I Have An Answer To My Question And A Solution To My Puzzle!

In my previous post, I talked about a puzzle that was triggered by a geometry problem that my daughter brought home from school. The puzzle is quite simple, and involves midpoint polygons. These are inscribed polygons that are derived by joining the midpoints of the sides of the exterior polygons. The puzzle was to find the ratio of the perimeter of the midpoint polygon to the perimeter of the outer polygon as a function of the number of sides of the polygon.Last week I stopped at hexagons because my techniques were not good enough to allow me to calculate the perimeter of a midpoint heptagon given a regular exterior heptagon of unit sides. But, I kept thinking about the puzzle on and off, and have now come up with a solution that should work for any regular polygon. In this post, I am going to explain how my method works, and present some results of my investigations.

The good news is that it is possible to express the ratio in terms of the number of sides for any regular polygon. The bad news is that the formula looks quite ugly and complicated. All the simple formulas like E=mc² are already taken!

The secret to working out the ratio starts with drawing a diagonal that connects two alternate vertices of the exterior polygon. By alternate vertices, I mean two vertices that are separated by one intermediate vertex. In the figure to the left, consider segments AB and BC to be part of the exterior polygon (I have not drawn the rest of the polygon because I want to derive a formula that applies to polygons with any number of sides. So, my figure only shows two sides of a polygon with an arbitrary number of sides). They are equal to each other because the polygon is regular. Let us assume they are each 1 unit in length.

Segment AC is a diagonal that connects two alternate vertices of the exterior polygon (B is the intermediate vertex that is skipped over by the diagonal). ABC is now an isosceles triangle. Angle ABC is an interior angle of the polygon, and it is a function of the number of sides the polygon has. In particular, we know that angle ABC is equal to (n-2)*180/n degrees, where n is the number of sides of the regular polygon. In the figure, I have represented this angle by x.

We also know that angles BCA and BAC are equal, and because they are part of triangle ABC, we know that each is equal to (180–x)/2. Since x = (n-2)*180/n degrees, we can substitute that in the expression above, and get the measure of angles BAC and BCA to be 360/2n degrees. In the figure, I have represented this angle by y.

Surprisingly, that is all the information we need to get the ratio we are after. The first step in calculating the ratio is the calculation of the length of segment AC. Why? Because, we know from my previous post that each side of the midpoint polygon is half the length of the diagonal that connects alternate vertices of the exterior polygon.

So, how do we go about calculating the length of segment AC? We apply the law of sines to triangle ABC to calculate the length of segment AC. We know that angle x is opposite to the segment AC whose length we want, and angle y is opposite the segment AB whose length is 1. Thus, 1/sin(x) = AC/sin(y). Therefore, AC = sin(y)/sin(x).

At this point, we can express x and y in terms of n, and we will have AC in terms of n. Divide the length of AC by 2, and that is the ratio we are after (the length of one of the sides of the midpoint polygon). Doing these substitutions, we have:

AC = sin(360/2n)/sin((n-2)*180/n)

Therefore, the ratio of the perimeter of the midpoint polygon to the perimeter of the regular exterior polygon (which is half of AC) would be:

R = sin(360/2n)/(2*sin((n-2)*180/n))

Well, it may not be pretty, or roll off the tongue quite like E=mc², but it is a function of n, the number of sides of the polygon, and that is what we started out seeking. So, pretty or not, we have achieved what we wanted to achieve, and anyway, half the fun of getting there is the traveling itself, right? So what if the destination was not stunning, the route was plenty scenic, right?!

Now, how do I know this is correct? Below is a table with the value of R calculated for regular polygons with different number of sides. The first 3 lines correspond to polygons that we dealt with in the previous post (squares, pentagons and hexagons). I had to start with squares because triangles do not have diagonals (unless you consider each side to be a diagonal also). As you can see, the value of the ratio as calculated using the above formula is identical to the value of the ratio which we derived using other methods in the previous post.

Number of Sides (n)	Interior Angle ABC (x) (in degrees)	Angle with Diagonal (y) (in degrees)	Length of Diagonal AC	Ratio of Perimeters (R)
4	90	45	1.414213562	0.707106781
5	108	36	1.618033989	0.809016994
6	120	30	1.732050808	0.866025404
7	128.5714286	25.71428571	1.801937736	0.900968868
8	135	22.5	1.847759065	0.923879533
9	140	20	1.879385242	0.939692621
10	144	18	1.902113033	0.951056516
100	176.4	1.8	1.999013121	0.99950656
1,000	179.64	0.18	1.99999013	0.999995065
10,000	179.964	0.018	1.999999901	0.999999951
1,000,000	179.99964	0.00018	2	1

It is interesting to note that, as predicted, the ratio does approach 1 as the number of sides goes up. I have added rows for 100, 1000, 10000 and 1000000 sides to the table above just to illustrate how the ratio gets closer to 1 as the number of sides goes up. There are so many 9's in the answers for the polygon with a million sides that I decided to do away with them, and put down the answer rounded to a billionth!

