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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.

Prove the triangle inequality: in any triangle \(ABC\) the side \(AB < AC+ BC\).

In certain kingdom there are a lot of cities, it is known that all the distances between the cities are distinct. One morning one plane flew out of each city to the nearest city. Could it happen that in one city landed more than \(5\) planes?

Let \(A=\{1,2,3\}\) and \(B=\{2,4\}\) be two sets containing natural numbers. Find the sets: \(A\cup B\), \(A\cap B\), \(A-B\), \(B-A\).

Let \(A=\{1,2,3,4,5\}\) and \(B=\{2,4,5,7\}\) be two sets containing natural numbers. Find the sets: \(A\cup B\), \(A\cap B\), \(A-B\), \(B-A\).

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 \(A\cup B\).

image

Given three sets \(A,B,C\). Prove that if we take a union \(A\cup B\) and intersect it with the set \(C\), we will get the same set as if we took a union of \(A\cap C\) and \(B\cap C\). Essentially, prove that \((A\cup B)\cap C = (A\cap C)\cup (B\cap C)\).

\(A,B\) and \(C\) are three sets. Prove that if we take an intersection \(A\cap B\) and unite it with the set \(C\), we will get the same set as if we took an intersection of two unions \(A\cup C\) and \(B\cup C\). Essentially, prove that \((A\cap B)\cup C = (A\cup C)\cap (B\cup C)\). Draw a Venn diagram for the set \((A\cap B)\cup C\).

Let \(A,B\) and \(C\) be three sets. Prove that if we take an intersection \(A\cap B\) and intersect it with the set \(C\), we will get the same set as if we took an intersection of \(A\) with \(B\cap C\). Essentially, prove that it does not matter where to put the brackets in \((A\cap B)\cap C = A\cap (B\cap C)\). Draw a Venn diagram for the set \(A\cap B\cap C\).
Prove the same for the union \((A\cup B)\cup C = A\cup (B\cup C) = A\cup B\cup C\).

For three sets \(A,B,C\) prove that \(A - (B\cup C) = (A-B)\cap (A-C)\). Draw a Venn diagram for this set.