Fibonacci's Cheat: 64 = 65?

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.