By the way, a polygon with 100 sides is called a hectogon. A polygon with 1,000 sides is called a chiliagon, one with 10,000 sides is called a myriagon, and one with a million (1,000,000) sides is called a megagon!

Since we know that the ratio of the area of an inscribed midpoint polygon to the area of the exterior regular polygon is the square of the ratio of perimeters, you can just take the numbers from the table in this post and square them to get the ratios for the areas without having to explicitly calculate the areas of either the exterior polygons or the inscribed midpoint polygons.

There exist formulas for the area of a regular polygon based on the number of sides, and the length of each side, so if you are curious, you can calculate the area of the exterior polygon using such a formula, calculate the area of the midpoint polygon using the ratio in the table to compute the length of each side, and then calculate the ratio of the areas. The amazing thing about mathematics is that you can derive general results (such as: the ratio of areas is the square of the ratio of the perimeters), and after that you don't have to do all the hard work involved in calculating specific results. But the tools exist to do so if you want to.

Well, I consider this another example of using very simple, well-known concepts, and stringing them together in just the right order to derive something a little more interesting and complicated. It is what makes and keeps mathematics so interesting and challenging. And hopefully, it will keep our brains young and fresh! Good luck in your mathematical explorations!!

An Interesting Geometrical Puzzle

My daughter recently brought home a puzzle that had me scratching my head for a while before being able to solve it. Actually, the solution was quite simple once I figured it out, but it spawned off a different puzzle in my head that may be a little trickier. In any case, I have not figured out a solution to the spawned-off puzzle yet, but let me not get ahead of myself. Let me first tell you about the puzzle that my daughter brought home.

You are given a regular pentagon, ABCDE (regular in this context means that all the sides are of equal length, and all the interior angles are equal to 108 degrees). You are also given the fact that the diagonal AC of this pentagon is 10 units long. Please refer to the figure on the left.

Now, inside this pentagon is another pentagon formed by joining the midpoints of each of the sides of the original pentagon. Call this inscribed pentagon FGHIJ. The puzzle is to use all this information to find the perimeter of the pentagon FGHIJ.

Initially, for some reason, I thought the solution would consist of finding the perimeter of ABCDE first and then somehow using that along with the length of AC to derive the perimeter of FGHIJ. That is what had me floundering for a few minutes. In fact, I even looked up the Wikipedia article on pentagons for some inspiration.

But the solution is actually much simpler (obviously, the fact that this puzzle was given to a middle-school student as an assignment is a big hint that this is not the geometric equivalent of Fermat's last theorem!). It is simply sufficient to realize that the diagonal creates a triangle (ABC in this case), and the line segment FG connects the midpoints of two of the sides of this triangle.

You can then apply the midpoint theorem on this triangle to conclude that side FG must be equal to 5 units in length because AC is 10 units long. And because FGHIJ is a regular pentagon, being the midpoint pentagon of a regular pentagon, its perimeter must be 5*5 = 25 units.

But, obviously, half the fun of solving a mathematical problem is coming up with related puzzles and questions. In particular, the puzzle I am now stuck with is as follows: what is the ratio of perimeters of midpoint polygons to the perimeters of the polygons they are inscribed in as a function of the number of sides of the polygon? Similarly, what is the ratio of areas of midpoint polygons to the perimeters of the polygons they are inscribed in as a function of the number of sides of the polygon?

Note that the polygons formed by joining the midpoints of the sides of a given polygon are called midpoint polygons, and have been studied by mathematicians in some detail. In particular midpoint polygons constructed from regular polygons are also regular, and are geometrically similar to the exterior polygon.

Let us take a regular triangle (also known as an equilateral triangle) first. Connecting the midpoints of the three sides to create an inscribed midpoint triangle inside the original triangle gives us a triangle with half the perimeter as the original. Note that in the figure below, triangle ADF is also equilateral, and since AD and AF are one half of the length of AB, DF is one-half of the length of AB. And since DE, EF and DF have the same length (each equal to half the length of the sides of the original triangle ABC), the new triangle has half the perimeter as the original triangle.

Also note that the original triangle ABC has been divided into 4 equilateral triangles of equal area by the addition of the inscribed triangle DEF. Thus, the area of triangle DEF is one quarter of the area of the original triangle ABC. Thus, the ratio of perimeters is 1/2, and the ratio of areas is 1/4.

