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Right Angles in Polygons


Date: 07/23/97 at 01:52:28
From: Heng Lok
Subject: Right angles in polygons

Dear Dr. Math,

This problem was given to me as a mathematics (11th grade) problem-solving
task. Two weeks later, no one (in my class) has solved it.

+---------+
|90     90|
|          \
|90         \
+------------+

This 5-sided polygon has 3 right angles, the maximum number when 
there are 5 sides.

All polygons have a maximum number of interior right-angles. A seven- 
sided polygon will have a maximum of 5.
              
From investigating, I have found that in a triangle, there is a 
maximum of 1 right angle, and in a quadrilateral, there are a maximum 
of 4. These polygons are not regular polygons of course. In a 5-sided 
shape, there can be a maximum of 3 right angles. Is there a relation 
between the number of sides in a polygon and the maximum number of 
right angles? Perhaps there is a formula. Please help me. 

Thank you very much.

Heng Lok


Date: 07/24/97 at 16:50:47
From: Doctor Rob
Subject: Re: Right angles in polygons

For convex polygons, the maximum is 3 for all numbers of sides > 4.
If you allow concavity, I can arrange for (n+4)/2 when n is even, and
(n+3)/2 when n is odd.  If you allow cross-overs, whenever n is a
multiple of 4, you can get n right angles, n-2 if n is odd, and n-1 
if n is 4*k+2.

I am not sure if these are optimal.  If you solve this problem, please
let us know.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 07/26/97 at 19:49:50
From: Doctor Ceeks
Subject: Re: Right angles in polygons

Hi,

For this, you may want to consider the following generalization:

Let k be an integer greater than or equal to 3. Then a regular k-gon
has interior angles measuring a=180(k-2)/k.

Let f_k(n) be the maximum number of interior angles measuring a in
a n-gon.

Determine f_k(n).

Your question is the case k = 4.

As a start, one can see that f_4(n)/n approaches 2/3 in the limit
as n tends to infinity as follows:

Suppose you have an n-gon with t right angles.

Then there can be no more than 90t+360(n-t) total interior angles.
Since every n-gon has 180(n-2) total interior angles, we have the
inequality:

90t+360(n-t) >= 180(n-2).

Rearranging yields:

3t <= 2n + 4

or

t <= (2n+4)/3.

On the other hand, you can make an n-gon that looks like a series of
squares which has, asymptotically, 2/3 of the interior angles right.
(It's hard to describe it precisely in words...if you can't find it,
please write back and I'll attempt to describe the shape precisely.)

-Doctor Ceeks,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 07/28/97 at 01:55:39
From: Heng Lok
Subject: Re: Right angles in polygons

Basically the only rules are that the shape must be closed and that
cross-overs are not allowed. They can be concave or convex. There is
meant to be a relation/formula to find this but I haven't found it 
yet.

Thanks for your help.


Date: 07/28/97 at 11:53:45
From: Doctor Ceeks
Subject: Re: Right angles in polygons

Hi,

I know the formula for 90 degree angles and if you want
me to tell it to you, please ask.

Otherwise there is a standard mathematical technique used
to tackle this sort of question:

1. Try to find a theoretical upper bound.

2. Construct families of polygons which realize the upper bound,
   and if you can't, try to see that the actual upper bound is
   smaller by the amount that you can see how to actually construct
   examples.

The upper bound established in my first message was based on the
fact that no angle exceeds 360 degrees, so this also provides
a clue as to how to realize the actual examples.

-Doctor Ceeks,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 12/18/2002 at 23:03:08
From: Doctor Peterson
Subject: Re: Polygons

Hi, Flea!

There are several formulas here, but none actually claims to be the 
formula answering the original question, "Is there a relation between 
the number of sides in a polygon and the maximum number of right 
angles?" You are apparently using Dr. Rob's formula

  (n+4)/2 when n is even, and (n+3)/2 when n is odd

but he says "I am not sure if these are optimal." Dr Ceeks gives

  t <= (2n+4)/3

for the number of interior 90-degree angles.

It appears to me that both of them are thinking only about right 
angles that turn inward, so that the interior angle is 90 degrees and 
not 270. The question mentions and may mean "interior right angles," 
although 270 degrees is also a right angle; but I wonder if you are 
instead counting right angles in either direction (perpendicular 
adjacent sides, whether going out or in). That question is more easily 
answered.

