<p>In the xy-coordinate plane, how many points are a distance of 4 units from the origin?</p>
<p>(A) One
(B) Two
(C) Three
(D) Four
(E) More than four</p>
<p>The answer is E. Can someone explain why this is so?</p>
<p>Think of a circle with a radius of 4 and its center at the origin. Any point on that circle is 4 units away from the origin. (infinity)</p>
<p>See this graph that I (horribly) made on paint.
<a href=“uploadpad.com is available for purchase - Sedo.com”>uploadpad.com is available for purchase - Sedo.com;
<p>Count on the green line and you’ll see that the blue dot is four units away from the origin. Then you will think that there are four so your choice might be (D). However, if you count the orange lines, you will also see that there are many more points then the four blue ones. That makes the right answer choice (E). Hope this helped :)</p>
<p>Not quite technically correct, Legatus. Distance is a straight line in static Euclidean geometry because displacement is not introduced without the concept of vectors, which are not considered in SAT mathematics. If we were talking about a more advanced mathematics or physics test, you would be right not to consider distance as displacement. Perhaps it’s worth telling the collegeboard to start specifying that they mean “displacement” when they say “distance,” but it seems like it could be inferred. As Jeffery correctly noted, they’re referring specifically to the infinite number of points on the circle satisfied by r=4.</p>
<p>Does the distance of 4 units limited to moving each unit at once? In any case, the answer is E.</p>
<p>I agree with JimboSteve that the question implies the distance as in change in position.</p>
<p>@Jumbo Hmm… thats strange. I remember getting it from somewhere >.< … but then again you seem to be correct. Thanks for putting me straight. :)</p>
<p>It would be an interesting thing to ask… is the infinity of points with distance four units from the origin greater than the infinity of points with displacement four units from the origin? I like the idea of comparing two uncountably infinite sets. It certainly seems like the answer is yes, but…</p>
<p>They are equal, I believe. If you mean what I think you mean (so that the displacement set would be shaped like a diamond) you can set up a one-to-one correspondence using the angle as measured from the origin.</p>
<p>But it would be more than the diamond, because it would include every point that you can reach by moving four units. For example, (0,0) to (0,0) is a distance of four units… if you move <2,0> and then <-2,0>. So it would contain the total area of the circle. But since both are defined on R^2 from [-2,2] and [-2,2] there is an infinite uncountable set associated with both (they don’t have cardinality with N) that implies an equal number of discrete points, which is counterintuitive. Weird stuff.</p>
<p>You are certainly opening up a can of (infinite) worms! </p>
<p>I haven’t worked it out yet but I would bet that the infinite set of points you can reach traveling a distance 4 units can be put in 1-1 correspondance with the points you can reach with a displacement of 4 units. I believe it comes down to this: are there more points on the interior of a circle than on the boundary? But I am pretty sure that I remember seeing a proof that a unit segment and a unit square can be put in 1-1 correspondance. This seems similar…</p>
<p>And of course…not on the SAT :)</p>
<p>Even with the whole region, you can still put the two sets in one-to-one correspondence. If we instead had the unit square and a unit segment, we could do it like this:</p>
<p>Take the coordinates of any point on a unit square (such as 0.0400000…, 0.386749687…). Combine them by alternating digits into one number (0.034806070409…). This is a number between 0 and 1! Every number can be matched with exactly one pair of coordinates (and vice versa) in this way, so the sets should be “equal”, whatever that means. </p>
<p>Since the circle and the circumference in this problem aren’t fundamentally different, the answer should be the same.</p>
<p>(you can also do more crazy stuff, such as matching the unit interval with all real numbers using tan(x), which matches [-pi/2, pi/2] with the reals and then scaling it down to [0,1]. and even crazier, you can draw a one-dimension curve which fills all the ‘space’ in a two dimensional shape! it’s called a peano curve:</p>
<p><a href=“http://en.wikipedia.org/wiki/Space-filling_curve[/url]”>http://en.wikipedia.org/wiki/Space-filling_curve</a></p>
<p>)</p>
<p>^ That was the proof! And it can easily be applied to this case as well. Every point on the interior of a circle can be expressed in polar coordinates as a distance and an angle (in radians.) The points on the circle can be specified with just a single number – either radian angle or arc length along the curve, whichever you prefer. Then, you can do the same “interweaving” of digits conquorer7 describes. </p>
<p>Strangely, this method works in three (or more) dimensions as well! So there are no more points in a sphere (or even in a hyper-sphere) than there are on a circle. It seems so wrong! The number of points on a segment 1 mm long is no more infinite than the number of points in all of space? No kidding, I am pretty sure that the mathematician who first worked this out (Cantor) spent some time in a mental hospital. So ponder this with care…(and you thought SAT math was driving you crazy!)</p>