectly fine strategy for the GMAT Quant. See: http://magoosh.com/gmat/2012/the-power- ... mat-quant/ If the answers are spread out, then estimation has a better chance of determining a unique answer. In what you suggested here, it was a totally lucky fluke that it wound up with the exact answer. Often, estimation does not produce an exact answer, but that's OK. Again, if the answer choices are widely space apart, then estimation is often enough to determine a single answer. Does all this make sense? Mike

I understand this, but I am trying to understand logically why it wouldn't give an exact answer all the time. If you have a symmetrical shape centered at a lattice point, how could the number of lattice points not equal the total area minus the lattice points on the shape?

I understand this, but I am trying to understand logically why it wouldn't give an exact answer all the time. If you have a symmetrical shape centered at a lattice point, how could the number of lattice points not equal the total area minus the lattice points on the shape?

Dear jakeqs, I'm happy to respond. There are a few different way to answer your intriguing question.

I guess one way is to say: I don't know how deeply you understand the circle. I don't know, for example, if you are familiar with the ancient problem of trying to square the circle, a problem that deeply frustrated the ancient geometers & Leonardo da Vinci (possible the smartest human being ever) and eventually was proven insoluble in the 19th century, because it turns out that pi is a transcendental number, not an algebraic number. In other words, the nature of pi and the nature of the circle are 100% completely incommensurate with our linear way of thinking. The more one appreciates the deeper realms of mathematics, the more one sees that the circle is not at all to be underestimated. Now, if the beginner knows the basic formulas, C = 2(pi)r and A = (pi)r^2, that beginner might really feel that he knows almost everything there is to know and appreciate about the circle, and in a profound way, this is a kind of hubris. There are many aspects of the circle that are very difficult to understand, and some things (such as squaring the circle) that are mathematically impossible to accomplish. Of course, all that is well beyond anything the the GMAT would ever ask, but your question touched deeply into these issue. Those two formulas are a very large part of what the GMAT expects you to know about the circle, but they are less than 1% of what there is to understand about the circle in the broader world of higher mathematics.

A more straightforward way to answer is to say: imagine a circle in the coordinate plan slowly increasing in radius. Think about the lattice points that are included as it increases:

Attachment:

circle increasing in Cartesian plane.JPG [ 40.01 KiB | Viewed 6744 times ]

The number of lattice points would not increase smoothly or linearly. Instead, at certain radii, a bunch of lattice points could be outside the circle, and then when the radius nudges a little bigger, suddenly a whole bunch get included. The jumps in the number of lattice points are discrete and irregular, and does not reflect the smooth increase in area at all. At a very profound level, there is absolutely no reason that the area of the circle (which is almost always an irrational number) would be equal to the number of lattice points contained (which is always a positive integer). There, we would be talking not only about the incommensurate nature of the linear vs. the circular, but also the incommensurate nature of of integers vs. irrational numbers. The span in incongruent logical realms is truly mind-boggling. Expecting those two to be exactly equal would be like expecting an octopus to full in love with a cockroach.

Now, even though there is essentially no way that these two numbers could be exactly equal, the area of the circle is an excellent general approximation for the number of lattice points, especially if the circle contains dozens of lattice points, and especially if the answer choices are spaced out.

Even though you don't need to know all the advanced stuff about the circle, it's very good to appreciate what things in mathematics you never should underestimate. The circle is one of those topics.

Does all this make sense? Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

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05 Oct 2015, 02:22

Easier way to look at the problem is:

Translate the figure to origin and try solving. It will be the same.

We got a figure which is symmetrical. I mean if u locate a lattice point in the circle randomly, there WILL be a corresponding point in some other quadrant for sure, except for the center(origin).

The random point will either have 3 corresponding points(4 points of a kind) or 1 corresponding point(2 points of a kind).

Which means the answer will be : 4*x + 2*y + 1(origin) = even number + 1 = odd number. Look for odd options.

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07 Jan 2016, 14:41

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mikemcgarry wrote:

jakeqs wrote:

Not sure if this works but seems the easiest way would be to find area of the circle=36Pi and subtract the 4 lattice points on the circle. Maybe it only works in this example and not always.

Dear jakeqs, As I expressed to tvrs09 above, what you are suggesting is an approximation method. Estimation is a perfectly fine strategy for the GMAT Quant. If the answers are spread out, then estimation has a better chance of determining a unique answer. In what you suggested here, it was a totally lucky fluke that it wound up with the exact answer. Often, estimation does not produce an exact answer, but that's OK. Again, if the answer choices are widely space apart, then estimation is often enough to determine a single answer. Does all this make sense? Mike

Mike,

This is a beautiful question. I approached this sort of like a permutations problem. We have x^2 + y^2 < 36 for points to lie inside the circle. Taking integer squares we can consider 0,1,2,3,4,5,-1,-2,-3,-4,-5. 11 in total. For 0,1,2,3,-1,-2,-3: we can have 11 possible pairs (including itself): 11*7 = 77 For 4,-4: we can't have 5 or -5: 2 * (11-2) = 18 For 5,-5: we can't have 5,-5,4,-4: 2 * (11-4) = 14 77 + 18 + 14 = 109. Hope this is also an alternate simple method.

