Tile a square with twenty congruent right-angled triangles. For each triangle, one leg is of length 1 and the other leg is of length 2.
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asked Feb 1 at 0:05
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3 Answers
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Note first that
the area of the square is 20 times the area of the triangle, namely 20, so the side length is $2\sqrt5$.
That's
twice the hypotenuse of the triangle
which immediately suggests the overall shape of the thing. With apologies for the horrific ASCII art:
. . . . . . .
..''|:
. . ..''.___. : . .
..''| | :
..''.___.___.___.___: .
: | | |:
. : .___.___.___.___. : .
:| | | :
. :___.___.___.___.___: . .
: | | ..''`
. . : .___.___..''. . . .
:| ..''
. . :.''. . . . . .
where
each of those 2x1 rectangles is of course divided into two triangles by one of its diagonals.
Here, have some slightly less horrific not-ASCII not-art:
answered Feb 1 at 1:29
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3$\begingroup$ I like this answer because it explains the math involved plus it gives two views of a solution in two different ways: ASCII art vs Graphic image AND Tilted vs Non-tilted square. I also appreciate the guide dots and the guidelines. $\endgroup$ Commented Feb 1 at 3:42
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1$\begingroup$ I upvoted this just for the ASCII art :) $\endgroup$ Commented Feb 1 at 13:03
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May I offer a more aesthetic tiling?
answered Feb 1 at 1:33
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1$\begingroup$ More aesthetic than ASCII at you mean?
:)$\endgroup$– M OehmCommented Feb 1 at 5:27 -
$\begingroup$ @MOehm Certainly - Gareth added the image about seven minutes after I posted. Even so, I find the mirror symmetry more pleasing to the eye. $\endgroup$ Commented Feb 1 at 5:48
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$\begingroup$ reminds me a bit of a rot13(fgne qrfgeblre) shape... (saw that as the first after unspoilering the solution) $\endgroup$ Commented Feb 1 at 18:43
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Using the solution of this classic puzzle, "If the dotted lines connect the midpoint of each edge with a vertex, find the fraction of the area of the yellow square to the large red one":
namely,
1/5 i.e. the same as each of the 4 red triangles.
Each red triangle and the yellow square can be divided into 4 of the triangles given in the question, resulting in this neat solution:
