#### (A new problem of the week)

Last week I mentioned “non-routine problems” in connection with the idea of “guessing” at a method. Let’s look at a recent discussion in which the same issues came up. How do you approach a problem when you have no idea where to start? We’ll consider some interesting implications for problem solving in general, with an emphasis on George Polya’s outline.

## First problem: mindless manipulation?

This came to us in March, from a student who identified him/herself as “J”:

Hi,

Recently I had to solve a problem

If (a + md) / (a + nd) = (a + nd) / (a + rd) and

(1 / n) – (1 / m) = (1 / r) – (1/n) , then

(d / a) = -(2 / n)

i.e. Given the two expressions above I need to prove the last equality.

I don’t understand problems like these. Basic Algebra books talk about problems like equation solving or word problems, but those are easy because

there’s always some method you can use. For example regarding equation solving you move x’s to the left, numbers to the right; word problems can be solved using equalities like distance = rate * time. But a problem like the one above it seems has no method;it seems like you’re supposed to just manipulate the symbols until you get the answer. For example I tried to solve it like this:Regarding the first expression, after multiplying numerator of the first fraction by the denominator of the second I get

(d / a) = ((m + r – 2n) / (n^2 – mr))

and 2mr = nm – nr then substitute for mr in the first expression.

I reached the solution by luck;I just manipulated the symbols andit took me a lot of time. So is there a more efficient way to solve problems like these? How to think about these problems?Am I supposed to just mindlessly manipulate the symbols until I get lucky?Finally are there any

books that deal with problems like these? Because like I mentioned it seems like most precalculus books talk about equation solving etc., problems which have a clear method. Thanks.

## The solution

Before we deal with the question, let’s look more closely at his solution.

We are given two equations:

$$\displaystyle\frac{a + md}{a + nd} = \frac{a + nd}{a + rd}$$

$$\displaystyle\frac{1}{n} – \frac{1}{m} = \frac{1}{r} – \frac{1}{n}$$

We need to conclude that

$$\displaystyle\frac{d}{a} = -\frac{2}{n}.$$

J gave only a brief outline of what he did; can we fill in the gaps?

My version is to first “cross-multiply” in each equation to eliminate fractions, and do a little simplification:

The first becomes $$(a + md)(a + rd) = (a + nd) (a + nd),$$ which expands to $$a^2 + rda + mda + mrd^2 = a^2 + 2nda + n^2d^2,$$ then $$rda + mda – 2nda = n^2d^2 – mrd^2,$$ which factors to yield $$(r + m – 2n)da = (n^2 – mr)d^2.$$ Dividing, we get $$\displaystyle\frac{d}{a} = \frac{r + m – 2n}{n^2 – mr}.$$

(You may notice here that in dividing both sides by *d*, we obscured the fact that the line before is true whenever *d* = 0. I’ll be mentioning this below.)

The second equation, multiplied by \(mnr\), becomes $$mr – nr = nm – mr,$$ which easily becomes $$2mr = nm + nr.$$ (J had a sign error here.)

Now, replacing \(mr\) with \(\displaystyle\frac{nm + nr}{2}\), we get $$\displaystyle\frac{d}{a} = \frac{r + m – 2n}{n^2 – \frac{nm + nr}{2}} = \frac{2(r + m – 2n)}{2n^2 – nm – nr} = \frac{2(r + m – 2n)}{-n(r + m – 2n)} = -\frac{2}{n}.$$

## How to solve it

Taking the question myself, I replied:

I tried the problem without looking at your work, and ended up doing almost exactly the same things. That took me just a few minutes. So probably it is

not your methoditself, butyour way of finding it, that needs improvement. In my case, I did the “obvious” things (clearing fractions, expanding, factoring) to both given equations,keeping my eyes open for points at which they might be linked together, and found one. It may be mostly experience that allowed me to find it quickly. That is, I didn’t “mindlesslymanipulate”, but “mindfullymanipulated”. And the more ideas there are in your mind, the more easily that can happen.So maybe just doing a lot of (different) problems is the main key.

