# CTK Insights

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18 Nov

### It Never Stops with Pythagoras

In the previous blog I described a discovery of Hirotaka Ebisui and an observation by Thanos Kalogerakis, both concerning what's known as Vecten's configuration. Vecten's configuration is a generalization of the famous Bride's Chair that underlies Euclid I.47, generally identified as the proof of the Pythagorean Theorem, although by now there are hundreds of them. In response to the previous post, Long Huynh Huu has expanded the two results, with the tools from linear algebra. To cut that introduction short, earlier today I was informed by Thanos Kalogerakis of a post by Carlos Hugo Olivera Días at the Peru Geometrico facebook group that adds another (and most fundamental at that) link to the just described chain of discoveries.

To find $M,$ and along the way determine that it does indeed exist, I'll employ the Law of Cosine, three times in the $\Delta ABC$ and then in each of the flank triangles. I'll use the well-known property of cosine: $=\cos (\pi-\alpha)=-\cos\alpha:$

$a^2=b^2+c^2-2\cos C$
$b^2=c^2+a^2-2\cos B$
$c^2=a^2+b^2-2\cos A$
$x^2=c^2+a^2+2\cos B$
$y^2=b^2+c^2+2\cos C$
$z^2=a^2+b^2+2\cos A.$

Adding all six up gives $(a^2+b^2+c^2)+(x^2+y^2+z^2)=4(a^2+b^2+c^2),$ i.e.,

$x^2+y^2+z^2=3(a^2+b^2+c^2),$

which not only finds that $$\displaystyle M=\frac{1}{3}$$ but actually proves that the expression $$\displaystyle M=\frac{a^2+b^2+c^2}{x^2+y^2+z^2}$$ is independent of the underlying triangle.

Thanos Kalogerakis made an additional observation. As we know, for the right triangle $x^2+y^2=5z^2,$ i.e., $x^2+y^2+z^2=6z^2,$ while, by the Pythagorean theorem, $a^2+b^2=c^2,$ i.e., $a^2+b^2+c^2=2c^2.$ Combining that with the latest revelation $6z^2=6c^2,$ i.e., $z^2=c^2$ which could be discerned from Vecten's diagram for a right triangle. That is exactly the case when a flank triangle is congruent to the base triangle.

11 Nov

### A Discovery of Hirotaka Ebisui And Thanos Kalogerakis

Today's communication from Thanos Kalogerakis brought to mind an insightful one page note by Alan Alda - a chapter in a collection This Explains Everything by John Brockman.

With every door into nature we nudge open, 100 new doors become visible, each with it own inscrutable combination lock.

On a rather small scale that Alan Alda's vision was paralleled in a message (and a short story that it conveyed) from Thanos Kalogerakis.

But again, as Alda puts it,

We need Einstein for GPS, but we can still get to the moon with Newton.

Who does not know or never heard of the Pythagorean theorem?

Early in the 19th century, a French geometer named Vecten studied a configuration of squares on the sides of a triangle. Vecten considered adding another layer of squares on the outer vertices of the Pythagorean squares, and then another one.

This configuration has many interesting properties, but Thanos Kalogerakis' message I mentioned at the outset, began with a statement, due to Hirotaka Ebisui. If we denote $A=a^2,$ $B=b^2,$ $C=c^2,$ then the Pythagorean theorem appears simply as $C=A+B.$

I shall denote the next layer of Vecten's squares by $A',B'C'$ and the next layer by $A'',B''C''$, as shown below:

Hirotaka Ebisui has found that in the case of the right triangle, $A'+B'=5C'$. On informing me of that result, Thanos added an identity for the next layer $A''+B''=C''.$

That was exciting enough for me to investigate. I can happily and proudly report that, for the next layer, $A'''+B'''=5C'''$.

Hence, I venture a general statement:

Let $A,B,C$ be the Pythagorean squares on the sides of a right triangle so that $A+B=C.$ Let's built recursively a sequence of triples of squares, $A_k,B_k,C_k$, with $A_0=A,$ $B_0=B,$ $C_0=C,$ and $A_{n+1},B_{n+1},C_{n+1}$ formed on the other vertices of $(B_n,C_n),$ $(C_n,A_n)$ and $(A_n,B_n).$ Then

$\begin{cases}A_n+B_n=C_n& n-even,\\A_n+B_n=5C_n& n-odd.\end{cases}$
17 Oct

### A pizza with a hole

The editorial in the Crux Mathematicorum (43(8), October 2017) posed an interesting problem; how to equally share a pizza with a hole.