This is a beautiful question. I approached this sort of like a permutations problem. We have x^2 + y^2 < 36 for points to lie inside the circle. Taking integer squares we can consider 0,1,2,3,4,5,-1,-2,-3,-4,-5. 11 in total. For 0,1,2,3,-1,-2,-3: we can have 11 possible pairs (including itself): 11*7 = 77 For 4,-4: we can't have 5 or -5: 2 * (11-2) = 18 For 5,-5: we can't have 5,-5,4,-4: 2 * (11-4) = 14 77 + 18 + 14 = 109. Hope this is also an alternate simple method.

Dear krazyvshank I'm happy to reply and I'm glad you appreciated the question!

I'm quite intrigued by your approach. It took me a little while to follow your reasoning. I believe you are saying that the x-values go from -3 to +3, and for each x value, there are 11 possible y-values. Essentially, you are making a vertical rectangle of lattice points, 7 x 11, with 77 points. Then you are attaching vertical strips on the sides of the rectangle in subsequent steps. Intriguing.

Just a piece of advice. When you are discussing your thinking in the coordinate plane, use x and y in your explanation, so others can visualize what you are visualizing.

Here's a related way that may be quicker. Think of all the x & y values between -4 and +4. These make a square of lattice points, 9 x 9, or 81 points, all inside the circle. This big square just has a single short strip on each side. Go to x = 5: this could have y = {-3, -2, -1, 0, 1, 2, 3}, seven possible point. Each side of the square has this extra strip of 7 points. That's 28 points, added to the square of 81, gives us 109.

You produced a very interesting approach.

Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

In the x-y plane, point (p, q) is a lattice point if both p [#permalink]

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07 Jan 2016, 17:56

mikemcgarry wrote:

krazyvshank wrote:

Mike,

This is a beautiful question. I approached this sort of like a permutations problem. We have x^2 + y^2 < 36 for points to lie inside the circle. Taking integer squares we can consider 0,1,2,3,4,5,-1,-2,-3,-4,-5. 11 in total. For 0,1,2,3,-1,-2,-3: we can have 11 possible pairs (including itself): 11*7 = 77 For 4,-4: we can't have 5 or -5: 2 * (11-2) = 18 For 5,-5: we can't have 5,-5,4,-4: 2 * (11-4) = 14 77 + 18 + 14 = 109. Hope this is also an alternate simple method.

Dear krazyvshank I'm happy to reply and I'm glad you appreciated the question!

I'm quite intrigued by your approach. It took me a little while to follow your reasoning. I believe you are saying that the x-values go from -3 to +3, and for each x value, there are 11 possible y-values. Essentially, you are making a vertical rectangle of lattice points, 7 x 11, with 77 points. Then you are attaching vertical strips on the sides of the rectangle in subsequent steps. Intriguing.

Just a piece of advice. When you are discussing your thinking in the coordinate plane, use x and y in your explanation, so others can visualize what you are visualizing.

Here's a related way that may be quicker. Think of all the x & y values between -4 and +4. These make a square of lattice points, 9 x 9, or 81 points, all inside the circle. This big square just has a single short strip on each side. Go to x = 5: this could have y = {-3, -2, -1, 0, 1, 2, 3}, seven possible point. Each side of the square has this extra strip of 7 points. That's 28 points, added to the square of 81, gives us 109.

You produced a very interesting approach.

Mike

Thanks Mike. I will take your advice! I wasn't visualizing any lattice points at all in my approach. My solution is based on the equation of a circle with radius of 6 - (X^2 + Y^2 =36). I was trying to find integer pairs (X,Y) that would satisfy the inequality X^2 + Y^2 <36 for those points to lie inside the circle and not on the circumference. The maximum possible integer value for either X or Y here would be a 5 with the other value 0 {(5,0) or (0,5)}. With this approach, we can see that the possible integer values for X and Y are -5 to +5.

Of these 11 values, -3 to +3 (7 values) can form an (X,Y) pair with each of the 11 and satisfy the inequality. Hence, the 7*11 = 77. -4 and +4 can form an (X,Y) pair with all but 2 of the 11 (pairing 4s with the 5s will not satisfy the inequality). 2*9 = 18. -5 and +5 cannot be paired with 4s and 3s. 2*7 = 14. Hence the total of 109.

Hope my explanation makes better sense now

Last edited by krazyvshank on 12 Jan 2016, 14:40, edited 1 time in total.

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12 Jan 2016, 13:40

mikemcgarry wrote:

In the x-y plane, point (p, q) is a lattice point if both p and q are integers. Circle C has a center at (–2, 1) and a radius of 6. Some points, such as the center (–2, 1), are inside the circle, but a point such as (4, 1) is on the circle but not in the circle. How many lattice points are in circle C? (A) 36 (B) 72 (C) 89 (D) 96 (E) 109

I spend good one and half minute on this question, and could not find any way to answer it in given two minute limit. So I took the complete logical guessing method in which i thought all the four section of a circle will have same number of points PLUS one for the origin. So it should a odd number. The only ODD number there is -- E. I guess it worked.

