Once in a while, it does me good to tackle a problem “out loud” (or at least ad-lib the typing), just so I can talk through my thought process unfiltered.

Today I’m going to look at question 102 from @ridermeister’s excellent collection of university entrance style problems:

What is the area of the region in the Cartesian plane whose points $(x,y)$ satisfy $\abs{x} + \abs{y} + \abs{x+y} < 2$?

### Sketch!

This yells out “I am a graph! Sketch me! Bigger than that!”. The question is, what does the graph look like?

My first approach, when I’m not sure about a graph, is to find some points – while also noting that swapping $x$ and $y$ changes nothing, so we can exploit the symmetry. If $(x,y)$ is in the region, so is $(y,x)$ and the graph is symmetrical in the line $y=x$.

I’ve kind of skipped past my DATAS framework here (domain, axes, turning points, asymptotes, shape) as I don’t see an easy way to apply anything much except the axes part. When $x=0$, we have $\abs{0} + \abs{y} + \abs{y} < 2$, which resolves to $\abs{y} < 1$ – so we know that $(0,1)$, $(1,0)$, $(-1,0)$ and $(0,-1)$ are on the boundary of the region.

What about other simple cases? How about $x=y$ and $x=-y$?

If $x=y$, then we have $\abs{x} + \abs{x} + \abs{2x} < 2$, which resolves to $\abs{x} < \frac{1}{2}$. That puts $\left(\frac{1}{2},\frac{1}{2}\right)$ and $\left(-\frac{1}{2},-\frac{1}{2}\right)$ on the boundary.

If $x = -y$, then we have $\abs{x} + \abs{-x} + \abs{0} < 2$, which gives us $\abs{x} < 1$. We now also have the points $(1,-1)$ and $(-1,1)$ – but we’re none the wiser about the general shape. It looks vaguely hexagonal, but I’m not about to pen that in just yet.

What about generalising slightly? Suppose $y=kx$ (which it must do, unless $x=0$, and we’ve already dealt with that case).

Then we have $\abs{x} + \abs{kx} + \abs{(1+k)x} < 2$, and we can simplify that to $\abs{x}\left(1 + \abs{k} + \abs{1+k}\right)$ – I’m pretty sure $\abs{ab} = \abs{a}\abs{b}$ for all real pairs. Yes, it comes from the sign laws. (Convince yourself of things you’re unsure of.)

There are three cases to consider here: $k$ is positive (where both terms inside the bracket’s absolute value signs are positive), $k$ is between 0 and -1 (the middle one changes sign) and $k$ is less than -1 (both of the values switch sign). Let’s deal with those separately.

If $k>0$, we’ve simply got $\abs{x} (2 + 2k) < 2$, so $\abs{x} < \frac{1}{1+k}$. That surprises me a little! But it gives us a parametrisation of $\left(\pm \frac{1}{1+k},\pm \frac{k}{1+k}\right)$, which (after a bit of thought – forgive me, I’m recovering from COVID as I write this) is the same as $y=1-x$ on the top side and $y = x-1$ on the bottom. That fills in two sides of the hexagon I had in my mind, which is a nice thing!

What about for $-1 < k < 0$? Then we have $\abs{x}\left(1-k + (1+k)\right)<2$, which turns out to be $\abs{x} < 1$. Neat! That’s a pair of vertical lines, and another pair of sides to something that’s looking more like a hexagon with every computation.

And if $k < -1$? Then we’ve got $\abs{x}\left(1-k - (1-k)\right)<2$, or $\abs{x}<\frac{1}{1-k}$. Parametrically, the edges are $\left(\pm\frac{1}{1-k}, \mp \frac{k}{1-k}\right)$… no, wait, I’ve confused myself with the signs. $y=kx$, even when $k$ is negative, so the edges are $\left(\pm\frac{1}{1-k}, \pm \frac{k}{1-k}\right)$. (The reason I checked was that the other version only gave one boundary line rather than two.)

When I actually sketch this out, I notice I’ve made another sign error – that whole last paragraph was wrong. My suspicions were raised because the boundary lines were not horizontal as I expected (we could actually fill them in by symmetry at this point). Double-check your hunches against your calculations and vice-versa.

I should have said that if $k < -1$, then we’ve got $\abs{x}\left(1-k-(1+k)\right)<2$, or $\abs{x} < -\frac{1}{k}$. A moment of panic about that minus sign, and a moment of relief as I remember that $k$ is negative, so it all pans out. (I initially put a 2 on top of the fraction also, but caught that in time). So, $\frac{1}{k} < x < -\frac{1}{k}$, which means $1 > y > -1$, giving us the missing two horizontal lines of the hexagon. Boom!

But we’re not done. Double check the question – we want the area of the region, not a sketch of it! There are two 1-by-1 squares and two triangles that add up to a 1-by-1 square, so the area is 3.

### Other approaches?

Now that I have a solution, it’s a good time to look for other approaches – is there a neater way to see the answer?

Another way to describe the region would be $\abs{x}<1$, $\abs{y}<1$ and $\abs{x+y} < 1$. (I’ll come back to this thought in a moment.)

I see another missed symmetry trick, or rather two of them: if I replace $x$ with $-x$ and $y$ with $-y$, then the original inequality is unchanged – which gives a half-turn rotational symmetry about the origin.

If we’ve got rotational symmetry, we only really need to think about two (non-opposing) quadrants – say the first ($x>0$ and $y>0$) and the second ($x<0$ and $y>0$).

If $x$ and $y$ are both positive, then we have $x + y + (x+y) < 2$, or $x+y<1$. That maps via rotation to the third quadrant as $x + y > -1$, which together give the $\abs{x+y}<1$ condition.

If $x$ is negative and $y$ is positive, then we have $-x + y + \abs{x+y} < 2$, with the absolute term differing depending on whether $\abs{x}$ or $\abs{y}$ is larger.

• If $\abs{x}$ is larger, then we have $-x + y + (-x-y) < 2$, or $-x<1$. That’s the left-hand line.
• If $\abs{y}$ is larger, we have $-x + y + (x+y)<2$, or $y<1$, which is the upper line.

Again, we can rotate that to the fourth quadrant to fill out the diagram.

I still feel like there’s some more insight to be had with this problem – but I’m going to call it solved.

Did you tackle it differently? Let me know about it!

* Edited 2023-01-31 to fix multiple formatting issues. Thanks to Andrew, Adam and Oscar for pointing them out.