## The 2018 South African Mathematics Olympiad — Problem 4

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 of the posts can be found here.

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

Let $ABC$ be a triangle with circumradius $R$, and let $\ell_A, \ell_B, \ell_C$ be the altitudes through $A, B, C$ respectively. The altitudes meet at $H$. Let $P$ be an arbitrary point in the same plane as $ABC$. The feet of the perpendicular lines through $P$ onto $\ell_A, \ell_B, \ell_C$ are $D, E, F$ respectively. Prove that the areas of $DEF$ and $ABC$ satisfy the following equation:

$\displaystyle \text{area}(DEF) = \frac{{PH}^2}{4R^2} \cdot \text{area}(ABC).$

Once again, we begin by creating a diagram. Again, since I already know how the solution plays out, I’ve drawn in the circle that passes through $P, E, D, H$, and $F$. We do know yet that these points are concylic, however, as it is not given directly in the problem statement.…

## 1.3 Relations: Equivalence relations

We know that a relation is called an Equivalence Relation when it is reflexive, symmetric AND transitive on some set A. Let’s look at some examples.

Example One: Let $a,b \in \mathbb{R}.$ Suppose we have that a is related to b (i.e. a ~ b) if $a - b \in \mathbb{Z}.$ We want to show that our relation ~ is an equivalence relation.

First, let’s unpack what the question requires us to prove: It wants us to show that the relation ~ on set A is an equivalence relation. Hence, we need to show that ~ is reflexive, symmetric and transitive.

The relation ~ (in this case) is defined as follows: IF any two real numbers, a and b, are related THEN we know that a – b is some integer.

It’s important to note that the order in which a relation is important! Always write your equations as they’ve been given in the question to avoid confusion and mistakes

Ok, let’s prove this.…

## 1.2 Relations: Properties

Note: Do not confuse binary operations (+, x, -, …) with relations. Recall the definition for a relation as:

A relation R on a set A is a subset R ⊆ A × A. We often abbreviate the statement (x, y) ∈ R as xRy.

For instance, the binary operation “x” has a numeric value: $3 \times 3 = 9 \text{ and } 40 \times \frac{1}{5} = 8.$ Yet a mathematical relation, for example “<“, has a True/False value: $3 < 3 \text{ and } 40 < \frac{1}{5}$ are both False expressions.

We want to look at some properties of relations. We will look at three properties for relation expressions:

Suppose A is a set with relation R, then

1. Relation R is Reflexive if $\forall x \in A, \text{ } xRx$
2. Relation R is Symmetric if $\forall x,y \in A, \text{ } xRy \rightarrow yRx$
3. Relation R is Transitive if $\forall x,y,z \in A, \text{ } xRy \wedge yRz \rightarrow xRz$

The first property tells us that if every element, x, in set A is related to itself, then the relation R acting on the set A is termed “reflective.”

The second property tells us that if “x is related to y” from set A implies that “y is also related to x,” then R is termed “symmetric.”

Lastly, if “x is related to y” and “y is related to z” implies that “x is also related to z,” then the relation R is termed “transitive.”

Let’s look at the following examples: $A = \{ 1, 2, 3, 4\}$ with relation $\le.$ Then:

$\forall x \in A, \text{ } x \le x$

In other words $\le$ is reflexive since every number in set A is equal to itself (i.e.…

## 1.1 Relations: Introduction

What are relations?

In every day life, a person is related to their parents, siblings, cousins, teachers, friends, etc. in some way. Similarly in mathematics, mathematical objects like numbers and sets are related to one another in some way. Many relations (symbols) will be familiar already:

1. $2 <3$
2. $\pi \approx 3.14$
3. $5 \in \mathbb{Z}$
4. $X \subset Y$
5. $a \equiv b(modn)$

Consider the following set $A = \{1, 34, 56, 78 \}.$ We can compare the numbers in A using the symbol “<” as follows: $1 < 56, 34 < 78$ etc. We can write this as a set in the following way: $R = \{ (1,34), (1,56), (1,78), (34,56), (34,78), (56,78) \}.$

Each pair in this new set R expresses the relationship x < y (where x and y are numbers from A).

In other words, $1<34, 1<56, 1<78, ...$ So if asked whether $34 < 78$ is true,  one only needs to look into our set R to find the pair $(34,78).$ If we didn’t find it, then the relation would be considered false for the given set. The above example is intuitive because we are already comfortable with the relation <. In more abstract cases, thinking of the relationship between mathematical objects in this way may be a little trickier!…

## 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).…