I added a few more thoughts about strategies:

There may be a better method for solving this, but finding it would take me a longer time than what I did. So

perseverance at trying things is necessary, regardless. Solutions to hard problems don’t just jump out at you (unless they are already in your mind from past experience);you have to explore. The ideas I describe for working out a proof apply here as well:I like to think of a proof as a bridge, or maybe a path through a forest: you have to start with some facts you are given, and find a way to your destination. You have to start out by looking over the territory,

getting a feel for where you are and where you have to go– what direction you have to head, what landmarks you might find on the way, how you’ll know when you’re getting close.(By the way, in my work I also found that d/a = 0 gives a solution, so that if d=0 (and a ≠ 0), the conclusion is not necessarily true. Did you omit a condition that all variables are nonzero?)

You are probably right that too many textbooks and courses focus on routine methods, and don’t give enough

training in non-routine problem solving. They may include some “challenge problems” or “critical thinking exercises”, but don’t reallyteachthat. One source of this sort of training is in books or websites (such as artofproblemsolving.com) that are aimed at preparation for contests. Books likePolya’sHow to Solve It(and newer books with similar titles) are also helpful.Here are a few pages I found in our archives that have at least some relevance:

What Is Mathematical Thinking?

Others of us may have ideas to add.

Some these have been mentioned in previous posts such as How to Write a Proof: The Big Picture and Studying Math: Want a Challenge?.

## Another problem: following Pólya

The next day, J wrote in with another problem, having already followed up on my suggestions:

Hi. I posted here recently asking about problem solving and algebra and I was recommended a book called

“How to solve it” by Pólya. I bought that book and now I am trying to solve some algebra exercises using it. Today I came across this problemIf bz + cy = cx + az = ay + bx and (x + y +z)^2 = 0 , then a +/- b +/- c.

(The sign +/- was a bit confusing to me since it’s not brought up anywhere in the book besides this problem, but Wikipedia says that a +/- b = 0 is a + b =0 or a – b = 0.)

In the book “How to solve it” Pólya says that first it’s important to

understand the problem and restate it.So my interpretation of a problem is this:

If numbers x, y, z are such that (x + y + z)^2 = 0 and bz + cy = cx + az and bz + cy = ay + bx, then the numbers a, b, c are such that a + b + c = 0 or a – b – c =0

Next Pólya says to

devise a plan. To do that he says you need to look at a hypothesis and conclusion andthink of a similar problemor a theorem.The best I could think of is an elimination problem, i.e. when you’re given a certain set of equations and you can find a relationship between constants. Can you think of any other similar problems which could help me solve this problem?

I first responded to the last question:

Hi again, J.

I would say that the last question you asked was “similar” to this, so the same general approach will help. That’s essentially what you said in your last paragraph, I think. I know that isn’t very helpful, but it’s all I can think of myself. You’d like to have seen a problem that is more specifically like this one, such as having (x + y + z)

^{2}= 0 in it, perhaps, so you could get more specific ideas.I only know that I have seen a lot of problems like this involving

symmetrical equations(where each variable is used in the same ways), and I suspect those problems can be solved by similar methods. But I don’t know one method that would work for this one.

I’ll get back to that question. But let’s focus first on Polya.

Here is what Polya says (p. 5) when he introduces his famous four steps of problem solving:

In order to group conveniently the questions and suggestions of our list, we shall distinguish four phases of the work. First, we have to

understandthe problem; we have to see clearly what is required. Second, we have to see how the various items are connected, how the unknown is linked to the data, in order to obtain the idea of the solution, to make aplan. third, wecarry outour plan. Fourth, welook backat the completed solution, we review and discuss it.

This process is then explained in more detail, and used as an organizing principle in the rest of the book. It can be amazing to see how many students jump into a problem before they *understand* what it is asking, or do calculations without having made any *plans*. On the other hand, it would be wrong to think of these four steps as a *routine* to be followed exactly; often you don’t fully *understand* a problem until you have started *doing something*, perhaps carrying out a half-formed plan and then realizing that you had a wrong impression of some part.

## Understanding the problem

And J has here a good example of a misunderstanding. This problem uses the plus-or-minus symbol (±) in a rare way, which in this case requires *asking* (not explicitly one of Polya’s recommendations, but valuable!).

The problem says this:

$$\text{If } bz + cy = cx + az = ay + bx \text{ and } (x + y + z)^2 = 0 \text{, then } a \pm b \pm c.$$

(No, that doesn’t quite make sense! We’ll be fixing that shortly.)

