TNDOT Friday 10: Algebra

October 11, 2024

9:27 am

I’m not sure last week’s TNDOT was given the attention it deserved, but no matter. I’ll update it shortly.

This week’s TNDOT is comprised of the first four questions from the 1911 pass Algebra Matric exam. The exam in total consisted of eleven questions, with an “either α or β” selection in two of the questions. (The exam was presumably to be done in three hours, but that’s only a guess.)

I’ve already posted two of the 1911 questions and their solutions, here, and I’m reproducing Q3 below, just for tidiness. I’ll post all eleven questions over the next three weeks.

Have fun.

1. Simplify

$\color{blue}\boldsymbol{\dfrac{(a-x)^2}{a(a-b)(a-c)}+\dfrac{(b-x)^2}{b(b-c)(b-a)}+\dfrac{(c-x)^2}{c(c-a)(c-b)}.}$

2. When is a series of quantities said to be in Harmonical Progression?

What are the first two terms of an H.P. whose third and forth terms are $\color{blue}\boldsymbol{\frac12}$ and $\color{blue}\boldsymbol{\frac74}$ ?

Find the relation between $\color{blue}\boldsymbol{a}$ , $\color{blue}\boldsymbol{b}$ , and $\color{blue}\boldsymbol{c}$ , that they may be respectively be the $\color{blue}\boldsymbol{p^{ \rm th}}$ , $\color{blue}\boldsymbol{q^{ \rm th}}$ and $\color{blue}\boldsymbol{r^{ \rm th}}$ terms of an H.P.

3(α) A certain quantity being divided into n equal parts, and also into n + 1 equal parts, the product of the n parts is n times that of the n + 1 parts. Find it.

(β) A farmer buys m sheep for £p and sells n of them at a profit of q per cent. At what price per head must he sell the rest to clear r per cent on his whole bargain?

4. Solve the equations

$\color{blue}\boldsymbol{\aligned &2(x+y+z) = a + b + c;\\[2\jot]&(c-a)x+(a-b)y+(b-c)z\\[2\jot]=\,&(b-a)x+(c-b)y+(a-c)z\\[2\jot]=\,&a^2 + b^2 + c^2 -bc - ca - ab. \endaligned}$

UPDATE (17/10/24)

Thanks to everyone for their contributions. We’ll take them one by one, including the already posted solution to 3, just for tidiness.

Q1. Red Five did the brute force, and Barry Hughes presented a very clever approach. Here, I’ll present a solution along the lines of Red Five, but probably a little more like what was expected in 1911. Note that the polynomial is symmetric in a, b and c, and students of the time would have been well practised on these. See the discussion on the (significantly harder) TNDOT 3.

The obvious first step is to obtain a common denominator,

$\color{Plum}\boldsymbol{D = abc(a-b)(a-c)(b-c)\,,}$

giving rise to the numerator (and being careful with the minus sings),

$\color{Plum}\boldsymbol{N = bc(b-c)(a-x)^2 + ac(c-a)(b-x)^2 + ab(a-b)(x-b)^2\,.}$

We can expand this out, but it is mostly easy to eyeball, coefficient by coefficient. To begin, the constant term is

$\color{Plum}\boldsymbol{a^2bc(b-c) + b^2ac(c-a) + c^2ab(a-b)= abc[a(b-c) + b(c-a) + c(a-b)]=0\,.}$

Similarly, the coefficient of the linear term is

$\color{Plum}\boldsymbol{ -2abc(b-c) -2bac(c-a) -2cab(a-b)= -2abc[(b-c) + (c-a) + (a-b)]=0\,.}$

Finally, the coefficient of the quadratic term is.

$\color{Plum}\boldsymbol{ bc(b-c) + ac(c-a) + ab(a-b)\,.}$

This takes a little thought, but not much (and it is simply listed as an “important identity” in Barnard and Child, a textbook of the time). The key is that setting a = b reduces the expression to 0, and similarly for b = c and c = a. It follows that a – b etc. must all be factors, and that only leaves the question of a minus sign.

