Problems

A chord of a circle is a straight line between two points on the circumference of the circle. Is it possible to draw five chords on a circle in such a way that there is a pentagon and two quadrilaterals among the parts into which these chords divide the circle?

We say that a figure is convex if a segment connecting any two points lays fully within the figure. On the picture below the pentagon on the left is convex and the one on the right is not.

Is it possible to draw \(18\) points inside a convex pentagon so that each of the ten triangles formed by its sides and diagonals contains equal amount of points?

Is it possible to cut an equilateral triangle into three equal hexagons?

• We call two figures congruent if their corresponding sides and angles are equal. Let \(ABD\) an \(A'B'D'\) be two right-angled triangles with right angle \(D\). Then if \(AD=A'D'\) and \(AB=A'B'\) then the triangles \(ABD\) and \(A'B'D'\) are congruent.

• It follows from the previous statement that if two lines \(AB\) and \(CD\) are parallel than angles \(BCD\) and \(CBA\) are equal.

We prove the other two assertions from the description:

• The sum of all internal angles of a triangle is also \(180^{\circ}\).

• In an isosceles triangle (which has two sides of equal lengths), two angles touching the third side are equal.

In the triangle \(ABC\) the sides are compared as following: \(AC>BC>AB\). Prove that the angles are compared as follows: \(\angle B > \angle A > \angle C\).

Consider a quadrilateral \(ABCD\). Choose a point \(E\) on side \(AB\). A line parallel to the diagonal \(AC\) is drawn through \(E\) and meets \(BC\) at \(F\). Then a line parallel to the other diagonal \(BD\) is drawn through \(F\) and meets \(CD\) at \(G\). And then a line parallel to the first diagonal \(AC\) is drawn through \(G\) and meets \(DA\) at \(H\). Prove the \(EH\) is parallel to the diagonal \(BD\).

Cut an arbitrary triangle into parts that can be used to build a triangle that is symmetrical to the original triangle with respect to some straight line (the pieces cannot be inverted, they can only be rotated on the plane).

On the diagram below \(AD\) is the bisector of the triangle \(ABC\). The point \(E\) lies on the side \(AB\), with \(AE = ED\). Prove that the lines \(AC\) and \(DE\) are parallel.

On the diagram below the line \(BD\) is the bisector of the angle \(\angle ABC\) in the triangle \(ABC\). A line through the vertex \(C\) parallel to the line \(BD\) intersects the continuation of the side \(AB\) at the point \(E\). Find the angles of the triangle \(BCE\) triangle if \(\angle ABC = 110^{\circ}\).