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The original puzzle :
But first, let's name some vertices
And instead of forcing the little square to be 1x1, let's give the little square a size of kxk, this means that the little square can be any size.
Step 1 : Giving the Large square a size.
Let's assume the side GH has a length offk where f is a scalar.
∵P0GHI is a square
∴GH = IH = fk
Step 2 : finding the other sides based on the assumptions
P0G = P0P3 + P3G
GH = k + P3G
fk = k + P3G
P3G = fk - k = FG
FG = P2P3 + EP2
fk - k = k + EP2
EP2 = fk - 2k = DE
DE = CP1 + P1P2
fk - 2k = CP1 + k
CP1 = fk - 3k = CB
Step 3 : stacking the height
Before we start tally up the heights, let's rename AI as h . And we can recap that
GH = fk
FG = fk - k
DC = fk - 2k
CB = fk - 3k
AI = h
And because all these sides are parallel, we can know that
AI + BC + CD = HG + GF
h + (fk -3k) +(fk - 2k) = fk + (fk - k)
h - 5k = -k
h = 4k
Step 4 : substitute k
In the original puzzle, k=1, so the original answer would be 4(1) = 4, but since we know that h=4k , we can always see what is k, h can be a result.
QED
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