In sum, we obtain

$\color{Plum}\boldsymbol{N = (a-b)(a-c)(b-c)x^2\,,}$

and finally,

$\color{Plum}\boldsymbol{\dfrac{N}{D} = \dfrac{(a-b)(a-c)(b-c)x^2}{abc(a-b)(a-c)(b-c)} =\dfrac{x^2}{abc}\,.}$

UPDATE (17/10/24)

Q2. Red Five has laid out the first part very nicely, and we’ll follow his solution. First of all, for those new to the expression, an harmonic(al) progression is a sequence whose reciprocals form an arithmetic sequence/progression. So, if the arithmetic sequence begins with $\boldsymbol{f}$ and has common difference $\boldsymbol{d}$ , then the corresponding harmonic progression begins,

$\color{Plum}\boldsymbol{\dfrac1{f},\ \dfrac1{a+f},\ \dfrac1{a+2f},\ \dfrac1{a+3f},\ \cdots \,.}$

So, the question tells us that

$\color{Plum}\boldsymbol{{f+2d=\dfrac{2}{1}},\quad {f+3d} = \dfrac{4}{7} \,,}$

and our problem amounts to determining $\boldsymbol{f}$ and $\boldsymbol{d}$ . That’s easy enough to do, but Red Five does it nicer, by writing the third and fourth terns with a “common numerator”, making out harmonic sequence.

$\color{Plum}\boldsymbol{?,\ ?,\ \dfrac7{14},\ \dfrac7{4}\,.}$

But then the first two terms come from simply adjusting the denominator by the common difference. (It’s easy to check that the common 7 in the numerator doesn’t interfere with this pattern.) So, the second and first terms are, in turn, $\boldsymbol{\frac{7}{24}}$ and $\boldsymbol{\frac{7}{34}}$ .

It is not exactly clear what is intended for the second part. We are given that

$\color{Plum}\boldsymbol{a=\dfrac{1}{f+(p-1)d},\quad {b} =\dfrac{1}{f+(q-1)d}, \quad {c} =\dfrac{1}{f+(r-1)d}\,.}$

That can easily be reciprocated to give,

$\color{Plum}\boldsymbol{\dfrac1{b}-\dfrac1{a} =(q-p)d,\quad \dfrac1{c}-\dfrac1{b} =(r-q)d\,.}$

We can get common denominators. Then, dividing the first equation by the second eliminates $\boldsymbol{d}$ , giving,

$\color{Plum}\boldsymbol{\dfrac{c(a-b)}{a(b-c)}=\dfrac{q-p}{r-q}\,.}$

I can’t see anything more interesting than that.

UPDATE (17/10/24)

Q3. (This is the solution posted as part of TNDOT 5, with some minor edits).

(α) If we let the quantity in question be Q, then we are told,

$\color{Plum}\boldsymbol{\left(\frac{Q}{n}\right)^n = n\left(\frac{Q}{n+1}\right)^{n+1}}\,.$

Dividing both sides by n, and then moving the Q-stuff to the right and the n-stuff to the left, we find,

$\color{Plum}\boldsymbol{Q = \frac{\left({n+1}\right)^{n}}{n^n}}\,.$

Or, we can write, $\boldsymbol{Q = \left({1 +\frac1{n}}\right)^{n}}$ , with obvious import.

(β) The main thing is to simply be methodical. To begin, the farmer sold m sheep at a profit of q%. Since he bought these sheep for £p each, he sold them for a total cost of

$\color{Plum}\boldsymbol{np(1+\frac{q}{100})}\,.$

Similarly, suppose he sold the remaining m – n sheep at a profit of s%. Then these would be sold for a total cost of

$\color{Plum}\boldsymbol{(m-n)p(1+\frac{s}{100})}\,.$

Altogether, this is supposed to yield a profit of r% on the sale of the m sheep. Which leads to the equation,

$\color{Plum}\boldsymbol{np(1+\frac{q}{100}) + (m-n)p(1+\frac{s}{100}) = mp(1+\frac{r}{100})}\,.$

Multiplying though by 100 and dividing by p,

$\color{Plum}\boldsymbol{n(100+q) + (m-n)(100+{s}) = m(100+ r)}\,.$

Rearranging, this gives,

$\color{Plum}\boldsymbol{ 100+{s} = \frac{m(100+ r)- n(100+q)}{m-n}= \frac{100(m-n) + mr - nq}{m-n}}\,.$

Then, finally, the price per head of the second batch of sheep is,

$\color{Plum}\boldsymbol{ p(1+\frac{s}{100}) = p\frac{100(m-n) + mr - nq}{100(m-n)}= p\left[1 + \frac{mr-nq}{100(m-n)}\right]}\,.$

UPDATE (24/10/24)

Q4. OK, belatedly, here is a solution for Q4. I’m honestly not sure what the examiners intended, but I think it probably had these kind of ingredients, if not this delivery. Thanks to everyone for your thoughts, which definitely guided my own.

