Hi everyone, it's Jenni with the How and Why of Mathematics and today I'm going to be going
over some of the hardest questions from the senior division of the Australian
Mathematics Competition, and these are for students around 17 years old so if
you're looking for something a bit easier then check out my previous video
on the junior division questions. So I'll start you off with something not too
hard; five percent of students got this one right so it's not too bad, and the
question is to find the area of the region defined by this inequality:
absolute value of x plus absolute value of y plus absolute value of x plus y
less than or equal to two, so you might like to pause the video here, have a go
at that, and then come back and see my solution. So we need to graph this
inequality and the idea with absolute values is that you want to just consider
one region where you know whether x and y are positive or not and x plus y, if
that's negative or positive, because then you can rewrite the equation without the
absolute values. So for example we could concentrate on the section where both x
and y are positive, up here, and then x plus y is going to be positive as well. Then
in this region the absolute value of x is just going to be equal to x and
that's just going to be equal to y and that's just going to be equal to x plus
y so the equation becomes that in this region, and then we can simplify that, so
that's 2x plus 2y and then let's divide both sides of the equation by 2
- both sides of the *inequality*, rather - that gives us x plus y less than or equal
to one. So that's a straight line, like you can rearrange this if you like
to be y less than or equal to -x + 1, and then it's a line with a y-intercept of 1
and a slope of negative one so that's going to be like that, where that's one
and that's one. Alright that's that bit. Now, if we go down here then y is gonna
be negative and x is going to be positive but x plus y could be negative
or positive, depending, so what we're going to have to do is consider like two
different cases: x plus y is greater than zero, or x plus y is less than or equal
to 0, and that's gonna split along that line because for example if x
plus y is greater than 0 that means y is greater than negative x so if you
graphed y equals negative x that would be this line. So this region is going to
be different from that region. In this region you've got x is positive so
absolute value of x is just x, absolute value of y is negative y because y is
negative there, and then if x plus y is positive in this region then that
absolute value of x plus y is just x plus y so then we just need to simplify
this. The y's will cancel and then we've got 2x so let's divide both sides of the
inequality by 2. We get x less than or equal to 1. So that is... the x = 1 is that
line there, and we just want to keep it within this region; we don't want to like
extend it that way or that way because this is only valid in this little
triangle thing here. All right, now we can move on to this section, so again
we've got x minus y but x plus y is negative
inside here so we have to do negative x plus y. And then we'll simplify that,
the x's will cancel, we get negative 2y, so let's divide both sides by negative 2
and that's gonna flip the inequality so it started as less than or equal, it's gonna
be greater than or equal to when you divide or multiply by a negative
number. So we get y greater than or equal to -1. Now the line y equals
negative 1 is along there, and again we don't want to extend it outside of this region. OK
and then over here, x and y are both negative and so is x plus y, so that
makes things easier. We've got negative x minus y minus the x plus y, that's less
than or equal to 2 and let's simplify this so we've got negative 2x minus 2y and
then let's divide both sides by the negative 2 so again that's gonna flip
the inequality, and then if you like you can rearrange this so it's negative x
minus 1, so that's a line with a slope of negative 1 and y-intercept of negative 1
so y-intercept of -1 and it's going like that. That point is at
negative 1 and that's at -1 as well. Now
we could continue with these regions but what I'm gonna do instead is just...
thinking about this equation, it's actually symmetric with respect to
swapping x and y. What I mean by that is that when you switch x and y in this
equation, so you like if I go like this, so just swapping x and y, it's actually
the exact same equation because the order doesn't matter when we add things.
Now what that means for the graph is that when you swap x and y in the graph,
it should be the same picture basically. Now, swapping x and y in the graph is the
same as reflecting along that line, so this bit I can just flip that over here
and it should be exactly the same. You can just go through the same process for
these two regions as well, writing out the equation, but this is a good shortcut.
OK so this is the region that we want, all of this in here, we just gotta work out
that area. Now you could do, like, you could split it into these two squares
and the two triangles and do like half base times height, or you could imagine
cutting this bit off and sticking it down there to make a full square, so we
get three full squares, so the area is going to be three. That's the answer.
Alright, that's it for this video so let me know in the comments if you're still confused
about anything about this question, and if you want to see more videos like this
then be sure to give me a thumbs up and subscribe and click the bell so you
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