To make the problem solvable, we need to assume a degree of abstraction. For example, if the hole makes it more difficult to divide a pizza, the assumption that it is possible to divide a a pizza without a hole into two equal parts needs to be accepted for a fact.

### A simplified problem

Now, to start with a simpler problem, how can we divided a rectangular pizza into equal parts with a single straight cut? We may find the midpoints of the opposite sides and join them with a cut.

There would be even less trouble dividing a circular pizza: any cut through the center of the pizza could be considered a fair division. Why so? Because circle has central symmetry such that the two halves of a circle fall on top of each other when the tray holding the pizza is rotated $180^{\circ}$ around the pizza's center. But then rectangle also has a center and any line through its center divides rectangle into equal parts - rectangle is also centrally symmetric.

### The real problem

Now we are in a position to tackle the problem from the Crux. Any line through the center of the rectangular pizza divides it into equal parts. One of these lines stands out. It's the one that divides into equal parts the circular hole, that's the line that passes through the hole's center. Thus to solve the problem we cut through the centers of the rectangle and that of the hole.

But that's not the only solution. In fact there are infinitely many more, although to find any of these in a constructive manner is rather difficult, if not impossible.

Draw a straight line in any direction outside the pizza. Now begin moving it towards the pizza perpendicular to its direction. The line will cross the pizza eventually. First the "front or remaining" portion of the pizza will be greater than the one already passed over. As the line proceeds with its movement, the former part decreases whereas the latter increases. With an appeal to continuity, we may conclude that at some point the two parts had the same worth. The key here is what is known as Bolzano's or the Intermediate Value Theorem. There are ways and ways to employ the theorem, e.g., by letting the line rotate around a fixed point.

The "Ham sandwich theorem" is the most famous statement of this kind, see, e.g., tippingpoint video on youTube.

25 May

### Pundits

Paul Meehl's (1954) book Clinical Versus Statistical Prediction: A Theoretical Analysis and a Review of the Evidence appeared 25 years ago. It reviewed studies indicating that the prediction of numerical criterion variables of psychological interest (e.g., faculty ratings of graduate students who had just obtained a Ph.D.) from numerical predictor variables (e.g., scores on the Graduate Record Examination, grade point averages, ratings of letters of recommendation) is better done by a proper linear model than by the clinical intuition of people presumably skilled in such prediction. The point of this article is to review evidence that even improper linear models may be superior to clinical predictions.

The most important development in the field since Meehl's original work is Robyn Dawes famous article.

Why are experts inferior to algorithms? One reason, which Meehl suspected, is that experts try to be clever, think outside the box, and consider complex combinations of features in making their predictions.

According to Meehl, there are few circumstances under which it is a good idea to substitute judgment for a formula. In a famous thought experiment, he described a formula that predicts whether a particular person will go to the movies tonight and noted that it is proper to disregard the formula if information is received that the individual broke a leg today. The name "broken-leg rule" has stuck. The point, of course, is that broken legs are very rare — as well as decisive.

### Reference

1. Robyn M. Dawes, The robust beauty of improper linear models in decision making, in Judgement under uncertainty: Heuristics and biases, Cambridge University Press; 1 edition (April 30, 1982)
2. Daniel Kahneman, Thinking, fast and slow, Farrar, Straus and Giroux (October 25, 2011)
28 May

### Biochemical Algorithms and School Algebra

One part of the book The History of Tomorrow by Yuval Noah Harari I happened to read under unusual circumstances. Harari's first book Sapiens: A Brief History of Humankind has been translated into English and is #1 best seller in anthropology on amazon.

One of the theses that Harari promotes in his second book is that a human is a collection of biochemical algorithms and, as such, can, in principle, be studied on a formal basis through data collection and relevant behavioral database mining. Part of this - he points out - is already taking place. On login, amazon.com baits you with a collection of items bought by others with your shopping habits and tastes. google offers a number of gadgets to monitor your vitals that, when it comes to making food selection, are prepared to give you advice or even make that choice for you. Such gadgets are continuously online (to be able to keep their anti-virus software up-to-date), collecting, sharing, and analyzing the data which is essentially you. The bottom line is that Harari foresees the situation in which a robot may know you better than you ever may hope to know yourself. And certainly it may perform good many tasks - mental tasks, in particular - much better than you and for you.

The author discusses societal repercussions: some occupations will disappear. E.g., already now much of the banking and brokerage goes online; computer programs trade autonomously on financial markets, drive cars and fly planes. Our dependence and, more importantly, reliance on the programs will progress, and - in time - the programs will proffer advice or even make decisions vicariously. Examples included seeking advice which movie to see, which occupation to pursue and even who to marry.

I read this and wondered when the current education system - with its uniform standards - would catch the wind and transform into something less rigid. There is simply no way it may remain useful, nay even functional, unless it starts following the trend.

I did not mean to either review or retail the story. I found the book fascinating and hope it will be translated into English in the near future. Now, I return to the circumstances of my reading this biochemical algorithms stuff.

Last week I had my appendix removed, with a cumbersome complication. Some of my organs did not wake up after the general anesthesia. (Everything is OK now.) It was an unpleasant experience. I decided to check the components of the mix used by the anesthesiologist. Two of the six were described as "paralytic." A quick search on google revealed that "paralytic" actually was a designation for "muscle relaxant." So I asked a doctor why my innards were squeezed rather than relaxed. The response was because we poked you during the surgery. Now talk of the biochemical algorithms.

A relaxed amoeba is sunning up on a stone when a passer-by pokes it with a sharp object. What does the amoeba do? No supercilious thinking takes place.The poking activates a suitable built-in biochemical algorithm and amoeba constricts into a small portion of its relaxed size. It will take her a few days to begin trusting the humans again. During these days the doctors were watching it closely and were instructing the nurses to remove the relief tubes one at a time. At one point, having removed a tube, an RN told me that I have eight hours to demonstrate the salubrious effect that the tube had on my organs. Said she, "It is 10 am now, you have eight hours to do it. It's 10, 11, ...," and she began counting on fingers, "11, 12, 1, 2, 3, 4, 5, 6 - you have until 6 pm to show that things had worked for you."

You know, the nurses there did not appear to lack in intelligence. They carry out intellectually demanding jobs, most of the time sitting in front of networked computers or handling mobile devices. I did not dare to ask her whether or not she took an algebra course either in high school or a college. It was obvious that, even if she did, the powerful algebraic tools that were (and are) being peddled to students for their own good by the education establishment had little effect on her thinking or the excellent manner in which she carried out her duties.

03 May

### No Need To Lose the Battle

In her recent post Tanya Khovanova bitterly complained of the difficulty the current spread and popularity of the Web and the social networks pose for successful teaching of students to think. A student nowadays can easily find online a solution to the puzzle he or she was supposed to rack the brain over and benefit from that exercise.

(In passing, I totally disagree with Tanya's thesis that "People who think make better decisions, whether they want to buy a house or vote for a president." That's factually wrong. This would be rather presumptuous to assume that the ones who disagree with one's choice of a president give their vote thoughtlessly. I am certain Tanya did not mean that.)

There are ways and ways to teach - I am reluctant to use the word "thinking" but rather - problem solving. Solving a problem starts first and foremost with posing a problem. Years ago I wrote about "mathematical droodles" - interactive activities that were supposed to lead the student to a formulation and a better grasp of a problem without explicitly articulating what it was about. In the same spirit, James Tanton just published two books, Without Words and More Without Words.

There is no much harm done by making solutions to puzzles available to students, though preferably not right away. There are even books of puzzles that come with solutions, e.g., Problem Solving Through Recreational Mathematics by Bonnie Averbach and Orin Chein.

It also useful to ask students to explain how they arrived at the solution - their train of thought that finally stopped when the solution took shape. A teacher may also demonstrate such process with all possible missteps and second thoughts. For example, Thimothy Gowers - A Fields Medalist - has recently demonstrated "live" his thought process by tackling one of IMO problems. Mike Lawler threw the gauntlet to other mathematicians to follow in Gowers' footsteps by offering to document their solving of a problem from the latest European Girls' Mathematical Olympiad. I even made an attempt to answer Mike's call.

And finally for this blog, it is worth remembering the last step of George Polya's paradigm for problem solving: Looking back. Is there another solution? Is it possible to modify the problem in a meaningful way?

Now I am going to quote Tanya's puzzle and solve it but only after offering a modified puzzle.

A sultan decides to give 100 of his sages a test. He has the sages stand in line, one behind the other, so that the last person in line can see everyone else. The sultan puts either a black or a white hat on each sage. The sages can only see the colors of the hats on all the people in front of them. Then, in any order they want, each sage guesses the color of the hat on his own head. Each hears all previously made guesses, but other than that, the sages cannot speak. Each person who guesses the color wrong will have his head chopped off. The ones who guess correctly go free. The rules of the test are given to them one day before the test, at which point they have a chance to agree on a strategy that will minimize the number of people who die during this test. What should that strategy be?

My modification is twofold. First, it's exactly the same problem as above, with the only difference in that the hats come in three (or more) distinct colors. I know one strategy - along the lines of the original puzzle - that reduces the number of possible victims to a reasonably small number. I am still pondering whether there is a better strategy.

Second, the puzzle can be modified in additional ways. For example, what if, instead of naming his hat's color, the sage was allowed to declare something more complex, like, say "red or blue"? Would that help to improve the strategy? To this I do not have a ready answer.

What if before doing his job the executioner was required to say loudly "Oops!" Could this help to save lives in the multicolor case?

Time for a solution to Tanya's puzzle:

Information helps a person make the correct determination of his hat's color. The more information one has, the better are his chances to stay alive. The only information sages have is what they see in front of them and the calls of the people behind them. If the order of calls is not random, it should follow a certain plan, probably sequential hat naming. The first one in the line has no information at all, the last one has all the information available. It is sensible to suggest that the process should start with the last fellow.

What information does he have? The number of hats of a certain color in front of him. The person in front of him has he same information minus one hat. If they agreed on a strategy of counting, say black hats, and the last one could say "27", the fellow in front of him would be immediately able to determine the color of his hat. If he sees 27 black hats in front of him, his hat is white. If he sees 26 black hats, his hat is black.

The "Aha moment" comes with the realization that there is no need for passing around the complete count. The decision, the sages make, are binary "either or". The last one may simply say "black" to indicate that the number of black hats in front of him is odd, and "white", if it's even. Assuming the last one calls "black", if the next to last sage sees an even number of black hats in front of him, he may conclude that his hat is black, otherwise it is white. The fellow in front of him gets this information, sees the number of black hats in front of him adds to that the binary 0 or 1 for each of the two calls from behind and determines the color of his hat: if the result is 1, his hat is black; if it's 0, the hat is white.

Full information is thus passing successively from the back to the front of the line. The only one who risks a beheading is the last one, and will have to be chosen by a draw.

26 Apr

### Not too easy - not too difficult

The other day, while driving my HS senior son to school (he could have taken a bus, but, for one, his time is at a premium; also, the drive gives us an opportunity for a small chat), we talked about how words with different basic meanings may mean the same thing in certain contexts. As an example I mentioned the expressions "not too late" and "not too early" both of which may mean "just in time." This conversation came to mind when I tried to characterize Problem #1 from the European Girls’ Math Olympiad which Mike Lawler posted on his recent blog:

Let $n$ be an odd positive integer, and let $x_1,x_2,\ldots,x_n$ be non-negative real numbers. Prove that

$\displaystyle\qquad\min_{1,2,\ldots,n}(x_{i}^{2}+x_{i+1}^{2})\le\max_{1,2,\ldots,n}2x_ix_{i+1},$

where $x_{n+1}=x_1.$

At first sight, the inequality appeared strange, if not erroneous because, for every two positive numbers $a$ and $b$, $a^2+b^2\ge 2ab$ since the latter reduces to $(a-b)^2\ge 0.$

Assuming, however, that the problem posed a meaningful question, it appears to be essential to take into account the fact that $n$ was an odd number. This leads to the assumption that $n\ge 3.$ The moment that the magnitude of $n$ came into the picture, the mathematical induction seemed like the one venue to follow to crack the problem.

To start with, suppose there are three non-negative real numbers $a,b,c$, listed in that order, and that $a^2+b^2\le b^2+c^2$ and $a^2+b^2\le c^2+a^2.$ These imply $a\le c$ and $b\le c.$ Without loss of generality, $a\le b\le c.$ Then, obviously, $bc$ is the maximum of the three products. We need to show that $a^2+b^2\le 2bc$. This follows from $a^2+b^2\le 2b^2\le 2bc$.

Now, the intuition for the inductive step: in passing from $n$ to $n+2$ the maximum on the right may only grow, whereas the minimum on the left may only decrease. This is naturally also true in passing from $n$ to $n+1$ such that the requirement that $n$ is odd may be a red herring, only stipulated as a disguised substitute for $n\ge 3.$

Well, trying to implement the induction I ran into one essential difficulty: increasing $n$ not only leads to additional terms among which to choose maximum or minimum, it also causes the terms $x_n^2+x_1^2$ and $2x_nx_1$ to be dropped from consideration. If neither $\displaystyle x_n^2+x_1^2=\min_{1,2,\ldots,n}(x_{i}^{2}+x_{i+1}^{2})$ nor $\displaystyle2x_nx_1=\max_{1,2,\ldots,n}2x_ix_{i+1}$ the underlying intuition worked just fine but bad things do happen and there was a need to account for the exceptional cases. The bright spot was that the two conditions can't complicate the proof simultaneously. If, for example, $\displaystyle x_n^2+x_1^2=\min_{1,2,\ldots,n}(x_{i}^{2}+x_{i+1}^{2})$ then, in particular $x_n^2+x_1^2\le x_1^2+x_2^2,$ implying $x_n\le x_2$ and, subsequently, $2x_nx_1\le 2x_1x_2$ so that the removal of $2x_nx_1$ could not affect the maximum on the right. Similarly, if $\displaystyle2x_nx_1=\max_{1,2,\ldots,n}2x_ix_{i+1}$ then $x_n\ge x_2$ from which $x_n^2+x_1^2\ge x_1^2+x_2^2$ and so the removal of $x_n^2+x_1^2$ would have no effect on $\displaystyle \min_{1,2,\ldots,n}(x_{i}^{2}+x_{i+1}^{2}).$

Beyond this, I had to consider several cases of where the dropped terms were smaller or greater of the added ones. I sought to simplify the situation and make it more formal. To this end, I replaced each number in the sequence $x_1,x_2,\ldots,x_n$ with $0$ or $1$. When $x_i\le x_{i+1}$ I replaced $x_i$ with $0$; otherwise it became $1$. (Of course $x_{n}$ was followed by $x_1$.) Now I had a sequence of $0$'s and $1$'s. From the base of my attempted induction it transpired that "good" sequences (i.e., the sequences that satisfied the required inequality) were reducing to sequences that contain at least two consecutive $0$s or two consecutive $1$s. The question became to insert one or two digits and to verify that, starting with a good sequence, the insertion would always lead to a "good" sequence.

This process made it obvious that my attempt at induction was misguided. Clearly, some insertions will work but others will not. For example, inserting $1$ between two zeros is liable to cause damage to the sequence unless the two zeros were a part of a triple $000$. If it was not, i.e., if we attempted to insert $1$ between the two zeros in $1001$, we'd get $10101$.

So what now? That stumbling step still had a positive effect: it led to a realization of what is a "bad" sequence. A bad sequence is the one that has no consecutive $1$s or $0$s. In other words, it's an intermittent sequence of $1$s and $0$s. I should have thought of this sooner without losing that much time. The idea of a bad sequence has highlighted another of my missteps. The fact is that for $n$ even there are bad sequences: $010101\ldots 1$ while, for $n$ odd, all sequences are good, for, in a bad sequence every $0$ may be paired with a preceding $1$ which makes its length even. Thus the induction from $n$ to $n+1$ did not have a chance to succeed. It also proved to be altogether unnecessary due to a simple statement:

For $n$ odd, all sequences of $1$s and $0$s are good.

