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Squares, Squares, Squares

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Rory Barnes

on 23 April 2010

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Transcript of Squares, Squares, Squares

Squares, Squares, and More Squares n = 1 Layer = n n = 2 n = 3 n = 4 Questions: 1) How many squares can be drawn in a 6x6 square lattice? 2) How many layers are needed in order to be able to draw at least one million squares? 3) What is the smallest square array of points needed in order to form at least 100 different sized squares? First, let's look at the upright squares. When n = 1, we have 1 unit square. When n= 2, we see 4 unit squares. When n= 3, there are 9 unit squares. And when n = 4, we have 16 unit squares. Does anyone see a pattern so far? Like... The number of unit squares in any given array is equal to n^2? Let's see if it is also true for the next size of upright square. We obviously cannot form a 2x2 squares in the first layer, so let's jump to the second layer. There is one 2x2 when n = 2. And when n = 3, we see 1... 2... 3... and 4 of the 2x2 squares. The pattern appears to be the same, but just to be sure we will look at what happens with 3x3 squares. Again... The same pattern.

Now that we know the look of the pattern,
let's count all the squares in a 4 layer square. 16 unit squares 9 - 2x2 squares 4 - 3x3 squares 1 - 4x4 square So there are 16 + 9 + 4 + 1 upright squares in a 4x4 array. Which is the same as (4)^2 + (3)^2 + (2)^2 + (1)^2. Now we have our formula for the upright squares.
So let's look at the slanted squares. We do not see the first slanted square until the second layer, and there is only one of them. In the third layer we have 4 of these slanted squares. Any guesses for the fourth layer??? 9!!! Ok, maybe not so exciting,
but you get the point... Now let us explore the next largest slanted squares. The next largest size of slanted squares occurs in the third layer, and there are two of them. And when combined with the smaller sizes we have this: So when n = 3, we have 4(1) + 1(1+1). Expressing these sums in terms of new sizes will come in handy later when we construct our final formula. But before we take this idea further, let us move to the next layer. = + We have 9 of the smallest slanted squares... and 4+4 of the next largest size... and 1+1+1 of our largest sizes. So we have 9 + (4 + 4) + (1 + 1 + 1)
= 9(1) + 4(1 + 1) + 1(1 + 1 + 1)

= (4-1)^2 (1) + (4-2)^2 (2) + (4-3)^2 (3) + (4-4)^2 (4)
By Rory Barnes and Richelle Etsitty After noticing this pattern in the slanted squares,
we were able to derive our formula... Now we must combine our two formulas. Which gives us... But we only need the sixth layer. The good news is we have already come up with a formula. Guess and check time! The number of upright square sizes is apparent... 1 new size when n = 1. Another new size when n = 2. Another new Size when n = 3. And another new Size when n = 4. We see that the number of upright sizes in n layers is simply n
Hopefully the slanted squares are this easy to count... One new size in the second layer. And again, only one new size in the third layer. Hmmm... But then there are two new sizes in the fourth layer! And, still, two new sizes in the fifth layer! Well it is not quite as simple as the upright squares, but the slanted squares still seem manageable. After taking this idea further, we see that the pattern continues and we arrive at another formula to count our slanted squares... Well it is a little ugly, but it works. Now let us combine this with our upright square sizes... Ewww... Why so ugly? Because it is not entirely true. Although it does find all of the new sizes in each new layer, it does not account for the cases where the slanted squares may be the same size as an upright square, or another slanted sqaure. For example: The formula leads us in the right direction,
but we need to subtract out the sums of squares that occur more than once. Using the formula and setting n = 19, we end up with 109 different areas,
but we must subtract out these second occurences... The key thing to notice is that we have 11 repeating areas in the first 19 layers.

109-11 = 98 And the question asks... So then we must need jump up to the 20th layer in order to have at least 100 different areas. Questions?
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