For simplicity let’s write,

$\color{Plum}\boldsymbol{S= a^2 + b^2 + c^2 - bc - ca - ab\,.}$

Then our equations are

$\color{Plum}\boldsymbol{\left\{\aligned &2(x + y + z) = a+b+c\,, \\[2\jot]&(c-a)x+(a-b)y+(b-c)z=S\,, \\[2\jot]&(b-a)x+(c-b)y+(a-c)z=S\,.\endaligned\right.}$

What we do first is make a change of variables:

$\color{Plum}\boldsymbol{x = X -a, \qquad y = Y -b, \qquad z = Z -c.}$

Why do that? Because it makes S disappear from our equations:

$\color{Plum}\boldsymbol{\left\{\aligned &2(X + Y + Z) = 3(a+b+c)\,, \\[2\jot]&(c-a)X+(a-b)Y+(b-c)Z=0\,, \\[2\jot]&(b-a)X+(c-b)Y+(a-c)Z=0\,.\endaligned\right.}$

(How do you think of doing that? Luck, perhaps, but also by asking the question: what is S doing there in the first place? Having asked that, it felt as if S maybe arose from a transformation, and so maybe a transformation would get rid of it. Then, with that idea, it wasn’t hard to see that x = X – a etc. spat out at least a lot of the right terms.)

This is now smelling very symmetric; note that the sum of the coefficients is 0 in the second and in the third equation. An easy way to prove this symmetry is to set

$\color{Plum}\boldsymbol{Y = X + s, \qquad Z = X + t.}$

Then the second and third equations become,

$\color{Plum}\boldsymbol{\left\{\aligned &(a-b)s+(b-c)t=0\,, \\[2\jot]&(c-b)s+(a-c)t=0\,.\endaligned\right.}$

Ignoring for now the possibility of dependence, the only solution is s = t = 0, implying X = Y = Z. Then the first equation gives us

$\color{Plum}\boldsymbol{X = Y = Z = \dfrac{a+b+c}2\,.}$

Substituting back, we have

$\color{Plum}\boldsymbol{x = \dfrac{b+c-a}2, \qquad y = \dfrac{a+c-b}2, \qquad z = \dfrac{a+b-c}2\,.}$

Finally, the dependence thing. It feels as if maybe students would not have been expected to consider this. But it’s not that hard to do. Checking the determinant on the 2 x 2 system (or on the original 3 x 3 system), it turns out that the equations are dependent exactly if S = 0. But then when can that happen? Well, by the Cauchy-Schwarz inequality,

$\color{Plum}\boldsymbol{|ab|\leqslant \dfrac{a^2}2 + \dfrac{b^2}2\,,}$

and similarly for the other terms. Also, we only obtain equality if a = b, and so on. It follows that the only way S can equal 0 is if a = b = c, in which case the second and third equations disappear.

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33 Replies to “TNDOT Friday 10: Algebra”

Red Five says:

October 11, 2024 at 2:01 pm

Question 1 is quite interesting. The cyclic nature of the expression is one that I have seen a lot in the algebra papers of the time and so I’m guessing students were well versed in how to factorise/solve them.

Question 4 seems to want a similar strategy.

Reply
Red Five says:

October 11, 2024 at 3:03 pm

I just found a nice trick to Question 2 – how soon is too soon to drop a hint?

Reply
Red Five says:

October 11, 2024 at 7:43 pm

Question 2 is an interesting one and I’m not entirely sure what type of answer is being sought in the third paragraph of the question.

The middle part is a nice play with numbers game (there is a shortcut I found by looking carefully at the numbers, but it doesn’t save that much time) and the final part seems to be an exercise in algebra.

