Fibonacci's Cheat: 64 = 65?

many people already know about it, but i just thought its a nice trick for those who dont.

by the way, its wrong. i have also proven why.

Tch, this looks like the triangles that are permutations of each other, yet the other triangle has a single unit square cut out of it from the bottom.

Without formal mathamancy, I'ma go out on a limb and say that Fibonacci's Cheat doesn't work because the fraction of squares in the middle don't align correctly. This "fudges" the actual squares in the whole figure, and the total sum of square parts that becomes "missing" in the middle is exactly 1.

For that matter, the shapes in the picture do not fit together exactly into a perfect rectangle.

There's a fucking gap between the blue + red line, and the green + orange line.

There's a pineappleing gap between the blue + red line, and the green + orange line.

yes, but can you geometrically prove why?

3/8 != 2/5

yes, but can you geometrically prove why?

Yes. Look at the gradients of the two lines on the blue and red shapes. They aren't the same.
Also, if you look at the top right corner of the "65 square" rectangle, they don't align properly.

yes, but can you geometrically prove why?

Yeah. I did it myself, and got a gap there. Isn't that proof enough?

Yeah. I did it myself, and got a gap there. Isn't that proof enough?

considering it takes exactly 30 seconds to prove it mathematically, no.

considering it takes exactly 30 seconds to prove it mathematically, no.

Exactly 30 seconds? o.O Can you really do it in exactly 30 seconds?

Take two piece-wise functions, f(x) and g(x), which define the slopes of the pieces arranged appropriately in the 5x13 rectangular diagram:

Code:

f(x) = | 0.4x, for x < 5
       | 0.125 + 0.375x, for x >= 5

g(x) = | 0.375x, for x < 8
       | -0.2 + 0.4x, for x >= 8

Then we do the following computation:

Code:

Integrate(function, x1, x2) = result

Integrate(f(x) - g(x), 0, 13)
= Integrate(0.4x - 0.375x, 0, 5)
    + Integrate(0.125 + 0.375x - 0.375x, 5, 8)
    + Integrate(0.125 + 0.375x + 0.2 - 0.4x, 8, 13)
= 0.025 * 5^2 / 2
    + 0.125 * (8 - 5)
    + 0.325 * (13 - 8) - [0.025 * ((13^2 - 8^2) / 2)]
= 0.3125 + 0.375 + 0.3125
= 1.

This tells us that there's exactly 1 square's worth of area, or lack thereof, enclosed by the four shapes.


Another proof is that any re-arrangement of all divisions from the original area will always have the same exact area as the original.

(I had nothing better to do atm)

Or just 3/8 != 2/5.

Or just 3/8 != 2/5.

^ This.

Exactly 30 seconds? o.O Can you really do it in exactly 30 seconds?

Take two piece-wise functions, f(x) and g(x), which define the slopes of the pieces arranged appropriately in the 5x13 rectangular diagram:

Code:

f(x) = | 0.4x, for x < 5
       | 0.125 + 0.375x, for x >= 5

g(x) = | 0.375x, for x < 8
       | -0.2 + 0.4x, for x >= 8

Then we do the following computation:

Code:

Integrate(function, x1, x2) = result

Integrate(f(x) - g(x), 0, 13)
= Integrate(0.4x - 0.375x, 0, 5)
    + Integrate(0.125 + 0.375x - 0.375x, 5, 8)
    + Integrate(0.125 + 0.375x + 0.2 - 0.4x, 8, 13)
= 0.025 * 5^2 / 2
    + 0.125 * (8 - 5)
    + 0.325 * (13 - 8) - [0.025 * ((13^2 - 8^2) / 2)]
= 0.3125 + 0.375 + 0.3125
= 1.

This tells us that there's exactly 1 square's worth of area, or lack thereof, enclosed by the four shapes.


Another proof is that any re-arrangement of all divisions from the original area will always have the same exact area as the original.

(I had nothing better to do atm)

thats how i did it -
consider there is the following diagram (excuse my mspaint skills again):

ADE is the green triangle, and DEBC is the orange quadrangle. i added DF which connects point D with line BC, and creates the angle BFD which is 90.

EDFB is a rectangle, and all his angles are 90, so EDF is 90.
ADC = ADE + EDF + FDC
ADC = ADE + 90 + FDC
ADE = arctan(8/3) = 69.443954780416535691706387252656
DF = EB = 5
CDF = arctan (2/5) = 21.801409486351811770244866086944
ADC = 69.443954780416535691706387252656 + 90 + 21.801409486351811770244866086944
ADC = 181.24536426676834746195125333959

so ABC is not really a triangle, because AC is not a line (ADC > 180). solved.
and yes, without writing everything and drawing the diagram it took 30 seconds.

Or you could just look at the gradient of Triangle ADE (3/8) and Triangle DFC (2/5) and see that they aren't the same. As pointed out several times in this thread. And far simpler than your method.

Or you could just look at the gradient of Triangle ADE (3/8) and Triangle DFC (2/5) and see that they aren't the same. As pointed out several times in this thread. And far simpler than your method.

yes i saw it appearing many times in this thread, and assumed it is simpler. but no one explained, you and rust were all like "2/5 != 3/8 kthxbai". so i chose to ignore it because i couldnt even understand what you two are speaking about (and i still dont understand, explanation anybody?)

yes i saw it appearing many times in this thread, and assumed it is simpler. but no one explained, you and rust were all like "2/5 != 3/8 kthxbai". so i chose to ignore it because i couldnt even understand what you two are speaking about (and i still dont understand, explanation anybody?)

Yes. Look at the gradients of the two lines on the blue and red shapes. They aren't the same.
Also, if you look at the top right corner of the "65 square" rectangle, they don't align properly.

If the gradients of the two lines are not the same, then the there isnt 1 line connecting the points AC, but 2. Therefore the shape ABC can't be a triangle, it has to be a quadrilateral.

If the gradients of the two lines are not the same, then the there isnt 1 line connecting the points AC, but 2. Therefore the shape ABC can't be a triangle, it has to be a quadrilateral.

whats a gradient >.>

whats a gradient >.>

The slope or steepness of a line. In your shape, the steepness of the line AD is 3/8 which means that for every 8 units you move across, the line moves up 3 units. If the was a continuous line from point A to point C, there gradient would be the same between any two points on the line.

So if I check the gradient of the line DC it should also be 3/8 if ABC is a triangle. However the gradient of the line DC is 2/5. Which means that for every 5 units you move across, the line moves up two units. That means that there is two separate lines connecting AC.
Therefore, the shape is a quadrilateral as it has 4 sides, not 3.

The slope or steepness of a line. In your shape, the steepness of the line AD is 3/8 which means that for every 8 units you move across, the line moves up 3 units. If the was a continuous line from point A to point C, there gradient would be the same between any two points on the line.

So if I check the gradient of the line DC it should also be 3/8 if ABC is a triangle. However the gradient of the line DC is 2/5. Which means that for every 5 units you move across, the line moves up two units. That means that there is two separate lines connecting AC.
Therefore, the shape is a quadrilateral as it has 4 sides, not 3.

ooooo
so if their gradients are different, its not a line...
naic, in what grade do they teach that :O

Its not a continuous line. It is still two straight lines.

I teach in in Years 9 - 12, depending on the course (in Australia that is. USA probably is a little different).