Mrs. Pythagoras' puzzle
Puzzle
When Mrs. Pythagoras
took counsel with her
spouse regarding the
best way of squaring a
two-square formed remnant of Athenian matting, the great
philosopher explained: "Knowing
that a square erected on the hypothenuse line of a right-angled triangle
is equal to the square of the base and
the square of the side combined, we
see that the square you require must
be equal to both squares, and therefore must be equal to the square of
the hypothenuse."
The truth of this great therum is
shown in the small illustration.
I shows a right-angled triangle; being three inches high, the square of
that side contains nine square inches;
the base, being four inches long, contains sixteen square inches, and the
five inch hypothenuse square contains as many square inches as both
of the other squares added together.
If we wish to make one square out
of two we merely place them together and draw a straight line from
the top of the smallest and draw the
line through the larger to the lower
corner, B to B, and it will form the
hypothenuse line of a triangle, giving the length of the square which
will be equal to both of the other
squares. A beautiful demonstration
is shown in three pieces by clipping
off the triangular pieces and fitting
them above as shown. It will always
make a perfect square from any two-square shaped piece of paper.
When this principle is thoroughly
understood any number of squares
or circles may be added together at a glance,
In the above illustration 1 and
2 would form a square equal to the
hypothenuse line of A. Squares 3,
4 and 5 are each added to the hypothenuse lines, showing that the
square E is as large as the five small
ones. All manner of forms, like
rhomboids, pyramids, crosses, etc.,
can be transformed into squares by
this method, which is invaluable in
the world of puzzledom. In the following small figure of a cross we see
that the lower square X would fill
the upper space X, thereby forming
a two-square form, which we know
how to transform into a square
which we can see must be of the dimensions of that dotted hypothenuse
line.
But to get back to Mrs. Pythagoras' matting puzzle, which the
great philosopher offered to solve
upon the principle of Euclid's 47 proposition known as pons asinorum.
"Now, Thag," says she, for she
always called him that in the house,
"I am feared these goods will fray
if they are cut on the bias, so I want
to get along without that hippopotamus line. Here is a plan which will
also do it in three pieces: Cut out
that long piece marked A, and stand
it on end at one side; then move the
piece C down one step, and it forms
a 13x13 square, all right, all
right.
"But, I don't like it altogether,
Thag; you see the pattern don't run
quite right on the squares in that
long piece. Can't you find a perfect
answer without giving any of the
squares that half turn? I know it
can be done."
There we have Mrs. Phythagoras'
new puzzle.
On the principle that every good
rule should work both ways, we will
now reverse the idea so as to produce two squares from one large
square. Place the point of the compasses at A and describe the arc B C
and any point on the arc will indicate the junction of the two smaller
squares, which are equal to the area
of the largest A B D C, and brings
us back to a two-square form of any
desired dimension.
It is easy to prove the truth of Euclid's famous proposition by erecting four tri- angles around the hypothenuse square C. The area of the large square being (A+B2) equals A2+2AB+B2; subtracting the contents of the 4 triangles 2AB proves A2+B2 of the side and base equals C2 of the hypothenuse.
Answer
Show answer
References
- Loyd, Sam [1914]. in Loyd, Sam, Jr.: Sam Loyd's Cyclopedia of 5000 Puzzles Tricks and Conundrums (in English). New York: Lamb Publishing company, page 66.
