A unit square is cut into rectangles. Each of them is coloured by either yellow or blue and inside
it a number is written. If the color of the rectangle is blue then its number is equal to rectangle’s
width divided by its height. If the color is yellow, the number is rectangle’s height divided by
its width. Let x be the sum of the numbers in all rectangles. Assuming the blue area is equal
to the yellow one, what is the smallest possible x?
Don't cut the square into more than one rectangle, I think. Regardless syntactically the question states that it's cut into rectangles, which I think automatically rules that answer out.
here's (roughly) your proof.
each blue rectangle has a minimum value of 1. There can be no blue rectangles, if cut into more than 1 shape.
play around with all-yellow thing.
Only one problem though.
> Assuming the blue area is equal
to the yellow one
Yeah that's what I've been working with
Gonna go for now
this way doesn't seem able to get me under 2.5 either, because i can't have an infinite amount of subrectangles right?
I mean if we're allowed to have infinite subrectangles then the answer is 0
Let m be the ratio between the area of the interior square and the area of the exterior square
x = sum(k=0)^inf 4(sqrt(m^k)-sqrt(m^(k+1)))/(sqrt(m^k)+sqrt(m^(k+1)))
x = sum(k=0)^inf 8/(sqrt(m)+1)-4
Interestingly enough the values for the 4 rectangles that make up each interior square don't change no matter how small the square gets. So x is infinite.
ofcourse, but not if you also decrease the width of all the blue rectangles and length of all the yellow rectangles.
define w as the width of the blue and the length of the yellow rectangles. Let's look at just the first "ring" of 4 rectangles. as w approaches 0, x approaches 4*w = 0. the square left inside is just the unit square scaled by some constant very close to one, so by induction and mathematical wizardry we can say x=0.
I suggest you actually look at the infinite sum of the ratios. Even as the widths of the blue rectangles and heights of the yellow approach zero, the sum approaches infinity. You can't just calculate the sum of an infinite amount of ratios that each approach zero, it only exists as a limit.
Alright, I'm gonna just lay out what I've done so others can take over.
A) We know the blue and yellow shades must cover spaces of 0.5 x 1 exactly. We must prove minimum values for both yellow and blue shades.
B) So let's prove that this must mean that if we divide the yellow rectangle into pieces with height <= 0.5 and width <= 1, the minimum value for x in yellow is 0.5:
If one of the subrectangles has height >= width/2, your ratio is obviously greater than 0.5. This case cannot work as you can already have a smaller division (0.5 x 1.0).
If no subrectangle has a ratio of greater than or equal to 0.5, it means that every division of the left side is smaller than half of every division of the right side. <insert proof of the number of every subrectangle here totalling a number >= 0.5>
C) Similarly the proof works out for the blue side as well.
I will explain in detail how it is calculated so you can see how your mathematical wizardy fails.
Let me be the area of the square created by the ring of rectangles. Then the sides of this square are sqrt(m). So to find the height of one rectangle we subtract this from the height of the unit square and divide by 2.
h = (1-sqrt(m))/2
And that must mean that the width of this rectangle is the difference between the width of the unit square and the width of the rectangle it touches. Since all four rectangles are the same shape, the width of that rectangle is equal to the height of this rectangle.
w = 1-(1-sqrt(m))/2 = (1+sqrt(m))/2
h/w = (1-sqrt(m))/2 / (1+sqrt(m))/2 = (1-sqrt(m)) / (1+sqrt(m))
Now let's look at the new interior square. This is identical to the unit square except scaled down by a factor of m, forming a new interior square of area m^2. So we can use the same logic to find
h = (sqrt(m)-sqrt(m^2))/2
w = sqrt(m)-(sqrt(m)-sqrt(m^2))/2 = (sqrt(m)+sqrt(m^2))/2
h/w = (sqrt(m)-sqrt(m^2)) / (sqrt(m)+sqrt(m^2))
hopefully by now you can see that the sum of the ratios of the 4 rectangles in the kth interior square will be
(sqrt(m^k)-sqrt(m^(k+1))) / (sqrt(m^k)+sqrt(m^(k+1)))
But this simplifies very nicely to 8/(sqrt(m)+1)-4
In other words, no matter which interior square we are in, the sum of the ratios will be the same. And since there are infinite interior squares, the sum of all the ratios approaches infinity even as m approaches 1.
Divide half the square into blue triangles with width 0. Divide the other half of the square into into infinite yellow rectangles with height 0.
Therefore the minimum value for x is 0
Maximum value is infinite
Let me elaborate on that poorly written post:
Divide the square horizontally. Top half of the square is cut into infinite blue rectangles with width 0. Bottom half is cut into infinite yellow rectangles with height 0. Blue area = yellow area, and x = 0
Something tells me one of you guys is going to school me and that I missed something really obvious cos the answer can't be that easy
Your taking 0 times infinity to be zero. You can't do this. You must examine the limit of x as the number of rectangles approaches infinity.
This goes for everyone who says the minimum is 0.