## The 2018 South African Mathematics Olympiad — Problem 3

The final round of the South African Mathematics Olympiad will be taking place on Thursday, 28 July 2019. In the week leading up to the contest, I plan to take a look at some of the problems from the senior paper from 2018. A list of all the posts can be found here.

Today we will look at the third problem from the 2018 South African Mathematics Olympiad:

Determine the smallest positive integer $n$ whose prime factors are all greater than $18$, and that can be expressed as $n = a^3 + b^3$ with positive integers $a$ and $b$.

In many number theory problems, it helps to consider the prime factors of the numbers involved, and in this problem we are in fact forced to do so because the question itself is about the prime factors of a number. When dealing with factors of a number or an expression representing some number, it of course helps to consider whether we can factorise the given expression.…

## The 2018 South African Mathematics Olympiad — Problem 2

The final round of the South African Mathematics Olympiad will be taking place on Thursday, 28 July 2019. In the week and half leading up the the contest, I plan to take a look at some of the problems from the senior paper in 2018, and have already written about the first problem

The second problem from the 2018 South African Mathematics Olympiad was

In triangle $ABC$, $AB = AC$, and $D$ is on $BC$. A point $E$ is chosen on $AC$, and a point $F$ is chosen on $AB$, such that $DE = DC$ and $DF = DB$. It is given that $\frac{DC}{BD} = 2$ and $\frac{AF}{AE} = 5$. Determine the value of $\frac{AB}{BC}$.

The first step of solving any geometry problem should always be to draw a sketch. This helps you to understand how different parts of the figure relate to each other, and an accurate sketch may help you to form conjectures. Sometimes having a deliberately inaccurate sketch on hand is also helpful as it may help to avoid circular reasoning.…

## The 2018 South African Mathematics Olympiad — Problem 1

The final round of the South African Mathematics Olympiad will be taking place on Thursday, 28 July 2019. In the two weeks leading up to the contest, I plan to take a look at some of the problems from the senior paper from 2018.

The first problem from the 2018 South African Mathematics Olympiad was

One hundred empty glasses are arranged in a $10 \times 10$ array. Now we pick $a$ of the rows and pour blue liquid into all glasses in these rows, so that they are half full. The remaining rows are filled halfway with yellow liquid. Afterwards, we pick $b$ of the columns and fill them up with blue liquid. The remaining columns are filled with yellow liquid. The mixture of blue and yellow liquid turns green. If both halves have the same colour, then that colour remains as is.

1. Determine all possible combinations of values for $a$ and $b$ so that exactly half of the glasses contain green liquid at the end.

## The definite integral

I realise now, in all the excitement of the FTC that I hadn’t written a post about the definite integral…that’s shocking! ok, here we go…the plan for this post:

• Look at our Riemann sums and think about taking a limit of them
• Define the definite integral
• Look at a couple of theorems about the definite integral
• Do an example
• Look at properties of definite integrals

That’s quite a lot, but we are more or less going to follow along with Stewart. Stewart just has a slightly different style to mine, so I recommend reading his for more detail, and mine for potentially a bit more intuition.

So, let’s begin…

We have seen in previous lectures/sections/semesters/lives that we can approximate the area under a curve by splitting it up into rectangular regions. Here are examples of splitting up one function into rectangles (and, in the last way trapezoids, but you don’t have to worry about this).…

## The South African Mathematics Olympiad

The South African Mathematics Olympiad is an annual mathematics competition for high-school students in South Africa. The competition is organised by the South African Mathematics Foundation, and comprises three rounds which increase in difficulty. The final round of the 2019 South African Mathematics Olympiad will take place on Thursday, 25 July, and the top ten junior (Grade 8 and 9) and senior (Grades 10—12) competitors will be invited to a prize-giving evening taking place on 14 September 2019. At the same time, the problem selection committee will meet to start setting the 2020 papers.

According to the SAMF, nearly 100000 students participated in the 2017 edition of the competition. The numbers for the 2019 competition are likely to be similar. These students all write the first round of the competition, which learners write at their individual schools in March every year. The papers are marked at the school, and any student with more than 50% is invited to participate in the second round of the competition.…

## The Fundamental Theory of Calculus part 2 (part ii)

OK, get ready for some Calculus-Fu!

We have now said that rather than taking pesky limits of Riemann sums to calculate areas under curves (ie. definite integrals), all we need is to find an antiderivative of the function that we are looking at.