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12 Jan 2016, 14:30

neeraj609 wrote:

mikemcgarry wrote:

In the x-y plane, point (p, q) is a lattice point if both p and q are integers. Circle C has a center at (–2, 1) and a radius of 6. Some points, such as the center (–2, 1), are inside the circle, but a point such as (4, 1) is on the circle but not in the circle. How many lattice points are in circle C? (A) 36 (B) 72 (C) 89 (D) 96 (E) 109

For a discussion of difficult Coordinate Geometry questions, as well as the full solution to this particular question, see magoosh. com/gmat/2014/challenging-coordinate-geometry-practice-questions/

Mike

I spend good one and half minute on this question, and could not find any way to answer it in given two minute limit. So I took the complete logical guessing method in which i thought all the four section of a circle will have same number of points PLUS one for the origin. So it should a odd number. The only ODD number there is -- E. I guess it worked.

But great solution by Karishma!!!

That is indeed a very good approach but there are two odd options and both of them are 1 more than a multiple of 4.

Re: In the x-y plane, point (p, q) is a lattice point if both p [#permalink]

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09 Sep 2016, 09:25

mikemcgarry wrote:

krazyvshank wrote:

Mike,

This is a beautiful question. I approached this sort of like a permutations problem. We have x^2 + y^2 < 36 for points to lie inside the circle. Taking integer squares we can consider 0,1,2,3,4,5,-1,-2,-3,-4,-5. 11 in total. For 0,1,2,3,-1,-2,-3: we can have 11 possible pairs (including itself): 11*7 = 77 For 4,-4: we can't have 5 or -5: 2 * (11-2) = 18 For 5,-5: we can't have 5,-5,4,-4: 2 * (11-4) = 14 77 + 18 + 14 = 109. Hope this is also an alternate simple method.

Dear krazyvshank I'm happy to reply and I'm glad you appreciated the question!

I'm quite intrigued by your approach. It took me a little while to follow your reasoning. I believe you are saying that the x-values go from -3 to +3, and for each x value, there are 11 possible y-values. Essentially, you are making a vertical rectangle of lattice points, 7 x 11, with 77 points. Then you are attaching vertical strips on the sides of the rectangle in subsequent steps. Intriguing.

Just a piece of advice. When you are discussing your thinking in the coordinate plane, use x and y in your explanation, so others can visualize what you are visualizing.

Here's a related way that may be quicker. Think of all the x & y values between -4 and +4. These make a square of lattice points, 9 x 9, or 81 points, all inside the circle. This big square just has a single short strip on each side. Go to x = 5: this could have y = {-3, -2, -1, 0, 1, 2, 3}, seven possible point. Each side of the square has this extra strip of 7 points. That's 28 points, added to the square of 81, gives us 109.

You produced a very interesting approach.

Mike

Hi Mike, I am struggling with a point. After getting 81 point I am going to X=5. But what is the guarantee that X=5 will take value from -3 to +3? (5,3) might be out of the circle. Will you explain this. Thanks in advance.

Hi Mike, I am struggling with a point. After getting 81 point I am going to X=5. But what is the guarantee that X=5 will take value from -3 to +3? (5,3) might be out of the circle. Will you explain this. Thanks in advance.

I'm happy to respond. Let's pretend that the center is (0, 0) and r = 6. All the points inside the circle must have a distance to the origin less than 6.

We agree that the lattice square, from -4 to +4 in both x & y, we have a square of 81 lattice points.

Now, go out to x = 5. How far is the point (5,3) from the origin? Well, we use the Pythagorean Theorem.

Attachment:

distance to (5, 3).JPG [ 15.23 KiB | Viewed 1251 times ]

The distance from the origin to (5,3) is the hypotenuse, and the x & y distances, 5 & 3, are the legs.

\(d^2 = 5^2 + 3^2\)

\(d^2 = 25 + 9 = 34\)

Incidentally, setting up a right triangle and using the Pythagorean Theorem is always the best way to find distance in the coordinate plane. We don't need to calculate an exact value of this distance. Whatever the square root of 34 is, it's less than the square root of 36, which is 6. Thus, the distance from (0, 0) to (5, 3) is less than 6. We can include a strip of seven point along the right side, from (5, 3) down to (5, -3). By symmetry, we could have such a strip on each side of the square: right, left, top, and bottom. A strip of seven on each of the four side gives us 28 extra points.

Attachment:

circle with lattice points.JPG [ 48.17 KiB | Viewed 1251 times ]

In that diagram, the dark purple points are the points in the square of 81 points. Along each of the four sides of the square, we have that strip of 7 extra points in bright green. Notice that (5, 3) and the other endpoints of those strips are just inside the circle, because sqrt(34) is just less than 6.

Does all this make sense? Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)