Reply
Banacek Spaces says:

October 12, 2024 at 11:49 am

Would someone be able to compare the difficulty of harder matriculation questions with
* intermediate olympiad from amt
** (real) olympiad
There was this joke-add in the newspaper years ago, “will swap post-war pdh for pre-war matriculation certificate”, (not necessarily in maths, it held on much longer than other sciences, esp social/humanities)
I also wonder what percentage of people of attempted / passed matriculation exams 100 or so years ago .

Reply
1. marty says:
  
  October 12, 2024 at 11:53 am
  
  I assume “pdh” is “phd”?
  
  Re (*), I think the AIMO is an interesting comparison. Pretty similar in difficulty (not style) it seems to me, maybe the old matric a little harder. I’m sure AMO and IMO are much harder.
  
  I’m not sure of the precise percentages a century ago, but I’m sure the approximate answers are “low” and “low”.
  
  Reply
  1. Red Five says:
    
    October 12, 2024 at 12:03 pm
    
    I suspect “pass” had a different meaning when there were not many universities and there was no VTAC or similar body students had to apply through.
    
    There are a few similarities between these old papers and the current AMT papers:
    
    1. The questions get to the point efficiently and effectively.
    2. The questions are about Mathematics.
    3. You often need to think a bit first or play around with the question before finding a path through it.
    
    I do not know for sure but could also proffer the following as possible similarities:
    
    4. The questions are not always in order of difficulty.
    5. Very few candidates will attempt all the questions on the paper.
    
    Your comment about humanities though is interesting. Look at the History or Science (Botany, Chemistry, Biology) exam papers from the same time as these TNDOTs and you may form the opinion I have that Mathematics has gotten easier while other subjects have gotten more difficult. The reasons are a complete mystery to me.
    
    Reply
CyberChalky says:

October 13, 2024 at 11:20 am

As one of the “New Dogs” (despite being definitely not “new” in years), I really appreciate these posts.
I am playing with 3-alpha, and other than the trivial solution of 4, with n=1 (which violates the “part*s*” statement, I’m only finding solutions in R, not Z. Is this intended to be the case, or is my logic faulty?

Reply
1. marty says:
  
  October 13, 2024 at 11:25 am
  
  Thanks, CC. The number of parts is an integer but the “quantity” need not be. Of course you can peek at the solution if you wish. Or, you can request a hint from the other new (and not so new) dogs, or write some details of your working and have others give you some guidance.
  
  Reply
Red Five says:

October 13, 2024 at 12:13 pm

Looking at Q4 – there is the “brute force” solution which works (eventually). I’m always wondering with these symmetric or symmetric-looking questions though that there is an elegant solution hiding somewhere.

I’ll keep trying for the rest of the day, but if anyone has found an elegant solution (that doesn’t require Cramer’s rule or some other brute forcing of a 3×3 system of equations) a hint would be greatly appreciated…

Reply
Red Five says:

October 13, 2024 at 8:13 pm

No one seems to be offering any answers, so I will offer my solution to Q1 for dissection.

If we make a common denominator of $abc(a-b)(a-c)(b-c)$ then the numerators combine to form the expression $(b^2c-bc^2)(a^2-2ax+x^2)-(a^2c-ac^2)(b^2-2bx+x^2)+(a^2b-ab^2)(c^2-2cx+x^2)$ .

After expanding, a bunch of terms cancel to leave $x^2(a^2b-ab^2+ac^2-a^2c+b^2c-bc^2)$

This then factorises to $x^2(a-b)(a-c)(b-c)$

So the entire expression is $\frac{x^2(a-b)(a-c)(b-c)}{abc(a-b)(a-c)(b-c)}$ which simplifies to $\frac{x^2}{abc}$

There are probably some algebraic short-cuts I’ve missed and the final factorisation I don’t think would be obvious to many students in the present, but maybe such things were taught a lot more broadly in 1911.

