Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss

Puzzle I can't solve.

22 views
Skip to first unread message

MajorOz

unread,
Jun 26, 2012, 1:18:03 PM6/26/12
to
1. ANY four sided figure

2. Inscribe a square on each of the four sides

3. Find the center of each square.

4. Draw two lines, joining the OPPOSITE centers

Prove:

The lines are perpendicular AND of equal length, but NOT necessarily
bisectors.



I have fussed with this for years, to no avail.

Rumor has it that the solution lies in a math classroom at the U of
WA.

???

Ken Pledger

unread,
Jun 26, 2012, 5:39:35 PM6/26/12
to
In article
<924c523d-0c10-484e...@v9g2000vbc.googlegroups.com>,
MajorOz <ozm...@gmail.com> wrote:

> 1. ANY four sided figure
>
> 2. Inscribe a square on each of the four sides
>
> 3. Find the center of each square.
>
> 4. Draw two lines, joining the OPPOSITE centers
>
> Prove:
>
> The lines are perpendicular AND of equal length, but NOT necessarily
> bisectors.
>
>
>
> I have fussed with this for years, to no avail....



Google for "van Aubel's Theorem."

Ken Pledger.

MajorOz

unread,
Jun 27, 2012, 11:32:34 AM6/27/12
to
On Jun 26, 4:39 pm, Ken Pledger <ken.pled...@vuw.ac.nz> wrote:
> In article
> <924c523d-0c10-484e-8f04-b94c11c16...@v9g2000vbc.googlegroups.com>,
>
>
>
>
>
>
>
>
>
>  MajorOz <ozma...@gmail.com> wrote:
> > 1. ANY four sided figure
>
> > 2. Inscribe a square on each of the four sides
>
> > 3. Find the center of each square.
>
> > 4. Draw two lines, joining the OPPOSITE centers
>
> > Prove:
>
> > The lines are perpendicular  AND of equal length, but NOT necessarily
> > bisectors.
>
> > I have fussed with this for years, to no avail....
>
> Google for "van Aubel's Theorem."
>
>    Ken Pledger.

Thank you so much.

I now sleep better.
0 new messages