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### Kitchen Tabletop

```
Date: 11/21/2001 at 13:10:49
From: Russell Kopp
Subject: Geometry

Hi:

I have a kitchen table whose functionality involves the use of math.
First I'll explain the table:

From a crow's eye view, in its folded state the table is a rectangle,
with adjacent sides of different length. (I suppose it could have been
square, but I'm not sure).

The tabletop has hinges on one of its longer sides - for reference
I'll call this its east side. Attached to these hinges is another
tabletop of equal size that sits on top of the original tabletop when
the table is in its smaller, folded state.

The bottom of the two tabletops is connected on its bottom or under
side to a pivoting device that allows the bottom tabletop (and of
course the top tabletop attached to it) to be rotated clockwise 90
degrees. Once this rotation is effected, the top of the two tabletops
may be unfolded from the bottom tabletop, revealing a table twice as
large in size as the original. In its unfolded state, the overlap of
the larger table top on its base is such that north and south table
ends overlap the base by equal amounts, and the west and east ends
overlap by equal amounts as well. The overlap of adjacent sides need
not be the same.

Were the table split into four quadrants, the pivot would clearly lie
in the upper right (northeast) quadrant.

I wish to know how to determine the correct pivot point such that it
meets all the guidelines discussed above.

I should note that the table's actual dimensions in its folded state
are 35.4375" by 24.4375". I've done a little research, and should note
for those with geometry backgrounds that this is not a golden
rectangle (a rectangle whose adjacent sides are in a ratio of
approximately 1:1.618.)  I say this because I thought a golden
spiral's limit might be the pivot point - though it appears not to be
the correct point.

Thanks,

Russ
```

```
Date: 11/21/2001 at 20:20:44
From: Doctor Rick
Subject: Re: Geometry

Hi, Russ, thanks for writing to Ask Dr. Math.

I like questions like this - discovering math in everyday things! I've
the question you have.

It turns out that the answer is simpler than you thought. Here is a
figure:

+---------------+
|               |
..........|...............|..........
:         |               |         :
:         |               |         :
:         |               |         :
:         |               |         :
:         |               |         :
:         |               |         :
:         |               |         :
..........|.......B.......A..........
:         |               |         :
:         |               |         :
:         |               |         :
:         |               |         :
:         |               |         :
:         |               |         :
:         |               |         :
..........|...............|..........
|               |
+---------------+

You have stated that the opened table is centered on the base; you
didn't state, but I think I can assume it, that the folded table is
also centered on the base. Thus I've drawn the folded (solid lines)
and opened (dotted lines) tabletops centered at the same point. The
point marked A, in the center of the hinged side of the folded table,
must move to point B, the center of the entire table, when the top is
rotated 90 degrees.

All we need now, to find the center around which the tabletop is
rotated, is to locate a point such that lines drawn from it to A and B
are the same length and form a 90-degree angle. That's easy:

+---------------+
|               |
..........|...............|..........
:         |               |         :
:         |               |         :
:         |               |         :
:         |               |         :
:         |               |         :
:         |           O   |         :
:         |         /   \ |         :
..........|.......B.......A..........
:         |               |         :
:         |               |         :
:         |               |         :
:         |               |         :
:         |               |         :
:         |               |         :
:         |               |         :
..........|...............|..........
|               |
+---------------+

We have here a right isosceles triangle OAB; OA and OB are equal, so
A can rotate to B, and AOB is a right angle, so the tabletop is
rotated 90 degrees. If the width of the folded table (perpendicular to
the hinged side) is W, then the pivot point O is W/2 to the east of
the center, and W/2 north.

The proportions of the table are irrelevant. We just need a base that
is at least half the width of the folded table.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Practical Geometry
High School Triangles and Other Polygons

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