Reply
1. Barry Hughes says:
  
  October 14, 2024 at 12:30 am
  
  Here is my take on Question 1.
  Let
  
  $f(x) = \frac{(a-x)^2}{a(a-b)(a-c)} + \frac{(b-x)^2}{b(b-c)(b-a)} + \frac{(c-x)^2}{c(c-a)(c-b)}.$
  
  Defining $f$ this way imposes the restrictions on $a$ , $b$ and $c$ that they are distinct and nonzero. Then
  
  $\begin{eqnarray*} f(a) &=& \frac{(b-a)^2}{b(b-c)(b-a)} + \frac{(c-a)^2}{c(c-a)(c-b)} = \frac{(b-a)}{b(b-c)} + \frac{(c-a)}{c(c-b)} \\ &=& \frac{c(b-a) -b(c-a)}{bc(b-c)} = \frac{a}{bc}. \end{eqnarray*}$
  
  Since the definition of $f$ is invariant under permutations of $a$ , $b$ and $c$ , we have
  
  $f(b) = \frac{b}{ca}\quad\textrm{and}\quad f(c) = \frac{c}{ab}.$
  
  Since $f$ is a quadratic expression in $x$ we can write
  
  $f(x) = \alpha x^2 + \beta x + \gamma.\qquad (*)$
  
  Define
  
  $g(x) = \frac{x^2}{abc}$
  
  The (possibly degenerate) quadratic equation
  
  $f(x) - g(x) = 0$
  
  for the abscissae of points of intersection of the curves $y=f(x)$ and $y=g(x)$ must be precisely one of the following:
  [1] a quadratic equation, with at most two real solutions;
  [2] a linear equation, with one real solution;
  [3] an equation of the form $\mathrm{constant}=0$ , where the constant is nonzero, which has no solutions for $x$ ;
  [4] the tautology $0=0$ for all $x$ , that is, $f(x) \equiv g(x)$ .
  
  The curves $y=f(x)$ and $y=g(x)$ have three distinct points in common, namely $(a,a/(bc))$ , $(b,b/(ca))$ and $(c,c/(ab))$ .
  Hence it is case 4 that applies and $f(x) \equiv g(x)$ .
  
  A more mundane approach is to proceed as follows immediately after equation ( $*$ ). We have three simultaneous equations
  
  $\begin{eqnarray*} \alpha a^2 + \beta a + \gamma &=& \frac{a}{bc},\\ \alpha b^2 + \beta b + \gamma &=& \frac{b}{ca},\\ \alpha c^2 + \beta c + \gamma &=& \frac{c}{ab}. \end{eqnarray*}$
  
  Subtracting the first pair of our simultaneous equations gives
  
  $\alpha (a^2-b^2) + \beta (a - b) = \frac{a^2 - b^2}{abc}$
  
  and dividing by the nonzero factor $(a-b)$ we have
  
  $\alpha (a+b) + \beta = \frac{a+b}{abc}.$
  
  Similarly, subtracting the last pair of our simultaneous equations gives
  
  $\alpha (b+c) + \beta = \frac{b+c}{abc}.$
  
  Thus we have a pair of simultaneous equations for $\alpha$ and $\beta$ .
  Subtracting the last two equations and dividing by the nonzero factor $(a-c)$ we find that
  
  $\alpha = \frac{1}{abc},$
  
  from which we can deduce at once that $\beta = 0$ . Using our calculated values of $\alpha$ and $\beta$ in any one of the three original simultaneous equations we find that $\gamma = 0$ .
  
  Reply
  1. marty says:
    
    October 14, 2024 at 12:35 am
    
    Thanks, Barry. Note your \begin{enumerate} didn’t work. I’m not exactly sure what does or does not work in comments, but I know \aligned works.
    
    Reply
  2. Red Five says:
    
    October 14, 2024 at 8:16 am
    
    I was following your argument right up to the point where you decided to define a new function $g(x)=\frac{x^2}{abc}$ . I see WHY it works and HOW it works, but I’m not sure where the idea to define this function comes from.
    
    It is just experience with problems of this type or is there something else happening that I’ve missed (that happens a lot…)
    
    Reply
    1. marty says:
      
      October 14, 2024 at 8:41 am
      
      Hi, RF. I’ll note, with all due respect, that Barry is an old dog. I’ll let Barry and others reply before wading in myself.
      
      Reply
Red Five says:

October 14, 2024 at 2:09 pm

Q2. Write $\frac{1}{2}=\frac{7}{14}$ and the sequence is then $t_1,t_2,\frac{7}{14},\frac{7}{4}$ .

