Example 1: Two triangles have sides of lengths 4 cm, 5 cm, and 6 cm. Are they congruent?
Solution: Yes, if both triangles have corresponding sides of these lengths, they are congruent by the SSS (Side-Side-Side) rule.
Example 2: Two triangles have two equal angles and the included side is equal. Are they congruent?
Solution: Yes, the triangles are congruent by the ASA (Angle-Side-Angle) rule.
The triangle inequality theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. The Midpoint Theorem (MPT) states that the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length. The Basic Proportionality Theorem (BPT) states that a line parallel to one side of a triangle divides the other two sides proportionally.
Example 1: Can a triangle have sides of lengths 4 cm, 5 cm, and 10 cm?
Solution: No, because 4+5=9, which is less than 10. Therefore, the sides do not form a triangle.
Example 2: In a triangle, a line parallel to one side divides the other two sides into segments of lengths 3 cm and 4 cm. What is the ratio of the divided sides?
Solution : By the Basic Proportionality Theorem, the sides are divided in the ratio 3:4.
The incenter is the center of the incircle (a circle inscribed in the triangle), while the circumcentre is the center of the circumcircle (a circle passing through all three vertices).
Example 1: Where is the circumcenter located in a right triangle?
Solution: The circumcenter is located at the midpoint of the hypotenuse.
Example 2: What is the incenter of a triangle?
Solution : The incenter is the point where the angle bisectors of a triangle intersect.
Colouring techniques are often used in graph theory or combinatronics to solve problems related to partitions, graph colourings, or pattern counting.
Example 1: In how many ways can the sides of a regular octagon be coloured using 3 different colours if adjacent sides cannot have the same colour?
Solution: This is a graph colouring problem, and using the chromatic polynomial gives the number of colourings as P=3(3−1)8−1=3×27=384P = 3(3-1)^{8-1} = 3 \times 2^7 = 384P=3(3−1)8−1=3×27=384.
Example 2: In how many ways can 4 rooms in a house be painted using 5 different colours if no two adjacent rooms can have the same colour?
Solution: This is a graph colouring problem, and depending on the adjacency of the rooms, the chromatic polynomial is used to count the valid colourings.
Parity refers to whether a number is even or odd, while invariance refers to a quantity that remains unchanged during a process.
Example 1: If 20 people are seated around a circular table, prove that there must be two people sitting next to each other who are either both men or both women.
Solution: By parity, if there are an odd number of men and women, there must be at least one pair of the same gender.
Example 2: Show that in a game where you flip a coin 100 times, there must be a run of consecutive heads or tails of length 6 or more.
Solution: By invariance, due to the law of large numbers, in a random sequence of 100 flips, a long run of consecutive heads or tails is highly probable.
This section involves complex combinatorics problems that require a combination of techniques like inclusion-exclusion, Pigeonhole Principle, and more.
Example 1: How many ways can 5 indistinguishable objects be placed into 3 distinct boxes such that no box is empty?
Solution: Use the inclusion-exclusion principle to account for the restriction that no box is empty. The number of ways is 6.
Example 2: In how many ways can 7 different books be divided among 3 students such that each student gets at least one book?
Solution: Using inclusion-exclusion, there are 540540540 ways to divide the books.