Moving on to a square, which is a regular quadrilateral, we can use the pythagorean theorem on triangle EBF to figure out that the length of EF is equal to (length of AB)/sqrt(2). And the area of EFGH is one-half of the area of ABCD. Thus, the ratio of perimeters is 1/sqrt(2), and the ratio of areas is 1/2.

The situation becomes more complicated in the case of a pentagon. In fact, it is not easy to figure out what the length of each side of an inscribed pentagon is if you know just the lengths of the sides of the original pentagon. But my research into regular pentagons on the Wikipedia site did have a positive side-effect: it turns out that the diagonals of a regular pentagon are in golden ratio to its side. In other words, if the length of a side is 1 unit, the length of a diagonal is eaual to the golden ratio. The value of golden ratio is 1.6180339887498948482045868343656.

From my solution of the puzzle which started this entire exploration, we also know that the length of each side of the inscribed pentagon is one half of the length of the diagonal. Thus, if the length of each original side is 1, then each diagonal is 1.6180339887498948482045868343656 units long, and the length of each inscribed side is 1.6180339887498948482045868343656/2 units long. Thus, the ratio of perimeters is 1.6180339887498948482045868343656/2.

The same Wikipedia article also tells me that the area of a regular pentagon with sides of length t is 1.720477401*t^2. Thus, the area of the inscribed pentagon would be 1.720477401*(1.6180339887498948482045868343656*t/2)^2. What we are interested in is the ratio of the areas, not the actual areas themselves. And that would simply be (1.6180339887498948482045868343656/2)^2.

What about a hexagon? Well, the analysis is a little more complicated in the case of a hexagon. First of all there are two formulas for the area of a hexagon that we can use to derive the length of a diagonal of the hexagon that connects two alternate vertices of the hexagon (note that a hexagon also has three diagonals that connect opposite vertices, and these diagonals are twice the length of each side. They are also longer than the diagonals we are interested in). You can find these formulas in the wikipedia article on hexagons.

The first formula for area is: area = 2.598076211*t^2. The second formula for area is: area = 1.5*d*t. Interestingly, d is the length of the diagonal connecting alternate vertices of the hexagon (it is the height of the hexagon when it is resting on one of the sides as its base). Using these two formulas, since the area of a hexagon is a single number, we can say that 2.598076211*t^2 = 1.5*d*t. From this, we can derive the value of d to be 2.598076211*t/1.5.

At this point, we can then use the midpoint theorem to say that each side of an inscribed hexagon would be one-half the length of each of these diagonals. Thus the ratio of perimeters would be 2.598076211/3. And the ratio of areas would be the square of that number since the area is directly proportional to the square of the length of a side.

So, that is as far as I have gotten so far. The area formula for heptagons does not allow me to calculate the length of a diagonal connecting alternate vertices of a heptagon. So, I can not calculate the length of a side of an inscribed midpoint heptagon either. You could say, I am stuck!

In table form, one can express this as below:

Number of Sides	Ratio of perimeter of midpoint polygon to perimeter of exterior polygon	Ratio of area of midpoint polygon to area of exterior polygon
3	0.50	0.25
4	0.70710678118654752440084436210485	0.50
5	0.8090169943749474241022934171828	0.65450849718747371205114670859138
6	0.86602540366666666666666666666667	0.75

Pretty basic observations follow from this.

• For regular polygons, area seems to be always proportional to the square of the length of a side. Thus, the ratio of areas of inscribed midpoint polygons to exterior polygons would be the square of the ratio of perimeters.
• The ratio of perimeters (and thus the areas) approaches 1, and will be 1 in the asymptotic case of a polygon with infinite number of sides (one could argue that the inscribed midpoint circle of a given circle is coincident with the given circle, and thus the ratio of perimeters as well as areas is indeed 1).
• Since the ratio of areas is the square of the ratio of perimeters, and the ratio of perimeters is less than 1.0, the ratio of areas is always less than the ratio of perimeters.
  

These observations though, do not tell me the answer to the question I started with: what is the ratio of the perimeter of an inscribed midpoint polygon to the perimeter of the exterior polygon? Is it possible to find an expression that tells me the value of this ratio for any type of polygon (obviously, such an expression would be a function of the number of sides of the polygon)? Any ideas? I would love to hear your thoughts on how to proceed with this investigation. Thank you, and good luck!