What does it mean when there are two of the same symbol? The Wikipedia page J found says, “In mathematical formulas, the ± symbol may be used to indicate a symbol that may be replaced by either the + or − symbols, allowing the formula to represent two values or two equations.” They give an example (the quadratic formula), where *either sign* yields a valid answer; then an example with two of the same sign (the addition/subtraction identity for sines) in which both must be replaced with the *same sign*; and third example (a Taylor series) where the reader has to *determine which sign* is appropriate for a given term. Later they introduce the minus-or-plus sign (\(\mp\)), which explicitly indicates the *opposite sign* from an already-used ±.

But here, we have two ±’s with no clear reason why they should be the same, or should be different. Is this a special case? J has assumed they are the same, so that it means “\(a + b + c = 0\) or \(a – b – c = 0\)“. This is the first issue I had to deal with:

First, though, did you mean to say that the conclusion is a ± b ± c

= 0? That wouldn’t quite mean what you said about it, because the two signs need not be the same. Rather, it means thateithera + b + c = 0, or a + b – c = 0, or a – b + c = 0, or a – b – c = 0:any possible combinationof the signs.

Now, how did I *know* that, when it goes against what Wikipedia seems to be saying? I’m not sure! There is actually some ambiguity; really, we just shouldn’t rule out this *possibility*. But I saw from the start that if the two signs are the same, then the problem has an odd *asymmetry*, requiring *b* and *c* to have the same sign in this equation, but not *a*. That simply seems unlikely, considering the symmetry elsewhere.

Sometimes we discover, as we proceed through the solving process, that we *have* to interpret the statement one way or another in order for it to be true – an example of my comment that understanding can come *after* doing some work. (That was actually the case here. But the problem really should have been written to make this clear!)

## Hints toward a solution

What this means is that

we don’t know the signsof the numbers. One thing that suggests is that we might be able to show some fact about a^{2}, b^{2}, and c^{2}, so that we would have to takesquare roots, requiring us to use ± before each of a, b, and c.It’s also interesting that they said that (x + y + z)

^{2}= 0, which means nothing more than x + y + z = 0. That also makes me curious, and at the least puts squares into my mind for a second reason.

Here I am just letting my mind wander around the problem, pondering what the givens suggest. This is part of both the understanding phase, and the “looking for connections” Polya talked about.

Not even being sure of the conclusion, I just tried manipulating the equations any way I could, just to make their meanings more visible; and then I solved x + y + z = 0 for z and put that into my derived equations, eliminating z. That took me eventually to a very simple equation that involved a, b, x

^{2}, and y^{2}. And that gave a route to the ± I’d had in mind.

We could say that my initial plan is, as I suggested at the top, to **explore**! We can refine the plan as we see more connections. (As I said, Polya has to be followed flexibly.)

There’s a lot of detail I’ve omitted, in part because much of my work was undirected, so you may well find a better way. But the key was to have some thoughts in mind before I did a lot of work, in hope of

recognizing a useful form when I ran across it. The other key wasperseverance, because things got very complicated before they became simple again! (I suspect that as I go through this again, I’ll see some better choices to make, knowing better where I’m headed.)I don’t think you told us where these problems came from; they seem like contest-type problems, which you can expect to be highly non-routine. As I said last time, until you’ve done a lot of these, you just need to keep your eyes open so that you are learning things that will be useful in future problems! I am not a contest expert, as a couple of us are, so I hope they will add some input.

Since we never got back to the details of this problem, let’s finish it now. Frankly, I had to look in my stack of scrap paper to find what I did in March, because I wasn’t making any progress when I tried it again just now. Clearly I could have given a better hint! I was hoping that just the encouragement that it could be done would lead to J finding a nicer approach than mine.