The required inequality holds for any sequence that reduces to a good one, i.e., for any sequence of odd length. Woof.

13 Apr

### 7 or 22

"Have you guessed the riddle yet?" the Hatter said, turning to Alice again.

"No, I give it up," Alice replied. "What's the answer?"

"I haven't the slightest idea," said the Hatter.

L. Carroll, Alice's Adventures in Wonderland

I have an instinctive dislike for the kind of questions that appear regularly on various forums and social networks. Here's an example that was posted at the CutTheKnotMath facebook page:

Without ever trying to answer such questions, I was always confident that the poster (if not the author) were smugly awaiting a definite reply, although, even with the most benevolent interpretation, the problem has to be considered ill-posed, like that of asserting the next term in a given sequence.

The question being posted at the CutTheKnotMath facebook page, I gave it some thought. The anticipated answer was likely to be

Since $(3+4=\;) 19 = 3 + 4 + 3\cdot 4$ and $(5+6=\;) 41 = 5 + 6 + 5\cdot 6,$ then $(1+3=) 1+3+1\cdot 3=7.$

Obviously, the author expected an algorithmic procedure (a formula most likely) that when applied to $3$ and $4$ led to $19$ but when applied to $5$ and $6$ produced $41.$ That algorithm had then to be applied to the pair $1,3.$

Bui Quang Tuan offered a different interpretation and wondered which is more natural:

$1 + 3 = (5 + 6)-(3 + 4) = 41-19 = 22$

That was an interesting approach to a question, most certainly overlooked by the author. (While the latter used the symbol of addition just to suggest a presence of an algorithm, Bui Quang Tuan cleverly accepted the two given identities as such and applied to them the regular operations of addition and subtraction. But even choosing a more orthodox interpretation, an algorithm that would bewilder the problem's author is not difficult to find.

Let's agree to denote the algorithm as a function of two variables. E.g., the above cold be described as $f(x,y)=x+y+x\cdot y$ with the common meaning of arithmetic operations. But here's another possibility: $g(x,y)=x\cdot y+y+y-x/x,$ such that $g(3,4)=19$ and $g(5,6)=41.$ With this interpretation $1+3=g(1,3)=1\cdot 3+3+3-1/1=8.$

Vu Xuan Hanh posted another example: $h(x,y)=x+y\cdot y.$ Indeed, $h(3,4)=3+4\cdot 4=19$ and $h(5,6)=5+6\cdot 6=41,$ implying that $1+3=h(1,3)=1+3\cdot 3=10.$

The latter example has caused a shift in my view of the problem. Perhaps, it is less like finding the next term of a sequence than expressing various integers with a fixed set of numbers using various arithmetic operations. For example, if the task is to place the symbols of arithmetic operations of parentheses between the digits $624663$ so as to get, say, $20$ as the result, we get the possibilities: $6\cdot 2+4+6+6/3=20,$ $(6-2)\cdot 4+6/6+3=20,$ $-(6-2)/(4-6)+6\cdot 3=20,$ $6-2+4+6\cdot 6/3.$ There bound to be more.

Why do I prefer the latter kind of problems to the one I considered at the beginning? The difference between them is in the veil of mystery with which one is presented. There is always an implicit stipulation that somehow the problem has a unique solution. It often comes with an enticing warning "99% of the population can't do that right. Can you?" But, obviously, it would be a rare circumstance where the problem of this kind has a unique solution. With this realization, the problem becomes less obnoxious and can be looked at from other angles. For example,

1. Find several formulas $f(x,y)$ for which $f(3,4)=19$ and $f(5,6)=41$. See what is $f(1,3).$
2. For a given formula $f(x,y)$ such that $f(3,4)=19$ and $f(5,6)=41,$ find which other integers can be represented as $f(x,y).$
3. For which integers $n$ there are $a,\;$ $b,\;$ and $f(x,y),$ such that $f(3,4)=19,$ $f(5,6)=41,$ and $f(a,b)=n.$
4. Are there integers $n$ provably not expressible as $f(x,y),$ where $f(3,4)=19,$ $f(5,6)=41,$ perhaps for a given $f.$
04 Apr

### Weekly report, the week of March 28, 2016

The week started with Gregoire Nicollier's posting where he applied his beautiful theory of spectral decomposition of polygons to quadrilaterals. I added a GeoGebra illustration to make the theory more accessible, not that it needed that.

