Pythagorean Theorem Puzzle
Mike Killianey
Pythagorean Theorem, but I can't remember exactly how the problem was
worded, or what the answer was, but here's the jist of it (If anyone
knows more, please let me know!):
Give a geometric construction argument for the Pythogorean Theorem.
The point is to take the three sides of the triangle and make three
squares out of them. Since a*a + b*b = c*c, the areas of the
small squares should exactly fit in the larger square. So given
any arbitrary right triangle, how can I calculate where to cut
the smaller squares in order to fit them in the larger square exactly?
If I remember right, the solution took no more that 8 or 12 cuts.
A diagram: /\
/ \ <---- okay, this is supposed to be a large square.
/ \
/ \
/ +-----------+
/ 2 / | |
\ c / | 2 |
\ / | a |
\ / | |
\ / | |
+-----------+-----------+
| | ^
| 2 | |
| b | +-- small square #1
| |<---- small square #2
| |
+-----------+
Jake
One that I saw once went like this:
(Unfortunately, it required drawing the same time in
several different positions...difficult to do with ascii
graphics, but I'll do my best)
b a
+-------+
| ^ |
a| / \ |b This square has area (a+b)^2;
| /c c\ | the four-sided figure on the inside is also
|< >| a square (easy to prove) with area c^2;
| \c c/ | and the four triangles' areas total 2ab.
b| \ / |a So we have (a+b)^2=c^2+2ab,
| v | Since (a+b)^2=a^2+2ab+b^2, we get a^2+b^2=c^2.
+-------+
a b
Hmmm... Actually, I remember the proof requiring another diagram...
it's 4:30 am, so it's entirely possible that I made a mistake above,
although it looks OK to me...
Up way past his bedtime,
jwgh
RING, DAVID WAYNE
>small squares should exactly fit in the larger square. So given
>any arbitrary right triangle, how can I calculate where to cut
>the smaller squares in order to fit them in the larger square exactly?
>
>If I remember right, the solution took no more that 8 or 12 cuts.
It only takes 3 cuts.
*** SPOILER ***
Arrange the squares in the usual way around the triangle. Shift
the largest one by a distance equal to its own edge, towards the other
two. Cut along the perimeter. I'll leave the rearrangement as an
exercize. Hint: Only the largest piece stays put.
Dave Ring
dwr...@zeus.tamu.edu
Carl J Lydick
>I seem to remember a geometric puzzle that had to do with the
>Pythagorean Theorem, but I can't remember exactly how the problem was
>worded, or what the answer was, but here's the jist of it (If anyone
>knows more, please let me know!):
>
>Give a geometric construction argument for the Pythogorean Theorem.
>The point is to take the three sides of the triangle and make three
>squares out of them. Since a*a + b*b = c*c, the areas of the
>small squares should exactly fit in the larger square. So given
>any arbitrary right triangle, how can I calculate where to cut
>the smaller squares in order to fit them in the larger square exactly?
>
>If I remember right, the solution took no more that 8 or 12 cuts.
But there's a much simpler geometric construction argument for the theorem:
Given a right triangle, construct three more identical triangles such
that their hypotenuses form a square. You end up with a square formed
by the hypotenuses, the four triangles, and a smaller square inside
that isn't covered by the triangles.
The area of the large square is, obviously, the square of the
hypotenuse.
The area covered by the triangles is 4*.5*the product of the sides.
The area of the small square is the square of the difference of the
sides.
The sum of the areas of the triangles and the area of the small square
is the area of the large square.
Hence: (a-b)^2+2ab = a^2 + b^2 = h^2
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CA...@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL
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