Poly-puzzle

An excellent explanation of a solution was presented by Andrei LAZANU Age: 12 School: 205, Bucharest, Romania. I have tried to add some diagrams to help his text, although I think that it might not be the case that this works for all triangles. Perhaps someone would like to think about exceptions?

Well done Andrei.

I used this month's clues to solve the problem. Triangle A from the triangle fits in the rectangle because:

• it has a right angle (the corner of the rectangle)
• I rotated triangle A around a point, I use the same method for triangle B see Fig. 1

Now, I see a rule, for fitting any triangle into a rectangle. It can be made in the following way:

• an altitude of the triangle is constructed
• a perpendicular from that altitude is constructed. Now, the altitude is kept only up to its intersection with the line joining the midpoints of the sides of the triangle
• the two triangles must be rotated and the rectangle is obtained.

Now, I finished with the rectangle. I search for a method to fit the pieces of the triangle into a square.

The area of the rectangle is 0.5h*b , which must be equal to s² . So, the side of the square is $s = \sqrt{{1\over2}h\times b}$.

Because I have already shown how it is possible to transform the triangle into a rectangle, now, for the transformation of the triangle into a square I use the previous transformation.

Because the clues helped me a lot, I shall use the figure.

First I demonstrate that the small pieces fit into the square, and for this I first have to show how triangles C, D and pentagon E have been obtained :

• The rectangle is oriented with the biggest side on the vertical
• From the top-left corner I measure a distance equal to s , the side of the square, and I join this point with the top-right corner: line L , defining triangle C.
• I made the same from the bottom-right corner, i.e. I measure a length equal to s. Then I construct a perpendicular on this side of the rectangle, defining this way triangle D and pentagon E. E is not always a pentagon!!

Now, the following step is to show that these 5 pieces, A, B, C, D and E fit into the square.

• I keep A and C in their places, I put aside triangle D.
• B and E, kept together, are made to slide along line L, so that the sides of triangle A and pentagon E are on the same line. This way, the other side of the square has been defined.
• triangle D is translated along the side of triangle C. It fits in the place because:
  • the length of the biggest side of triangle C is the sum of the lengths of the hypotenuses of triangle D, and one side of the pentagon E, in reversed order in the square in respect to the rectangle
  • the length of the side common to triangle D and pentagon E is added to the smallest side of the rectangle to form the side of the square.

The general method to obtain the square from a triangle is the following:

• an altitude of the triangle is constructed
• a perpendicular on that altitude at its half (that is parallel to other side) is constructed. Now, the altitude is kept only up to its intersection with the median line (I'm not so sure about the English term for this line). So, two triangles, A and B, were obtained, and a trapezoid.
• the two triangles must be rotated as shown before, and the rectangle is obtained.
• The side of the square is calculated. A length equal to this side is measured from the opposite corners, along the longest side of the rectangle, and triangles C and D and pentagon E are formed.
• I slide (E + B) and D in opposite senses on the side common with triangle C up to the moment the sides of A and E and D and B respectively are one in continuation of the other.
• The square is ready!!