Submissions and Requests
As a puzzlesite owner, I get an occasional request for help on one problem
or another. When they're especially interesting, I usually turn them into
a POTW quickly. Though some may not be completely applicable to a POTW,
they're still interesting. Here are a few questions I've received,
followed by a few puzzle submissions I've received.
Solutions are included within the puzzles below. Solvers are listed
at the end.
Requests for help

An interior designer has an intriguing dilemma. She has to move an amazingly
heavy armchair, but the only possible movement is to rotate it through ninety
degrees about any of its corners. Can it be moved so it is exactly beside its
starting position and facing the same way? [Update 9/18: One reader suggested
this problem should be attacked without requiring the angle of rotation
always be 90 degrees. Consider this as part (b) of this problem.]
Solution:

a) No. To see this, paint the floor in a checkerboard pattern with the chair on
a white square and, without loss of generality, assume the chair faces North.
Every time the chair is moved, it moves to a square of the opposite color
and its orientation toggles between North/South and East/West. So it faces
North or South when on white squares and East or West when on black. The
squares adjacent to the chair's starting position are black.
 b)
Rotate 30 counterclockwise about B.
60 clockwise about C and finally,
30 counterclockwise about B.

An angus is anchored to a 6 meter rope. The rope is attached to the outside
corner of a shed measuring 4 meters by 5 meters in a rolling grassy meadow with
varying elavations. What area of grass can the cow graze in 1.5 hrs?
Solution:
The animal can reach 3/4 of a circle of radius 6m, plus 1/4 of a circle of
radius 1m (going around the corner on the 5m side), plus 1/4 of a circle
of radius 2m (going around the corner on the 4m side).
Total is 27pi + 0.25pi + 1pi = 28.25pi.

Prove that: Within a set of triangles having a constant base and constant
perimeter,
the isosceles triangle has the maximum area.
Solution:
Let us consider Heron's formula for area:
S^2=p(pa)(pb)(pc).
p and pa are constant and
(pb)(pc)=(ab+c)(a+bc)=[a^2(bc)^2)]/4.
Hence S^2=k[a^2(bc)^2], where k is a constant.
a is also constant, so the area is maxim when (bc)^2=0, that is b=c.
Also, if an ellipse is drawn so that the endpoints of the base of the
triangle are the foci and the locus of the third point of the triangle
is the ellipse itself, then the solution is clear from the diagram.

What is the missing number ?
30
26 34
22 ? 26
11 18 27 31
Solution:
The missing number is 36. The rule to get a number of the pyramid is:
sum the digits of the two adjacent numbers below it and multiply the result
by 2.
For example, if we take 11 and 18, the result is 2*(1+1+1+8)=2*11=22.
To get the missing number we calculate 2*(1+8+2+7)=2*18=36.
All the numbers in the pyramid follow this rule.
This problem or a problem quite similar to this can be found in the book
"The mathmagician and the pied puzzler".
[Solution from Pedro Romero 10/9/02.]

You have 6 pool balls, all identical in size though differently numbered.
5 balls are of the same weight, the 6th is different (i.e. heavier or
lighter).
You have a set of balancing scales and, using the scales twice only, you
have to identify the rogue ball.
Solution:
It can't be done. A good analysis of the ballweighing puzzles can be found
at:
http://recpuzzles.org/sol.pl/logic/weighing/balance. This page shows
that with two weighings, a rogue ball can only be found among 4.
Small Submissions

S     S
Enter a 4 letter word, divisible by 2, to get odd numbers.
Solution:
EVEN, creating SEVENS

14 people are eating at their hotel's restaurant. They are seated at
two tables which hold 8 and 6 people. If they are seated at these tables
for every meal, how many meals are needed to assure that every pair of
people sit at the same table at least once? What if the tables instead
hold 9 and 5 people?
Solution:
One three meal solution for 8/6:
1: ABCDE FGH : IJK LMN
2: ABCDE IJK : FGH LMN
3: ABCDE LMN : FGH IJK
One four meal solution for 9/5:
1: 1,2,3,4,5,6,7,8,9
2: 1,2,3,4,10,11,12,13,14
3: 5,6,7,8,10,11,12,13,14
4: 9,10,11,12,13,14,1,2,3

a/b + b/c + c/a = 1
Does a solution exist for a, b, and c belonging to the set of
negative and positive integers?
Does any solution exist? If so, show a simple example.
Solution:
There is no solution with integers.
For nonintegers, a simple solution is:
a = 1, b = 1, c = sqrt(2)1.

Simplify the infinite product
(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^16)..., given x < 1.
Solution:
Multiply by (1x) to get
(1x)(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^16)..., which is
(1x^2)(1+x^2)(1+x^4)(1+x^8)(1+x^16)..., which is
(1x^4)(1+x^4)(1+x^8)(1+x^16)..., which is
(1x^8)(1+x^8)(1+x^16)...
And we can see that, continued infinitely, multiplying the infinite
product by (1x) yields 1. Thus, given x < 1, the infinite product
simplifies to 1/(1x).
Source: Many sources. Submissions from Denis Borris and Ravi Subramanian.
Solutions were received from:
Al Zimmermann, Denis Borris, Claudio Baiocchi, Jeremy Galvagni,
Dane Brooke, Adrian Atanasiu, Saw L. B., Ross Millikan, Clinton Weaver,
Jayavel Sounderpandian, Paul Botham, J.S. Yoder, Kevin Costello,
Pedro Romero,
and Susan Hoover.
Mail to Ken