As a reminder, to calculate the definite integral of a continuous function, we have:

$\int_a^b f(x)dx=F(b)-F(a)$

where $F$ is any antiderivative of $f$

Remember that to calculate the area under the curve of $f(x)=x^4$ from, let’s say 2 to 5, we had to write:

$\int_2^5 x^4 dx=\lim_{n\rightarrow \infty}\sum_{i=1}^n f(x_i)\Delta x=\lim_{n\rightarrow \infty} f\left(2+\frac{3i}{n}\right)\frac{3}{n}=\lim_{n\rightarrow\infty}\frac{3}{n}\left(2+\frac{3i}{n}\right)^4$

And at that point we had barely even started because we still had to actually evaluate this sum, which is a hell of a calculation…then we have to calculate the limit. What a pain.

Now, we are told that all we have to do is to find any antiderivative of $f(x)=x^4$ and we are basically done.

Can we find a function which, when we take its derivative gives us $x^4$?…

## The Fundamental Theory of Calculus part 2 (part i)

OK, now we come onto the part of the FTC that you are going to use most. We are finally going to show the direct link between the definite integral and the antiderivative. I know that you’ve been holding your breaths until this moment. Get ready to breath a sign of relief:

The Fundamental Theorem of Calculus, Part 2 (also known as the Evaluation Theorem)

If $f$ is continuous on $[a,b]$ then

$\int_a^b f(x) dx=F(b)-F(a)$

where $F$ is any antiderivative of $f$. Ie any function such that $F'=f$.

————-

This means that, very excitingly, now to calculate the area under the curve of a continuous function we no longer have to do any ghastly Riemann sums. We just have to find an antiderivative!

OK, let’s prove this one straight away.

We’ll define:

$g(x)=\int_a^x f(t)dt$

and we know from the FTC part 1 how to take derivatives of this. It’s just $g'(x)=f(x)$. This says that $g$ is an antiderivative of $f$.…

## The Fundamental Theorem of Calculus part 1 (part iii)

So, we are now ready to prove the FTC part 1. We’re going to follow the proof in Stewart and add in some discussion as we go along to motivate what we are doing. What we are going to prove is that:

$\frac{d}{dx} \int_a^x f(t) dt=f(x)$

for $x\in [a,b]$ when $f$ is continuous on $[a,b]$.

Proof:

we define $g(x)=\int_a^x f(t)dt$ and we want to find the derivative of $g$. We will do this by using the fundamental definition of the derivative, so let’s look at calculating this function at $x$ and $x+h$ – ie. how much does it change when we change $x$ by a little bit?

$g(x+h)-g(x)=\int_a^{x+h}f(t) dt-\int_a^x f(t) dt$

But remember that the definite integral is just the area, so this difference is the area between a and x+h minus the area between a and x. Which is just the area between x and x+h. Using the properties of integrals, we can write this formally as:

$g(x+h)-g(x)=\int_a^{x+h}f(t) dt-\int_a^x f(t) dt=\left(\int_a^{x}f(t)+\int_x^{x+h}f(t)\right)-\int_a^{x}f(t)=\int_x^{x+h}f(t)dt$

and we can write, for $h\ne 0$:

$\frac{g(x+h)-g(x)}{h}=\frac{1}{h}\int_x^{x+h}f(t)dt$

Restated, we can think of this as the area between x and x+h divided by h.…

## The Fundamental Theorem of Calculus part 1 (part ii)

OK, so up to now we can’t actually use the FTC (Fundamental Theorem of Calculus) to calculate any areas. That will come from the FTC part 2.

For now, let’s take some examples and see what the FTC is saying. I’ll restate it here:

The Fundamental Theorem of Calculus, part 1

If $f$ is continuous on $[a,b]$ then the function $g$ defined by:

$g(x)=\int_a^x f(t) dt$,     for $a\le x\le b$

is continuous on $[a,b]$ and differentiable on $(a,b)$, and $g'(x)=f(x)$.

——

Let’s look at some examples. We’re going to take an example that we can calculate using a Riemann sum. Let’s choose $f(x)=x^2$.

If we integrate this from $0$ to some point $x$ – ie. calculate the area under the curve, we get:

$\int_0^x t^2 dt=\frac{x^3}{3}$.

Make sure that you can indeed get this by calculating the Riemann sum.

So, what does the FTC part 1 tell us? It says that if we take the derivative of this area, with respect to the upper limit, $x$, then we get back $f(x)$.…