# Australian Intermediate Maths Olympiad

**5 hours**left at this price!

- 4 hours on-demand video
- 42 articles
- 14 downloadable resources
- Full lifetime access
- Access on mobile and TV

- Certificate of Completion

Get your team access to 4,000+ top Udemy courses anytime, anywhere.

Try Udemy for Business- Mathematics
- Number theory (Arithmetic and modular arithmetic)
- Geometry (Triangles and circles)
- Proofs (Direct, induction, contraposition, contradiction)
- Algebra

- Concentrate for 2 hours
- Determined to succeed

First longer time limit competition for students in Australia to display their maths skills over 4 hours. We use this as a concrete way to teach students long term maths skills rather than just focus completely on the test. It will be a great way to improve one's concentration length, logical reasoning and overall problem solving skills. Come join us unravel the beauty of mathematics.

- Students keen to improve their maths wanting to go to the IMO
- Students qualified for the AIMO through school or AMC

Topics in Olympiads typically are A G N C Algebra, Geometry, Number Theory, Combinatorics.

We will focus on the first 3 and leave the rest to problem solving.

I will break each into two parts, one easier and one more advanced (depends more on earlier stuff)

Algebra I- factoring a sum of powers

Algebra II- proof and logic

Geometry I- cyclic quads

Geometry II- incircles and excircles

Number Theory I- modular arithmetic

Number Theory II- fundamental theorem of arithmetic.

Rate your commitment from 1 to 3. Say its 2.

Allocate two 4 hour blocks for continuous study (one can be the monday class) and another can be self study.

Topics Number 1, 2, Geometry 1,2, Algebra 1,2.

Geometry

1. cyclic quads from circumcircle and associated circle geo, geometric angle bisector, incircles and area trick.

2. extended sine rule and basic trig as in 1.1 revisited, angle bisector theorem, sine rule as area trick

Number

1. Mod arithmetic and divisibility rules

2. Fundamental theorem of arithmetic, primes

Algebra

1.Distributative law

2. Proofs

The first component is more problem solving the second component is more proofs.

I taught these notes at James Ruse in the year just after I had visited the Easter selection school. The difficulty seemed too much for many but the few who got through it really excelled. I may have been influenced by the level of the students at the selection school of a similar age. The notes are modified in 2019 to be more appealing to younger less experienced students.

Solution and working out process summarised from Kavan's presentation described below

**SPOILER**

Essentially a case bash to find N:

First reason that N cannot be 2 or 3 digits (this reasoning can be very precise because it will come in handy as part of the final proof). Trying out numbers will give that 1449 works. Now we prove this is the smallest.

Write proof below with the above in the back of ones mind but not counting towards the actual written answer:

1449 has digit sum 18. 2 times 1449 is 2898 with digit sum 27.

Now prove there is no smaller with this property:

Case1 N has 2 digits: must be 99 as use of smaller digits will not reach the sum of 18. However, 99 doubled is 198 fails to have digit sum 27.

Case2 N has 3 digits: 2N is less than 2000. If 2N<1000, then 2N=999 or its digit sum would be less than 27, which is not even as 2N must be. Therefore 2N must start with a 1, which means the other digits must be two 9s and one 8 to have digit sum 27 and since it is even must be 1998. But this would mean 999 is N which does not have digit sum 18. Hence N has 4 digits. (and starts with 1)

Now if N has 4 digits and not exceeding 1449 then 2N has 4 digits starting with 2 and therefore must end with either 7,9,9 or 8,8,9. To be even 2N=2988 or 2898, the latter being our answer with the first larger.

Can you see steps in this proof can be refined further by removing redundancies? Here is a summary only and details are omitted for clarity:

Step1 2N>=2000 as the only even number less than 2000 with digit sum 27 is 1998 but half of this does not have digit sum 18.

Step2 2N>= 2898 since only this along with 2988 are even with 4 digits, first digit 2 and digit sum 27.

Step3 2N=2898 since half of this is 1449 with digit sum 18.

Most of the laws in algebra stating familiar results but once these are stated it becomes clear that they come up many times in many similar ways. For example the rules for multiplication are almost the same as those for addition. One interesting law that relates the two (and removes their symmetry) is the distributative law: x(y+z)=xy+xz. Such a law was not really needed when numbers were involved with the values clearly stated because you would just add the numbers in the brackets first then multiply the result by the outside number. However when the numbers are represented by letters this is not often possible but we still need to re express the terms to better relate with other terms.

Going from left to right in this law is called EXPANDING. Going from right to left in this law is called FACTORISATION. Much of the first laws of algebra one learns falls under one of these two processes.

Associativity of an operation which operates on two inputs relates the ways in which the operation can operate on three inputs. The statement itself says that given three inputs, operating on the first two and using the result to operate with the third is the same as operating on the last two and then combining the result (as the second input) with the first input. This exercise and similar exercises show that this means that when adding three inputs it does not matter how the inputs are paired and added the answer will be the same.

Very fundamental part of geometry from which basically all results follow. A more relaxed version of congruence is similarity which allows the triangles to be proportional to each other rather than exactly the same. The conditions for similarity are the same but with S now denoting the pair of sides be in the same ratio as any other pair of corresponding sides, rather than being exactly equal.

A cyclic quadrilateral is one whose 4 vertices lie on a circle. We show here that such a quadrilateral must have its opposite angles adding up to 180 degrees.

The converse is also true but not examined here : anytime a quadrilateral has opposite angles adding up to 180, its vertices lie on a circle.