Anti-Reflective Coatings Have Come A Long Way

Anti-reflective coatings are used on optical lenses to increase the collecting of light, and to prevent stray light from bouncing around inside systems with multiple lenses. They have been around for a while now, having been invented in Germany in the 1930's. Anti-reflective coatings on corrective glasses make them look better, and also reduce glare, especially in high-contrast situations like night-time driving.

When I bought my daughter's latest pair of eyeglasses, I decided to go with Zenni's latest anti-reflective coating on the lenses. This coating is the latest generation, and is technically advanced. Not only does it reduce glare and make the glasses less likely to stand out in photographs, but it also boasts several advanced features.

This new generation of AR coatings is hydrophobic and oleophobic. This means that the coating repels not only water, but also oils. Repelling oil is significant because most smudges, fingerprints, etc., on lenses are caused by the oils on your hands and fingers. By repelling most oils, the oleophobic lenses stay cleaner longer, and are much easier to clean.

The differences between her old pair of glasses and the new pair of glasses has made a believer out of me as far as this coating is concerned. My daughter gets crystal clear vision without having to deal with glare from table lights, her computer screen, overhead lights, etc. Since more light passes through the lenses instead of being reflected away, things appear brighter, and with better contrast. The lenses have been much easier to clean and keep clear of dirt, smudges and fingerprints. And, best of all, she likes the fact that she does not have remove her glasses before posing for photographs!

Quick Review Of The GE Disposall GFC720T 3/4 HP Garbage Disposer

Last week, after a long and laborious struggle, my old garbage disposer finally died. It was an Insinkerator Badger model which had a 1/2 HP motor. It was installed about 5 or 6 years back, and it was never a great disposer. It worked OK, but it was loud, and got stuck quite frequently. It required my wife and I to unclog it pretty regularly with a plunger.

Eventually, one fine day last week, the disposer simply refused to work. I could hear the motor humming when I turned it on, but the motor would not turn. I used the provided allen wrench to turn the motor externally thinking something was stuck. But I could turn the motor externally quite easily, but the motor never wanted to turn by itself. I even turned the motor this way and that way externally while it was on, but apart from humming and becoming warm, the motor refused to turn.

It was time to do some shopping. I called and fixed up an appointment with my usual plumber to replace the garbage disposer. Then, I went out and shopped around for a new garbage disposer. I checked reviews on various sites also and found that the best ones were very expensive (well over $250). When I went to the stores, I found the best-reviewed garbage disposers selling for well over $300. The best-reviewed garbage disposer is one made by Insinkerator, by the way. I also found the disposer I eventually purchased, the GE Disposall GFC720T selling for just over $100.

The specs of this disposer are quite reasonable. It has a 3/4 HP motor, and a 2-stage grinding mechanism. I had used Insinkerator brand garbage disposers for the past 13 years, and had gone through 3 of them in that time period. I decided it was perhaps time to try a different brand. It also helped that the GE Disposall was about $200 cheaper than the Insinkerator brand disposer of the same power rating.

Like all 3/4 HP disposers, the GE Disposall is quite large around. At least larger than 1/3 or 1/2 HP disposers. It seems to be slightly smaller than the corresponding 3/4 HP disposers from Insinkerator. It should fit under most sinks without much of a problem, but if you have an extra small under-sink space, you might want to check the clearances before buying a bigger disposer.

I did not do the installation. That was what I paid the plumber over $300 to do! It took him well over 2 hours to do it. It took about 15 minutes for him to test and remove the old garbage disposer. After I showed him how the motor never turns even though it hums, he concurred with me that it was possible the motor was partially burnt out. He also tried turning it manually and found that there was no obstruction to the motor turning. The motor had just given up, and it was time for me to give up on the disposer.

Next it was time to put the new disposer in place. The fittings that attach this disposer to the sink are a little different from what were used for the Insinkerator. This is not surprising since every brand probably uses proprietary, and perhaps patented, fittings to attach it firmly to the sink and prevent leaks. Insinkerator is by far the best-selling garbage disposer, so plumbers are much more familiar with installing those than any other brand. The plumber had to refer to the installation directions that came with my new disposer several times during the installation to get it all squared away.

Next came the process of hooking up the drains. This took a long time because the outlet of this disposer was lower than the outlet of my previous disposer. This meant that all the under-sink plumbing had to be adjusted to make sure that the drains worked as before. He had to extend one section of pipe (well, actually that simply meant he had to replace the existing pipe with a slightly longer section of pipe), and some others had to be cut down to size. I am not going to hold this against the new disposer because I would have probably had the same problem simply by going from my original 1/2 HP disposer to a 3/4 HP disposer, even if I had stuck with the same brand.