Since this a Harmonic progression, it is of the form $\frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d},\frac{1}{a+3d}$ . We can then work backwards to get $t_2=\frac{7}{24}$ and $a=t_1=\frac{7}{34}$

From here it is easy enough to write $\frac{1}{d}=-\frac{7}{10}$ and therefore $d=-\frac{10}{7}$ .

For the last part I think they were after something akin to:

If $b$ is the harmonic mean of $a$ and $c$ then $q$ will be the arithmetic mean of $p$ and $r$ .

Reply
1. marty says:
  
  October 14, 2024 at 2:11 pm
  
  Thanks, RF. I might wait a day or two before summing up people’s ideas and updating the post.
  
  Reply
  1. Red Five says:
    
    October 14, 2024 at 4:01 pm
    
    All good. Q3 has been answered before and I doubt my solution to Q4 is much use to anyone. Still thinking of a way to streamline it (based on Barry’s method) but no progress yet.
    
    Reply
marty says:

October 17, 2024 at 6:01 pm

I’ve updated the post with solutions to Q1 and Q2 (and Q3).

I haven’t yet thought about how to do Q4 nicely and/or as might have been intended. I’ll leave it for now, and people can of course still make suggestions.

Reply
Red Five says:

October 18, 2024 at 5:00 pm

OK. I’ve had a further thought about Q4 (but it may not go anywhere).

What about, as an intermediate step in solving for $x,y,z$ we try solving for $a,b,c$ in terms of $x,y,z$ .

Two (moderately) obvious solutions come to mind:

$a=y+z,b=x+z,c=x+y$ and (a somewhat trivial solution) $a=b=c=\frac{2}{3}(x+y+z)$ .

Whether either of these are useful, I do not yet know, but maybe someone else can find a way forward from here…?

Reply
random_commenter says:

October 18, 2024 at 9:26 pm

Re:Q4

I eventually solved this one (and the working seemed to come out nicely) after making the assumption that x, y and z must each be some linear combination of a, b and c. Though I’ve had trouble with a mathematical justification for this (if there even is one). It just seemed right and happened to work out.

Reply
1. marty says:
  
  October 18, 2024 at 9:35 pm
  
  Thanks r_c. I’ll look tomorrow and/or feel free to include some details.
  
  Reply
2. Red Five says:
  
  October 18, 2024 at 9:39 pm
  
  Yes… I started with ideas such as $a=x+y,b=x+z,c=y+z$ to see where that leads and got somewhere but haven’t been able to finish it off nicely yet.
  
  The final part can be re-written as $(b-a)(c-a)+(a-b)(c-b)+(a-c)(b-c)$ but again, I’m not sure where this leads just yet (if anywhere).
  
  Reply
3. random_commenter says:
  
  October 19, 2024 at 1:13 am
  
  Let
  
  $\begin{align*} x=\alpha_{1}a+\beta_{1}b+\gamma_{1}c\\ y=\alpha_{2}a+\beta_{2}b+\gamma_{2}c\\ z=\alpha_{3}a+\beta_{3}b+\gamma_{3}c\\ \end{align*}$
  
  Then from 2 equation we get
  
  $\begin{align*} 1+\alpha_{1}=\alpha_{2}=\alpha_{3}\\ \beta_{1}=1+\beta_{2}=\beta_{3}\\ \gamma_{1}=\gamma_{2}=1+\gamma_{3}\\ \end{align*}$
  
  So
  
  $\begin{align*} x=\alpha a+(1+\beta)b+(1+\gamma)c\\ y=(1+\alpha)a+\beta b+(1+\gamma)c\\ z=(1+\alpha)a+(1+\beta)b+\gamma c\\ \end{align*}$
  
  Then from equation 1
  
  $(1+2\alpha)a+(1+2\beta)b+(1+2\gamma)c=0$
  
  $\alpha=\beta=\gamma=-\frac{1}{2}$
  
  Therefore
  
  $\begin{align*} x=\frac{-a+b+c}{2}\\ y=\frac{a-b+c}{2}\\ z=\frac{a+b-c}{2}\\ \end{align*}$
  
  This was a pain to do on mobile (and from memory) but at least it gave me something to do and should give some idea of what I did. Still have no idea if it’s actually valid or if I just got lucky.
  