Gregoire showed how his theory supplied one-line proofs to the problems considered in the previous week: squares on the sides of a parallelogram and squares on the sides of an arbitrary quadrilateral. The whole episode serves a signal example of the unifying, illuminating power of general theory.

Next came a posting about emergence of a parallelogram in a trapezoid, with one sides passing through the apex of an isosceles triangle with the base the opposite side of the trapezoid. The problem, especially when furnished with an illustration, appeared quite intuitively simple. However, several remarks made online by the visitors showed that the intuition alone may be quite misleading.

One solution employed analytic geometry, another disguised indirect reasoning based on the uniqueness of the solution to the famous Heron's problem. But the third solution stemmed from the observation (by Gregoire Nicollier) that the whole configuration is just a special case of Pappus' theorem.

On April 1st I indulged myself in a doodling with a problem that appeared to beg for a generalization and seemed to hold a promise of a non-trivial result. I am afraid that the promise remained illusionary. But the weekend brought a very satisfying compensation.

Leo Giugiuc and Dan Sitaru shared on the CutTheKnotMath facebok page an elementary problem from their article at the Gazeta Matematica.

The purpose of the article was to introduce an application of Linear Algebra to proving various inequalities. The one they posted at the facebook was the simplest example of such an application. The posting engendered a stream of responses with elementary solutions. As of this writing, ten different proofs have been added to their original one. The present collection supplies a perfect illustration to the fact that even the most simplest of the problems may be looked at from various angles. I wish that the authors of school textbooks that often include only answers or "solutions to the odd-numbered problems" paid more attention to the possibility of alternative view points.

01 Apr

### Doodling on April 1st

I came across the following problem several days ago but hesitated to write about it until April 1st. It is simple, practically trivial, and still, after doodling with it for some time, I was left with an open question. If it appears too trivial, even unworthy of mention, do please make an allowance for the date, selected in the hope of forgiveness.

Here's the problem:

Given that $(1+\sin\alpha)(1+\sin\beta)(1+\sin\gamma)=\cos\alpha\cos\beta\cos\gamma.$ Find an alternative expression for

$\qquad(1-\sin\alpha)(1-\sin\beta)(1-\sin\gamma).$

Do give it a thought before reading further.

To be on safe side, assume that none of the factors $(1-\sin\alpha),$ $(1-\sin\beta),$ $(1-\sin\gamma)$ vanishes, for, otherwise, $0$ would be an alternative form. Under this assumption, we may multiply $(1+\sin\alpha)(1+\sin\beta)(1+\sin\gamma)=\cos\alpha\cos\beta\cos\gamma$ by $(1-\sin\alpha)(1-\sin\beta)(1-\sin\gamma):$

$\qquad(1-\sin^2\alpha)(1-\sin^2\beta)(1-\sin^2\gamma)=\cos\alpha\cos\beta\cos\gamma(1-\sin\alpha)(1-\sin\beta)(1-\sin\gamma)$

or $(\cos^2\alpha)(\cos^2\beta)(\cos^2\gamma)=\cos\alpha\cos\beta\cos\gamma(1-\sin\alpha)(1-\sin\beta)(1-\sin\gamma)$ and, assuming $\cos\alpha\cos\beta\cos\gamma\ne 0,$ we obtain

$\qquad(1-\sin\alpha)(1-\sin\beta)(1-\sin\gamma)=\cos\alpha\cos\beta\cos\gamma.$

So, under the assumption $\cos\alpha\cos\beta\cos\gamma\ne 0,$ that product is equal to both $(1+\sin\alpha)(1+\sin\beta)(1+\sin\gamma)$ and $(1-\sin\alpha)(1-\sin\beta)(1-\sin\gamma).$ Clearly the angles must be special.

