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A rectangle the length of which is 8 inches and the width of

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A rectangle the length of which is 8 inches and the width of which is 6 inches is made up 48 1 inch by 1 inch squares. Through how many of the squares does a diagonal of the rectangle pass?

6
1
14%
8
2
29%
10
1
14%
12
3
43%
16
0
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Total votes: 7
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by MartyMurray » Fri Jun 03, 2016 5:07 am
OA: D
Last edited by MartyMurray on Thu Jun 09, 2016 9:11 pm, edited 1 time in total.
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by 800_or_bust » Fri Jun 03, 2016 11:43 am
Marty Murray wrote:OA by Sunday.
I came up with 12. Not 100% sure. I drew the rectangle such that the diagonal ran from the lower left-hand corner to the upper right-hand corner. I then partitioned the rectangle into four equal smaller rectangles - each 3x4. Two of these - the one in the upper left and the one in the lower right - would not have any portion of the diagonal pass through; rather, it would intersect them at the center of the figure. So that knocked out 24 of the 48 possible squares right off the bat.

Then I zoomed in on one of the smaller 3x4 rectangles, and simply drew the figure counting the number of squares the line would pass through (recognizing for each movement of 1 unit to the right, the line would move up 3/4 of a unit). I counted six. Recognizing that this would be symmetrical with the upper right triangle, the final tally was 12.
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by MartyMurray » Fri Jun 03, 2016 12:30 pm
800_or_bust wrote:
Marty Murray wrote:OA by Sunday.
I came up with 12. Not 100% sure. I drew the rectangle such that the diagonal ran from the lower left-hand corner to the upper right-hand corner. I then partitioned the rectangle into four equal smaller rectangles - each 3x4. Two of these - the one in the upper left and the one in the lower right - would not have any portion of the diagonal pass through; rather, it would intersect them at the center of the figure. So that knocked out 24 of the 48 possible squares right off the bat.

Then I zoomed in on one of the smaller 3x4 rectangles, and simply drew the figure counting the number of squares the line would pass through (recognizing for each movement of 1 unit to the right, the line would move up 3/4 of a unit). I counted six. Recognizing that this would be symmetrical with the upper right triangle, the final tally was 12.
Man, what a great hack, eliminating 24 and then working with 1/4 of the figure.

Yes, the diagonal goes up, or down, 3/4 for every 1 it goes across.

NAILED that puppy!
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by regor60 » Sat Jun 04, 2016 10:42 am
An estimated approach could be to recognize that the maximum number of blocks would be 14,based on having travel that number worst-case. A minimum would be simply the length of the diameter, here 10.

Then average the two, which equals 12.

I googled the analytical solution and this agrees reasonably well for various sizes.

You can draw it and derive the solution but that wouldn't be the best approach on a test
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by MartyMurray » Sat Jun 04, 2016 11:07 am
regor60 wrote:An estimated approach could be to recognize that the maximum number of blocks would be 14,based on having travel that number worst-case. A minimum would be simply the length of the diameter, here 10.

Then average the two, which equals 12.

I googled the analytical solution and this agrees reasonably well for various sizes.

You can draw it and derive the solution but that wouldn't be the best approach on a test
I tend to agree that drawing the entire thing may not be the best approach, but if you think about it, 10 is not really the minimum.

10 inches is the length of the diagonal, but you don't need 10 squares to get 10 inches, because you are going through the squares diagonally. So the minimum is really more like 8, because to get from one side to the other you do have to go through 8 columns of squares.

In fact, if the figure were 8 x 8, the diagonal would be longer than 10, and you would go through exactly 8 squares.
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by regor60 » Sat Jun 04, 2016 12:55 pm
I understand. But if you apply the approach described and average the 11.2 diagonal in your 8x8 example with 16,the result rounds to 14,which agrees with the exact approach. So perhaps more work needed to explain why works.
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by MartyMurray » Sat Jun 04, 2016 5:41 pm
regor60 wrote:I understand. But if you apply the approach described and average the 11.2 diagonal in your 8x8 example with 16,the result rounds to 14,which agrees with the exact approach. So perhaps more work needed to explain why works.
The exact number of squares traversed by the diagonal in the 8 x 8 example is actually 8. The diagonal of the 8 x 8 is made up of the diagonals of 8 of the smaller squares, just to be clear.

Think about it. The diagonal of the 8 x 8 bisects two of the 90 degree corner angles, and does the same with the corner angles of 8 of the squares.
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by regor60 » Sun Jun 05, 2016 5:47 am
I get it that 8 is the actual number traversed by the duagonal, I'm just remarking that the approach I described seems to work as a way of estimating if you don't already know the exact approach
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by Matt@VeritasPrep » Tue Jun 07, 2016 10:39 pm
We could also approach this analytically. (My visual instinct sucks, so my geometry solutions are always slipping into algebra; I wouldn't have been able to hack it in ancient Greece.)

Fix the rectangle in the coordinate plane with one vertex at the origin and its opposite at (6, 8). The diagonal is the line y = (4/3)x, so any point on the diagonal has coordinates (x, (4/3)x).

Now consider what happens as we move (x, (4/3)x) from (0,0) to (6,8). Every time either x or (4/3) passes a new integer value, we'll be in a new square. (For instance, when we hit (3/4, 1), we'll be on the border of a new square, and when we hit (1, 4/3), we'll be on the border of another.)

That means we need only count the number of times x and (4/3)x pass new integer values INSIDE THE RECTANGLE (i.e. not at (0,0) or when we arrive at (6,8)). x will do this 5 times (at 1, 2, 3, 4, and 5) and y will do this 7 times (at 1, 2, 3, 4, 5, 6, and 7), so we'll pass through 5 + 7 = 12 squares.
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by Matt@VeritasPrep » Wed Jun 08, 2016 8:42 am
For the sake of simplicity, I left one annoying detail out of my first solution, but if you were to try to approach this for any problem, it's worth noting that you have to correct for any overcounting that results when the line passes through a lattice point: a point in the plane for which both coordinates are integers.

For instance, when we reach the first point at the border of a square (3/4, 1), we count one square: the one through which we just passed, from (0,0) to (3/4, 1). When we reach the next border (1, (4/3), we count a second square: the one that contains the space between (3/4, 1) and (1, 4/3). This gets us two squares, by counting x = 1 and y = 1.

But when we pass through the point (3, 4), we've got a problem, since we count it TWICE (for x = 3 and y = 4) even though it only represents passage through ONE square: the square that contains the space between (9/4, 3) and (3, 4).

In our problem it isn't an issue, since we also have to count the last square (from (21/4, 7) to (6, 8)), and the last square and the overcounted (3, 4) offset. But in a general solution you'd have to note this and correct for it more carefully.
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