  Reply
  1. Red Five says:
    
    October 19, 2024 at 8:44 am
    
    I think it is quite a brilliant idea. I’m not completely sure how you went from $(1+2\alpha)a+(1+2\beta)b+(1+2\gamma)c=0$ to $(1+2\alpha)=0$ though.
    
    Can you elaborate/explain the logic here?
    
    Reply
    1. random_commenter says:
      
      October 19, 2024 at 9:48 am
      
      $\alpha$ , $\beta$ , $\gamma$ must be some specific values that works for any/all values of a, b, c. Only way that’s true is if $1+2\alpha=0$ etc.
      
      As I mentioned earlier, I’m not entirely convinced by my solution. I’d still like to better justify my first step. It just seemed like it might be right after a little thinking and a little trial and error.
      
      Reply
      1. Red Five says:
        
        October 19, 2024 at 11:34 am
        
        OK, I see (sort of).
        
        Trial and error certainly works (it took until my fourth attempt to guess the correct permutations of +,- signs though) because of the cyclic nature of the terms/coefficients and yet I’m convinced (without proof, yet) that there is a nice solution lurking somewhere.
        
        I think your idea is a very good one. Whether it is what the author of the question had in mind is anyone’s guess.
        
        Reply
        
        marty says:
        
        October 19, 2024 at 11:41 am
        
        Thank you, both. I haven’t had time to look at this but will try to today or tomorrow.
        
        Reply
  2. random_commenter says:
    
    October 19, 2024 at 12:44 pm
    
    Had some further thoughts.
    
    My initial observations were that $x=-a$ , $y=-b$ , $z=-c$ and $x=b+c$ , $y=a+c$ , $z=a+b$ were obvious solutions to equation 2, but also obviously neither alone sufficient for equation 1. Hence the idea to try a linear combination of a, b, c for each x, y, z. Which eventually worked out alright.
    
    I’ve just noticed that the right solution is the average of these two solutions and, likewise, sub in both these solutions to equation 1 and you get $-2(a+b+c)$ and $4(a+b+c)$ , which averages correctly.
    
    Perhaps there’s something there, perhaps not.
    
    Reply
Red Five says:

October 20, 2024 at 12:18 pm

Here is my final offering on problem 4:

Define $S=a^2+b^2+c^2-ab-ac-bc$

Then transpose the last two equations to become:

$(c-a)x+(a-b)y=(c-b)z+S$ and $(b-a)x+(c-b)y=(c-a)z+S$ .
Solve these for $x,y$ and we get $x=c-a+z$ and $y=c-b+z$ .

Substitute these into $2(x+y+z)=a+b+c$ and we get $6z=3a+3b-3c$ and hence $z=\frac{a+b-c}{2}$ .

From there the results for $x,y$ follow.

There was still a lot of tough algebra to get this solution but it feels neater than my other approaches.

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1. marty says:
  
  October 20, 2024 at 12:41 pm
  
  Thanks, RF. That looks nice. I still haven’t had time to look (or at TNDOT 11), but hopefully tomorrow.
  
  Reply
random_commenter says:

October 20, 2024 at 12:47 pm

One final thought.

Let $b=0$ and $c=0$

Then it’s very easy to get $x=-\frac{a}{2}$ and $y=z=-x$

Likewise $a=0$ and $c=0$

Then $y=-x=-z=-\frac{b}{2}$

And $a=0$ and $b=0$

$z=-x=-y=-\frac{c}{2}$

Combine for final answer.

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1. Red Five says:
  
  October 20, 2024 at 2:28 pm
  
  I think this approach has a lot of promise. Because the values of $a,b,c$ permute, setting each one to zero in turn and asking what happens leads to the result.
  
  Nice.
  
  Now, why does that work?
  
  EDIT – I think it is valid if we allow $a,b,c$ to be ANY numbers or say the equations hold for ANY choices of $a,b,c$ but this is not entirely clear from the question.
  
  Reply
marty says:

October 24, 2024 at 10:25 pm

OK, I’ve updated the post with a solution to Q4. On to TNDOT 11 …

Reply

UPDATE (17/10/24)

UPDATE (17/10/24)

UPDATE (17/10/24)

UPDATE (24/10/24)

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