For any even n, it is possible to make an n-gon in which ALL angles 
are right angles:

    +---+
    |   |
    |   +---+
    |       |
    |       +---+
    .           .
    .           . . +
    |               |
    |               +---+
    |                   |
    +-------------------+

For k steps, this gives 2k right angles.

For odd n, you can get n-2 right angles:

    +---+
    |   |
    |   +---+
    |       |
    |       +---+
    .           .
    .           . . +
    |               |
    |               +---+
    |                   |
    |                   +
    |                 /
    +---------------+

I've added one extra vertex, and all but two of the angles are right 
angles. It's easy to see that this is the best you can do: each right 
angle turn will go from a vertical to a horizontal edge, so sides will 
alternate between the two orientations. With an even number of sides, 
the last side would come back parallel to the first, which is 
impossible (unless you allow polygons with vertices inserted into a 
straight edge). The last side therefore has to be at an angle to the 
first, and the angle before that must be its complement. These two 
angles therefore can't be right angles, but all the rest can.

So the answer to this question is

    n   when n is even
    n-2 when n is odd

And that fits your example for n=9.

But maybe you are counting only interior right angles, after all. I 
just played around a bit more and found this 9-gon with 7 interior 
right angles:

    +---------------+
    |               |
    |           +   |
    |         / |   |
    |       /   +---+
    |     /
    |   +---+
    |       |
    +-------+

That does improve on Dr. Rob's formula, showing that it is non-optimal, as
he had allowed. I'll have to think more about this. Do you have any ideas
about a formula for the real maximum?

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/
    


Date: 12/18/2002 at 16:07:56
From: dead_flea
Subject: Re: Polygons

I looked at Doctor Ceeks' formula for right angles in a polygon, and tried
it out on some examples. It worked for even but not for odd (n + 3)/2. For
example, 9 has 7 right angles, so you see it can't work.

Please can you look at this again?


Date: 08/16/2017 at 20:40:25
From: Doctor Peterson
Subject: Re: Polygons

Hi.

Actually, Dr. Ceeks did not restrict his answer to odd numbers of sides;
it applies to any n. Also, he misstated the inequality, which really
should have been t < (2n + 4)/3, since the non-right angles must each be
LESS than 360 degrees, not "no more than."

And the construction he alludes to demonstrates that the greatest integer
that satisfies his inequality can actually be attained, leading to an
actual formula for t, not just an inequality.

I believe that Dr. Ceeks' claim is that his formula does give the actual
maximum, in this way: The inequality shows that t must be an integer less
than (2n + 4)/3, and a construction can yield a polynomial that actual
attains the greatest integer less than (2n + 4)/3, so that is the actual
maximum.

We can use the floor function (the greatest integer less than or equal to
a number), by reducing the bound to (2n + 3)/3 to ensure that the floor
will always be less. The answer is

   t = floor(2n/3 + 1)

So all that is left is to actually find polygons that have this many right
angles.

I found two discussions of the problem that illustrate what you have to
do.

This PDF includes a quick derivation of the formula with a little less
justification than I would like, and without actually showing that the
claimed values can be attained:

  Getting It Right, by B S Beevers
  Mathematics in School, March 1996
    http://www.m-a.org.uk/resources/Vol-25-No2-Mar_1996_Getting_it_right.pdf

This book reports an open-ended discussion and includes a table that
represents an inaccurate stage in their thinking, but shows a dodecagon
with 9 right angles that can serve as the key for the needed construction,
and is undoubtedly what Dr. Ceeks meant by "an n-gon that looks like a
series of squares":

  Mathematical Knowledge for Primary Teachers
  Andrew Davis, Maria Goulding, Jennifer Suggate
  https://books.google.com/books?id=7T8lDwAAQBAJ&pg=PA42

From that, you can show that whenever n is a multiple of 3, t = 2n/3 + 1
is attainable.

Incidentally, in searching for those, I found another page that deals with
the problem as I interpreted it:

    http://mathcentral.uregina.ca/qq/database/qq.09.09/h/bruce1.html

I've left a few things that you can pursue, such as finishing the proof
that the formula can be attained.

I hope you enjoy it. I've had fun digging into this. 

- Doctor Peterson, The Math Forum at NCTM
  
/pre
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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