Problems
From the examples above, we see that we often need to pick \(k\) objects from \(n\) objects where the order of the \(k\) objects is ignored. The number of ways to pick them is notated with the special symbol \(\binom{n}{k}\), pronounced “\(n\) choose \(k\)". Following a similar line of reasoning as the examples, we can write down a general formula:
\[\binom{n}{k} = \frac{n\times (n-1) \times \dots (n-k+1)}{k\times (k-1) \times \dots 1} = \frac{n!}{k!(n-k)!}.\]
\(n!\) is a shorthand for \(n\times (n-1)\times \dots \times 1\), pronounced “\(n\) factorial". This is another useful expression and allows us to write down many formulas succinctly.
Today we will solve some geometric problems using the triangle inequality. This is an inequality between the lengths of the sides of any triangle, or between the distances of any three points.
The shortest path between any two points \(A\) and \(B\) is a straight segment - every other path is longer. In particular, a path through another point, \(C\), is equal or longer. \[AC + BC \ge AB\] The triangle inequality says that the sum of lengths of any two sides of a triangle is always larger than the length of the third side. The inequality only becomes an equality if \(ABC\) is not actually a triangle and the point \(C\) lies on the segment from \(A\) to \(B\).
Even though it is a simple idea, it can be a really helpful tool in problem solving.
Today we will solve some logic problems. This time, we are visiting a strange planet. This planet is inhabited by two kinds of aliens, Cricks and Goops. The physical differences between them are not enough for a human being to distinguish them, but they have another remarkable feature. They can only ask questions. Cricks can only ask questions whose answer is yes, while Goops can only ask questions whose answer is no.
A set is a collection of objects of any specified kind, the objects are called elements or members, the objects in one set cannot repeat, namely \(\{1,2,3\}\) and \(\{1,2,2,2,3\}\) are identical sets. We denote a set by a capital letter \(A\), or \(B\) and write \(x\in A\) if \(x\) is an element of \(A\), and \(x\notin A\) if it is not. The notation \(A=\{a,b,c,...\}\) means that the set \(A\) consists of the elements \(a,b,c,...\). The empty or void set, \(\emptyset\), has no elements. If all elements of \(A\) are also in \(B\), then we call \(A\) a subset of \(B\) and we write \(A\subseteq B\). It is an axiom that the sets \(A\) and \(B\) are equal \(A=B\) if they have the same elements. Namely, \(A\) is a subset of \(B\) and \(B\) is a subset of \(A\) at the same time.
For any sets \(A\) and \(B\), we define their union \(A\cup B\), intersection \(A\cap B\), and the difference \(A-B\) as follows:
the union \(A\cup B\) is the set of all elements that belong to \(A\) or \(B\);
the intersection \(A\cap B\) is the set of elements that belong to both \(A\) and \(B\);
the difference \(A-B\) consists of those \(x \in A\) that are do not belong to \(B\).
Sometimes it is useful to draw sets as Venn diagrams, on the diagram below the pink circle represents the set \(A\), the yellow circle represents the set \(B\), the orange part is the intersection \(A\cap B\), the pink part is \(A-B\), the yellow part is \(B-A\), and the whole picture is the union \(A\cup B\).
There are various ways to prove mathematical statements. One of the possible methods which might come in handy in certain situations is called proof by contradiction. To prove a statement we first assume that the statement is false and then deduce something that contradicts either the condition, or the assumption itself, or just common sense. Due to the contradiction, we have to conclude that the first assumption must have been wrong, so the statement is actually true.
A closely related method is called contrapositive proof. An example should make the idea quite clear. Consider the statement “if the joke is funny, then I will be laughing". Another completely equivalent way of saying it would be “if I am not laughing, then the joke is not funny". The second statement is known as the contrapositive of the first statement.
We can often prove a statement by proving its contrapositive. Many statements are proven by deriving a contradiction. However, one can often rewrite them as either a direct proof or a contrapositive proof.
Let’s take a look at both of these techniques.
Let’s play some games today! We will play a classic game known as nim, which is thought to be one of the oldest games.
Typically people play nim using matchsticks, though stones and coins are popular too. There are a few heaps of matchsticks in nim. Players take turns to remove matchsticks from a heap of their choosing. The player can remove any number of matchsticks they wish from that heap. Whoever has no matchsticks left to take loses.
This following position will be written as \(\text{Nim}(3,3,3)\):
As another example, this is \(\text{Nim}(1,2,3,4)\):
We will omit heaps of size zero, so \(\text{Nim}(3,0,3,0,3)\) is the same as \(\text{Nim}(3,3,3)\).
Nim is important because a large class of games are equivalent to it despite its simple appearance. The interested reader should look up "Sprague-Grundy Theorem".
Let us introduce a few terms that will be helpful for analyzing games. A game \(G\) consists of some positions and a set of rules. A position \(g\) in the game \(G\) is called a winning position if the player starting this turn has a winning strategy. This means as long as the player starting this turn continues to play optimally, the second player has to lose. Conversely, a position \(g\) is a losing position if the player starting this turn has no winning strategy.
A useful common problem-solving strategy is to divide a problem into cases. We can divide the problem into familiar and unfamiliar cases; easy and difficult cases; typical and extreme cases etc. The division is sometimes suggested by the problem, but oftentimes requires a bit of work first.
If you are stuck on a problem or you are not sure where to begin, gathering data by trying out easy or typical cases first might help you with the following (this list is not exhaustive):
Gaining intuition of the problem
Isolating the difficulties
Quantifying progress on the problem
Setting up or completing inductive arguments
Let us take a look at this strategy in action.
In an \(n\times n\) table, two opposite corner squares are black and the rest are white. We wish to turn the whole \(n\times n\) table black in two stages. In the first stage, we paint black some of the squares that are white at the moment. In the second stage, we can perform the following two operations as much as we like. The row operation is to swap the colours of all the squares in a particular row. The column operation is to swap the colours of all the squares in a particular column. What is the fewest number of white squares that we can paint in the first stage?
An example of the row operation: let W stand for white and B stand for black and suppose that \(n=5\). Also suppose that a particular row has the colours WWBWB. Then performing the row operation would change this row to BBWBW.
We wish to paint the \(15\) segments in the picture below in three colours. We want it such that no two segments of the same colour have a common end. For example, you cannot have both \(AB\) and \(BC\) blue since they share the end \(B\). Is such a painting possible?
A circle is inscribed into the triangle \(ABC\) with sides \(BC=6, AC=10\) and \(AB= 12\). A line tangent to the circle intersects two longer sides of the triangle \(AB\) and \(AC\) at the points \(F\) and \(G\) respectively. Find the perimeter of the triangle \(AFG\).
