2009-04-07, 03:46 PM
(This post was last modified: 2009-04-07, 03:51 PM by KajitiSouls.)
haha01haha01 Wrote: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 >= 8Then 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)

