A Geometry Puzzle That Looks Impossible At First

Comments · 118

• AN

  @andreydeev4342 · 6 days ago

  Instead of calculating the last (most left) rectangle, you could just get height of bottom rectangle as 8a / (8a / 5) = 5, so the side of square is 30 + 5 = 35

  31

• IT

  @ItalixPubg · 1 day ago

  You had the right idea at the beginning then you overcomplicated everything horribly...<br>So the little rectangle has a size of 8xa (I will always put the vertical size first).<br>If you add the two rectangles above you triple the area and since the second dimension is identical it means that the new rectangle has a size of 24xa<br>When you add the rectangle on the left, the area grows by a factor of 4/3 and since this time the first dimension is unchanged it means that the second dimensions grows by the same factor, so you obtain a 24x(4/3a) rectangle.<br>And you continue like this: when you add the rectangle on the top you get a 30x(4/3a) rectangle then when you add the rectangle on the right you get a 30x(8/5a) rectangle then when you add the seventh rectangle at the bottom you get a 35x(8/5a) rectangle and then you know the area of the square: 35²=1225.<br><br>But you can also be even smarter and realise that you only need to work on the vertical dimension.<br>The first expansion triples it from 8 to 24 then adding the fourth rectangle leaves it unchanged then adding the fifth rectangle will multiply it by 5/4 from 24 to 30 then finally adding the seventh rectangle will make it grow by a factor of 7/6 from 30 to 35.

  2

• AD

  @adamwho9801 · 6 days ago

  I definitely made it harder than it needed to be.

  12

• SU

  @SuryaSurya-jl5gf · 6 days ago

  It&apos;s like a spiral theorem : from little inside to bigger outside😊👍.

  3

• NA

  @nakamakai5553 · 6 days ago

  This is very clever and highly satisfying! Such a good puzzle and video, thanks.

  3

• MS

  @MsNosis · 5 days ago

  I like the usage of coloured lines and simple animations to break this seemingly impossible puzzle down.

  4

• JO

  @jonahansen · 6 days ago

  I did it entirely differently, incrementally, by starting with the area of each smaller rectangle as 1/8 the total &quot;A&quot;, calling it &quot;a&quot; so A = 8*a. Then, starting with the rectangle with known height 8, one can get it&apos;s length as a/8, which transfers to the two smaller ones above, and just keep finding lengths or heights by division in turn.

  3

• RA

  @rasmuslille1570 · 6 days ago

  Saying that all the rectangles are different got me confused in the beginning, thankfully its explained differently in Scientific American.

  9

• ST

  @steve_weinrich · 6 days ago

  Thank you so much for including the units!

• HV

  @hvnterblack · 6 days ago

  Clean solutuion and really big + for remebereing about units. That is standard, that should be keeped.

  6

• KN

  @KnakuanaRka · 6 days ago (edited)

  I’ve seen a puzzle like this before. <br>You don’t need to consider the area of the whole area you’ve figured out before, just that of the current rectangle. For example, the two rectangles solved first are congruent and their width adds to that of the original rectangle, so their width must be half the original, and thus their height must be twice it for 16. Then the next rectangle has a height of 8+16=24, 3 times the original height, so the width must inversely be a third or x/3, and so on.

  3

• JA

  @jaspermcjasper3672 · 1 day ago

  Says too many times that all of the rectangles have the same area. No. Only the rectangles that aren&apos;t composed of (or containing) smaller rectangles have the same area.

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