Then came the process of hooking up the disposer to the electric system. Here, I think GE could have definitely done a better job of things. The problem was that the new disposer did not come with all the hardware to hold the electrical connections inside the body of the disposer. The old disposer had come with a nut that holds the electrical connections in place. This disposer did not, and the old nut would not fit on the new disposer. The plumber had to use a lot of electrical tape to simulate what the nut did by taping the electrical connections securely to the disposer so that they don't eventually come apart.

Finally, it was all done, and it was time to test the disposer. We did not have actual garbage to test it with, but we filled the sink with water (the disposer comes with a sink drain plug that can be used to plug up the drain and therefore fill up the sink), and then pulled the plug while running the disposer. There were no leaks anywhere, and the disposer itself ran quite well. Like all garbage disposers I have used, this one also has motor overheat protection, that trips a fuse when the motor gets overheated. You can reset it after the motor cools down by pressing in a small switch on the side of the garbage disposer opposite the water outlet.

It is much quieter than the old disposer. Most of the time, all you hear is a faint humming sound from the electric motor. The actual grinding mechanism is well-insulated and you don't hear it most of the time. However, the problem this disposer has is that if the motor is slowed down from its usual speed because of the load it is under, the whole disposer starts vibrating violently. This vibration is then transmitted to the sink, and from there to the entire counter-top.

Most of the time that I use the garbage disposer, I am very happy with its performance. But when it gets a little overloaded and it goes into this vibration mode, it produces quite a racket with everything in the sink, and most of the stuff on the counter-top jumping around because of the vibration. Once the disposer is able to get back to its normal speed, it quiets down, but in the meantime, it is quite scary. I don't think it is healthy for the sink and counter-top either to be so violently shaken every time this happens. If you have breakable stuff on the counter-top, better move them well inside from the edges of the counter-top before switching on this garbage disposer.

The user manual suggests never to let garbage get into the disposer before it is switched on. That way, the disposer deals with the garbage while it is spinning, and never gets into a situation where it tries to start up under load. And I have noticed that this slowdown and vibration occurs only on start-up, not after the disposer has started spinning at speed. So, if you are very disciplined about making sure that your disposer is not clogged with garbage before you switch it on, you may never have to deal with this vibration problem.

The next thing to note is that this garbage disposer does not come with an allen wrench to turn the motor externally. If the motor is stuck, the advice in the user's manual is to use a broom stick inside the garbage disposer to try and get it unstuck. Personally, I think the allen wrench to turn the disposer from outside is a much better, more practicable idea.

Also, the garbage disposer has a very narrow opening from the sink. I am not a large man, and I have smaller than average hands, but I can't put my hands into the disposer through the opening in the sink. I used to be able to do this quite easily with the old disposer, and regularly used to do so when something was stuck inside, and had to be taken out to the garbage rather than being disposed off down the drain. My wife and kids, who have smaller hands than I do, can reach in, but the disposer seems to be designed to prevent most normal-sized hands from being inserted into the body of the disposer.

What do you do if something that does not belong in the disposer does get stuck inside the disposer? The user's manual's suggestion is to use long tongs to fish out such offenders from the body of the disposer. Well, so far, I have been lucky that things like spoons or forks have not fallen into the disposer. But I can tell you that I don't like the prospect of fishing for things like that with tongs. Can you imagine manipulating a spoon or fork using tongs carefully enough to retrieve it through the narrow opening of the disposer? My advice would be to be very careful about what goes into the garbage disposer so that you never have to take anything out.

So, what is the bottomline? For the cost, this is a very solid garbage disposer. It is powerful, and quiet. It has never jammed or had problems disposing of any garbage so far. However, installation was a bit of a problem, not only because plumbers are unfamiliar with this brand, but also because it does not have the correct hardware included for securing the electrical connections.

Even though it is quiet when running at its normal speed, any slowdowns caused by extra load result in excessive vibration that could be damaging to the sink, the counter-top and the plumbing under the sink. There is no way to get the disposer unstuck by using an allen wrench from the outside of the disposer. I am not entirely sure how successful poking around the inside of the disposer with a broom handle will be. And more importantly, the sink opening of this disposer is designed to prevent most adults from being able to reach inside the disposer to retrieve stuff that does not belong in there. The user's manual suggests using long-handled tongs to do this. I would say, good luck doing so!

But it is early days still, and I don't know how durable it is going to be. If it works like it does right now three years or five years from now, I would have no complaints about it whatsoever. I will update you on any problems that it develops down the road. For now, there are no insurmountable drawbacks to this disposer except for the vibration under load. I would therefore give this a 3.5 or 4 rating out of 5. If the vibration problem were fixed, it might merit a 4.5 out of 5.