But here’s what I find in my incomplete notes from then. First, I rewrote the equality of three expressions as two equations, and eliminated c; I’ll use a different pair of equations than I did then, with that goal in mind: $$cx + az = ay + bx\; \rightarrow\; c = \frac{ay-az+bx}{x}$$ $$ay + bx = bz + cy\; \rightarrow\; c = \frac{ay-bz+bx}{y}$$

Setting these equal to eliminate c, $$\frac{ay-az+bx}{x} = \frac{ay-bz+bx}{y}$$

Cross-multiplying, $$ay^2-ayz+bxy = axy-bxz+bx^2$$

Solving \(x + y + z = 0\) for *z* and substituting, $$ay^2-ay(-x-y)+bxy = axy-bx(-x-y)+bx^2$$

Expanding, $$ay^2 + axy + ay^2 + bxy = axy + bx^2 + bxy + bx^2$$

Canceling like terms on both sides, $$2ay^2 = 2bx^2$$

Therefore, $$\frac{x^2}{a} = \frac{y^2}{b}$$

We could do the same thing with different variables and find that this is also equal to \(\frac{z^2}{c}\). So we have $$\frac{x^2}{a} = \frac{y^2}{b} = \frac{z^2}{c} = k$$

Now we’re at the place I foresaw, where we can take square roots: $$x = \pm\sqrt{ak}$$ $$y = \pm\sqrt{bk}$$ $$z = \pm\sqrt{ck}$$

Therefore, since \(x+y+z=0\), we know that $$\pm\sqrt{ak}+\pm\sqrt{bk}+\pm\sqrt{ck}=0$$

and, dividing by \(\sqrt{k}\), we have $$\pm\sqrt{a}\pm\sqrt{b}\pm\sqrt{c}=0$$

In March, it turns out, I stopped short of the answer, thinking I saw it coming. But in fact, I didn’t attain the goal! **I hoped that a, b, and c would be squared** before we have to take the roots. We seem, however, to have proved that

**they must all be positive**, which makes the conclusion impossible!

I’m wondering if the problem, which was never quite actually stated, might have been different from what I assumed. In fact, armed with this suspicion, I tried to find an example or a counterexample, and found that if $$\begin{pmatrix}a & b & c\\ x & y & z\end{pmatrix}= \begin{pmatrix}1 & 4 & 1\\ 1 & -2 & 1\end{pmatrix}$$ satisfies the conditions, with $$bz + cy = cx + az = ay + bx = 2,$$ but no combination of signed *a*, *b*, and *c* add up to 0. So the real problem must have been something else …

## Remembering how to solve a problem

At this point J abandoned that path, and closed with a side issue:

Hi Doctor. I have one more question about problem solving.

I spent some more time on the problem we discussed then I skipped it and decided to focus on other problems instead. I managed to solve a few of them but then I took a long break when I came back

I couldn’t remember the solutions without looking at my work.I don’t know if you read

How to solve itby Pólya. I ask since at the beginning of that book Pólya gives an example of a mathematical problem. The problem in question is this:Find the diagonal of a rectangular parallelepiped if the length, width, and height are known.

He asks the reader to consider the auxiliary problem of finding the diagonal of the right triangle using Pythagoras theorem. I am telling you this because the solution to this problem is very clear; I can recall it even long after I finished reading. I do not feel the same about algebra problems. I solve them, do the obvious things, and then I almost immediately forget. Does that happen to you? If not

how do you remember the solution?I just want to know if you find these algebra problems as unintuitive as I do.

My memory is as bad as anyone’s! I replied,

I wouldn’t say that I remember every solution I’ve done, or every solution I’ve read. The example you give is a classic that stands out, particularly the overall strategy. Others are more ad-hoc and don’t feel universal (in the sense of being applicable to a large class of problems), so they don’t stick in the memory.

I don’t have my copy of Polya with me (I’ve been meaning to look for it), but I recall that one of his principles is to

take time after solving a problem to focus on what you didand think about how it might be of use for other problems. This is something like looking around before I leave my car in a parking lot to be sure I will recognize where I left it when I come back from another direction.I want to fix the good idea in my mind and be able to recognize future times when it will fit.But even though I do have that habit, there are some problem types that I recognize over and over, but keep forgetting what the trick is. (Maybe sometimes it’s because I’ve seen two different tricks, and they get mixed up in my mind.)

So you’re not alone. For me, though, it’s not such much being

unintuitive, as just not beingmemorable, or being too complex for me to have focused on them enough to remember.

So Polya recognized the likelihood of forgetting (failing to learn from what you have done), and the need to make a deliberate effort there!