First, the problem is generalized in an obvious way: Assuming $\displaystyle\prod_{i=1}^{n}\cos\alpha_i\ne 0,$ that product is equal to both $\displaystyle\prod_{i=1}^{n}(1-\sin\alpha_i)$ and $\displaystyle\prod_{i=1}^{n}(1+\sin\alpha_i).$

That looks quite right, but for what $n$? For $n=1$ there is no product but the question still makes sense: if $1+\sin\alpha=\cos\alpha$ and $\cos\alpha\ne 0$, what is $1-\sin\alpha$? As before, $1+\sin\alpha=\cos\alpha=1-\sin\alpha$. The two equations give $\sin\alpha=0$ and $\cos\alpha=1.$

Next, consider the case $n=2$: $\cos\alpha\cos\beta\ne 0$ and $(1+\sin\alpha)(1+\sin\beta)=\cos\alpha\cos\beta.$ In this case, $(1-\sin\alpha)(1-\sin\beta)=\cos\alpha\cos\beta.$ The difference of the two equations $2(\sin\alpha+\sin\beta)=0$ and we get $\displaystyle\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}=0$, meaning $\alpha+\beta=2\pi k,$ $k$ an integer, or $\alpha-\beta=\pi (2k+1).$ Thus we see that, for example, $\alpha=-\beta$ is a suitable pair of angles. The given equation becomes $1-\sin^2\alpha=\cos^2\alpha$ which is true.

For $n=3$, we obtain the equation

$\qquad\sin\alpha+\sin\beta+\sin\gamma+\sin\alpha\sin\beta\sin\gamma=0.$

With the choce of, say, $\alpha=0,$ we reduce the problem to the case of $n=2.$ Are there other solutions that bind all three variables? There bound to be. For example, suppose $\alpha+\beta+\gamma=\pi.$ Then $\sin 2\alpha+\sin 2\beta+\sin 2\gamma=4\sin\alpha\sin\beta\sin\gamma,$ meaning that, for $\alpha+\beta+\gamma=2\pi,$ $\displaystyle\sin \alpha+\sin \beta+\sin\gamma=4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2},$ which converts our equation to

$\qquad 4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}+\sin\alpha\sin\beta\sin\gamma=0$

or, via the double argument formulas, and assuming that none of the sines vanishes,

$\qquad\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2}=-\frac{1}{2}.$

A solution is plausible; I'll have to look into that later on.

Now, for the general case, $\displaystyle\prod_{i=1}^{n}(1+\sin\alpha_i)=\prod_{i=1}^{n}\cos\alpha_i$, with the latter not being $0$.

Immediately, there are special values of $\alpha\text{'s}$, for which our derivation will work. For example we may split all factors into single angle expressions or pair up some angles and solve equations with one or two variables. Let's call such sets of angles reducible. In general, for a given $n$, the set $\{\alpha_i\},\,i=1,\ldots,n$ is said to be reducible if it may be split into two non-empty sets, say, $I_1$ and $I_2$ such that $I_1\cup I_2=\{1,\ldots,n\}$ and $\displaystyle\prod_{i\in I_1}(1+\sin\alpha_i)=\prod_{i\in I_1}\cos\alpha_i=\prod_{i\in I_1}(1-\sin\alpha_i)$ and $\displaystyle\prod_{i\in I_2}(1+\sin\alpha_i)=\prod_{i\in I_2}\cos\alpha_i=\prod_{i\in I_2}(1-\sin\alpha_i)$.

The original problem did not require solving equations, however, the question for which angles the stipulation of the problem made sense, was reasonable and natural. So lets call such sets of angles "solutions." In this terminology, we just defined "reducible solutions." Thus the questions pops up, "Are there irreducible solutions?"

For $n=2$, $\alpha=\beta=0$ should be considered "reducible" but, say, $\displaystyle\alpha=-\beta=\frac{\pi}{4}$ is "irreducible". Thus the question of the existence of the irreducible solutions for $n\gt 2$ appears quite legitimate: "Are there irreducible solutions?" If that is too difficult, we may try answering an easier question: "Assuming all solutions are reducible, in how many ways the problem may be reduced?"