How To Really Solve This Tricky Algebra Problem (XI)

Welcome to the eleventh entry in the tricky algebra problem series. As we near the end of this year (2022) and prepare for the next (2023), I came across a fitting algebra problem (quite literally).

I found multiple sources that claim that this problem is from the Vietnamese International Maths Olympiad. But despite my efforts, neither could I confirm this nor could I narrow down on the original source of the problem.

With the housekeeping done, let’s start with the mathematics; this will be fun. To begin, we have the following equations:

Given these equations, your task is to figure out the solution to the following expression:

Do you think you can solve this problem?

Spoiler Alert

If you wish to solve this problem on your own, then I suggest that you pause reading this essay at this point and do so now. Beyond this section, I will be explicitly discussing the solution to this problem.

After you are done with your attempt, you may continue reading this essay to compare our approaches. All the best!

This essay is supported by Generatebg

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Setting Up the Tricky Algebra Problem

Let us start by labelling our starting equation-set as follows:

The very first thing to note about the starting equation-set is that none of the variables: a, b, or c can be equal to zero. If any of them were indeed equal to zero, equation-2 would be impossible (due to division by zero).

Although this is a subtle point, this will play a key role throughout our solution. Now, let try to find a way to relate equation-1 with equation-2. One easy way to do this is by simply reciprocating equation-1 as follows (note that this is possible because (a + b + c) ≠ 0):

Let us now algebraically manipulate equation-3 such that we have all the terms on the left-hand side:

We can further simplify the fractions as follows:

As we continue to simplify, we can eliminate the denominator (note that (a*b*c)≠0 based on equation-2):

With equation-4, we are almost ready to arrive at our solution. But there are a few things we still need to consider.

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Solving the Problem Further

What equation-4 in combination with equation-1 tells us is that one of the following equations/conditions is true:

(a = −b) or (b = −c) or (c = −a)

Let us now assume that (a = −b). If we plug this into equation-1, then we arrive at the following result for ‘c’:

Now, we still need to solve for ‘a’ and ‘b’. To do this, let us consider equation-2:

With this equation, we arrive at the solution to our problem.

The Solution for the Tricky Algebra Problem

Based on our algebraic simplification so far, the solution is as follows:

Recall that we originally assumed that (a = −b). We still have two more possibilities to consider: (b = −c) or (c = −a).

Since we have a symmetric equation, regardless of which equation we consider, we will have two terms taking each other out and one term remaining.

So, the final generalised solution to our trickly algebra problem would take the following form:

I hope you enjoyed this problem as much as I did. Sadly, this is the last algebra problem from me this year.

But worry not! I’ll be back with more algebra problems next year. So, if you are into that sort of thing, keep an eye on this space!

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Further reading that might interest you:

• How To Really Use Mathematical Induction?
• The Three Prisoners Puzzle: How To Really Solve It?
• How To Really Solve The Dartboard Paradox?

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