paradox

mun · **Joined:** Fri, 28 Dec 2007 12:01:53 UTC **Posts:** 1261

Make a equilateral triangle ABC of side 10cm BC AS BASE
Now for going from b to c there are two ways, one b-c and another b-a-c.
b-a-c has length of 20 cm
if we divide AB, BC and AC into halves say D, E & F respectively, then way b-d-e-f-c has length 20 cm.
now we divide BD, BE, DE, EF, FC& EC into 4 halves and join those points in similar manner. The way still has length 20 cm
we keep on dividing it infinitely and the path has length of 20 cm every time.
But BC has length of 10 cm and while infinite divisions it is near to BC.
SOLVE THE PARADOX.

Soroban · **Posted:** Wed, 8 Feb 2012 19:05:59 UTC

Hello, mun!

Quote:

Make a equilateral triangle ABC of side 10cm; BC as base.
Now, for going from B to C, there are two ways: one B-C and another B-A-C.
B-A-C has length of 20cm.

if we divide AB, BC and AC into halves say D, E & F respectively,
. . then path B-D-E-F-C has length 20 cm.

Now we divide BD, BE, DE, EF, FC & EC into 4 halves and join those points in similar manner.
The path still has length 20 cm.

We keep on dividing it infinitely and the path has length of 20 cm every time.
But BC has length of 10 cm and while infinite divisions it is near to BC.

Solve the paradox.

No matter how many times we compute the zig-zag path, it looks like this:

Code:

          *       *       *                   *
        /   \   /   \   /   \               /   \
    B *-------*-------*-------*-- . . . --*-------* C
      : - - - - - - - -  10 - - - - - - - - - - - :

Even at the molecular (or even atomic) level,
. . the length of the zig-zag path is always 20 cm;
. . the length of BC is always 10 cm.

Years ago, I saw a similar problem involving the diagonal of a square.
Suppose we have a square with side 4.

Code:

    A o---------------o B
      | *             |
      |   *           |
      |     *         |
    4 |       *       |
      |         *     |
      |           *   |
      |             * |
    D o---------------o C
      : - - - 4 - - - :

We see that $AB + BC \,=\,8.$

Bisect

and

and form this zig-zag path.

Code:

              D
    A o-------o
      | *     |
      |   *   |
      |     * |
    4 |       o-------o F
      |      E  *     |
      |           *   |
      |             * |
    D *---------------o C
      : - - - 4 - - - :

We see that the zig-zag path $\text{A-D-E-F-C}$ has length 8.

Divide the horizontal and vertical segments again and we have:

Code:

          G 
    A o---*
      | * |
      |   *---* I
      |  H  * |
    4 |       o---* K
      |      J  * |
      |           *---* M
      |          L  * |
    D *---------------o C
      : - - - 4 - - - :

We see that the zig-zag path $\text{A-G-H-I-J-K-L-M-C}$ has length 8.

As we continue this process, the zig-zag path approaches the diagonal.

Therefore: . $AC \,=\,8$

Denis · **Posted:** Wed, 8 Feb 2012 20:07:19 UTC

Nope Soroban: a paradox is 2 doctors.

Soroban · **Posted:** Wed, 8 Feb 2012 22:13:59 UTC

You're thinking of paramedics, Denis.

A paradox is what you park your yacht between.
. . (Or is that paramour?)

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