Problems
Let \(ABCD\) be a parallelogram. The segment \(EF\) is parallel to the diagonal \(BD\), and the segment \(EG\) is parallel to the diagonal \(AC\). Show that the areas of the triangles \(\triangle EFD\) and \(\triangle EGC\) are equal.
Today we will solve some problems about finding areas of geometric figures. All you need to know in order to solve every problem in this set is: to calculate the area of a triangle we multiply the length of a side by the length of a height to that side and divide by \(2\), namely:\(\frac12 AB \times CD\), as for rectangle we just multiply two adjacent sides (\(EF \times GF\)), and when we have a circle we calculate the area by \(\pi r^2\), where \(r\) is the radius of the circle.
In a trapezium \(ABCD\), the side \(AB\) is parallel to the side \(CD\). Prove that the areas of triangles \(\triangle ABC\) and \(\triangle ABD\) are equal.
On the left there is a circle inscribed in a square with side \(1\). On the right there are \(16\) smaller, identical circles, which all together fit inside a square of side \(1\). Which area is greater, the yellow or the blue one?
A circle is inscribed in a square, and another square is inscribed in the circle. Which area is larger, the blue or the orange one?
In a square, the midpoints of its sides were marked and connected to the vertices of the square. There is another square formed in the centre. The side length of the large square is \(10\). Find the area of the smaller square. (That is, the red one)
In a parallelogram \(ABCD\), point \(E\) belongs to the side \(AB\), point \(F\) belongs to the side \(CD\) and point \(G\) belongs to the side \(AD\). We know that the marked red segments \(AE\) and \(CF\) have equal lengths. Prove that the total grey area is equal to the total black area.
In a regular hexagon of area \(72\), some diagonals were drawn. Find the area of the red region.
Three semicircles are drawn on the sides of the triangle \(ABC\) with sides \(AB=3\), \(AC=4\), \(BC=5\) as diameters. Find the area of the red part.
Today we will be solving problems using the pigeonhole principle. What is it? Simply put, we are asked to place pigeons in pigeonholes, but the number of pigeons is larger than the number of pigeonholes. No matter how we try to do that, at least one pigeonhole will have to contain at least 2 pigeons. By “pigeonholes" we can mean any containers and by “pigeons" we mean any items which are placed in these containers. This is a simple observation, but it is helpful in solving some very difficult problems. Some of these problems might seem obvious or intuitively true. Pigeonhole principle is a useful way of formalising things that seem intuitive but can be difficult to describe mathematically.
There is also a more general version of the pigeonhole principle, where the number of pigeons is more than \(k\) times larger than the number of pigeonholes. Then, by the same logic, there will be one pigeonhole containing \(k+1\) pigeons or more.
A formal way to prove the pigeonhole principle is by contradiction - imagine what would happen if each pigeonhole contained only one pigeon. Well, the total number of pigeons could not be larger than the number of pigeonholes! What if each pigeonhole had \(k\) pigeons or fewer? The total number of pigeons could be \(k\) times larger than the number of pigeonholes, but not greater than that.
