Squares puzzle

Replies

• 

  Kaustubh Katdare

  skipper

  A square of paper is 8 by 8 inches. Section the square so that the sections can be reassembled into a rectangle of 65 in[sup]2[/sup], one more than 8x8 in[sup]2[/sup].
  
  Where does the extra square inch come from?😕

  Interesting. I'm not even sure if that's possible.
• 

  Saandeep Sreerambatla

  I tried this but i am always getting <=64 but not 65.
  
  iS this really possible??
  
  I have tried in forming the rectangle with 13*5.
  
  I am able to form 12.5 * 5 and 1.5X1 rectangle is remaining.
  
  So i guess i need two more .5 *1 pieces which adds up to 65 but we have only 64.
  
  So i am not able to solve this 😔
• 

  skipper

  The 13*5 factor is the key, yes.
  
  You first make a cut straight across the square at a height of 5". Then section the larger piece again, by cutting at an angle from a point 3" along one [longer] side to the opposite side, at a point 5" along. Then cut the remaining rectangular section (which is 3"x8") diagonally from corner to corner.
  
  Now the 4 pieces can be arranged into a 13"x5" rectangle, with one more square inch or area. Sectioning and translating the square "creates" the extra unit, because of the arithmetic sectioning function, which leaves a solution that says 8x8 = 13x5.
  Note: the RHS is a pseudoprime...
• 

  skipper

  This extra square "function" kind of does your head in. You start with a square (which can be a unit square) and divide it into two unequal parts, in a 5/3 ratio, then you divide each of these two, into equal halves.
  
  You start with 64 square 'units' and go to 40 + 24, then 2x(20 + 12) by dividing equally.
  Somehow this gets from (3 + 5)(3 + 5) to (3 + 5 + 5)(0 + 5), you add a 5 to the first 'side' and remove a 3 from the other, of the rectangle (the 3/5 symmetry thingy). But the 13x5 figure has a diagonal that means there are 2x32 square areas as right triangles, joining the hypotenuses gives you 2x32 + 1 units (??)
  The inverse, when you start with a rectangle with a certain ratio of sides, makes the unit(s) vanish.
  
  ... how to fold up a hypotenuse inside a square, so a unit vanishes when you section a 13x5 rectangle.
• 

  gohm

  nice one Skipper!
• 

  skipper

  
  
  I found this, which shows the illusion up close.
  I should add I've been a bad robot; you can't really do this in Euclidean space, even with paper because the triangles are all different - it just looks like they're the same in both figures, or are similar triangles as they say.
• 

  skipper

  OK I've been and made a fool of myself at another site, and so the outcome is that you can't make an extra unit of paper, or hide one.
  
  It's possibly easier to see why if you start with the RHS figure the 13x5 rectangle, call this R. The square on the left is made up of two rectangles R' and r, a larger and smaller. This "paradox" has a connection to the least/most distance function over an area (let's call this a gravitational surface, for instance).
  
  Anyways, in R a straight diagonal from (0,0) to (13,5) will miss the points (5,2) and (8,3) inside it, where the vertical cuts are made from the upper and lower edges. If Instead, you make the inner diagonal cut from (5,2) to (8,3) then cut from these points to the nearest corner you get two trianguloids or quadrilaterals, the "almost" triangles have slightly convex hypotenuses; the two parts of R' from R will make a 5x8 rectangle however.
  
  The extra unit (the 65th) is in the extra area above the real diagonal of each 3:8 triangle on the right. These overlap when you "slide" the trianguloids together to make the corners of r.
  
  Now this is all in 2 dimensions, if you ignore the physical fact that paper has a thickness. In 3 dimensions the least distance function is a lot more complicated, and gravity which "acts centrally" means a shortest distance on a sphere with gravity, is a great circle path.
  
  The extra unit that folds up inside the overlap in r, is least when the cuts from the corners to the central points are straight, the inner cut (1:3) doesn't extend into the diagonal (2:5); extending it makes a gap between the halves of R'. This missing paper goes into the overlap in r
• 

  Saandeep Sreerambatla

  Sorry I couldnt understand , whether we could make or not ?? the extra inch??
• 

  skipper

  
  
  Here you can see where the extra unit is; rearranging the triangles (which appear to be similar, but aren't) makes the unit appear and disappear. Can you see where it goes to or comes from?
• 

  Saandeep Sreerambatla

  Thanks for the explanation.
• 

  skipper

  Right, if you look closely at the colored triangles, you can see where the thick black line on the right side of the larger red triangle in the top figure lines up with the bottom figure. The heights in both figures are slightly different.
  The extra unit is all in the hypotenuses of the triangular sections; if you go from the square to the rectangle, there is 1 extra unit to allow for starting with 64. Then there's a slight gap between the two halves of R - a parallelogram appears that is 1 unit in area, but with a length equal to the diagonal of R. The width can be calculated since the length and area are knowns.
  
  Going from the rectangle to the square, there's an extra unit to "fold" in; this appears as an excess area or overlap between the triangles in the smaller part (r). Same deal with the area, length and width. As long as straight cuts are made along the "diagonals" that fit in R', the larger part, this will form an 8x5 rectangle; if the middle cut from (5,2) to (8,3) extends into the cuts across R, there will be a gap in R, and a greater excess in r
• 

  bottomline99

  The coloured triangle with the least area has a concave hypotenuse (check the gradient of each triangle). However the triangle with extra area has a convex hypotenuse.
• 

  skipper

  Here's the algebraic part: each half of the R' figure is a product of a triangle and a rectangle. There is a 2:5 triangle (expressed as a ratio of sides) on top of a 3x5 rectangle in each half, on the LHS, on the RHS (in R the 5x13) the triangles are slightly deformed 2:5's but equivalent. Both cases lead to two halves A and B or A' and B' (the deformed pair).
  
  If we call ab the product of one half of R', then ab = 2:5+3x8; the first term is a scalar (a scalene triangle), the second is like a vector product (a rectified ||'gram). When you use the RHS for ab you need the deformation product.
  
  Now AB = A'B' and ab is the left-hand algebra, for R, R', in AB. On the right ab has an inner "skew" that vanishes when you join A' and B'.