Problems

It is easy to construct one equilateral triangle using three identical matches. Is it possible to construct four equilateral triangles by adding just three more matches identical to the original ones?

Look at the following diagram, depicting how to get an extra cell by reshaping triangle.

Can you find a mistake? Certainly the triangles have different area, so we cannot obtain one from the other one by reshaping.

This problem is often called "The infinite chocolate bar". Depicted below is a way to get one more piece of chocolate from the \(5\times 6\) chocolate bar. Do you see where is it wrong?

Consider the following "proof" that any triangle is equilateral: Given a triangle \(ABC\), we first prove that \(AB = AC\). First let’s draw the bisector of the angle \(\angle A\). Now draw the perpendicular bisector of segment \(BC\), denote by \(D\) the middle of \(BC\) and by \(O\) the intersection of these lines. See the diagram

Draw the lines \(OR\) perpendicular to \(AB\) and \(OQ\) perpendicular to \(AC\). Draw lines \(OB\) and \(OC\). Then the triangles, \(RAO\) and \(QAO\) are equal, since we have equal angles \(\angle ORA = \angle OQA = 90°,\) and \(\angle RAO = \angle QAO,\) and the common side \(AO\). On the other hand the triangles \(ROB\) and \(QOC\) are also equal since the angles \(\angle BRO = \angle CQO = 90°\), the hypotenuses \(BO = OC\) the legs \(RO = OQ\). Thus, \(AR = AQ,\) \(RB = QC,\) and \(AB = AR + RB = AQ + QC = AC.\) Q.E.D.

As a corollary, one can show that all the triangles are equilateral, by showing that \(AB = BC\) in the same way.

Let’s prove that any \(90^{\circ}\) angle is equal to any angle larger than \(90^{\circ}\). On the diagram

We have the angle \(\angle ABC = 90^{\circ}\) and angle \(\angle BCD> 90^{\circ}\). We can choose a point \(D\) in such a way that the segments \(AB\) and \(CD\) are equal. Now find middles \(E\) and \(G\) of the segments \(BC\) and \(AD\) respectively and draw lines \(EF\) and \(FG\) perpendicular to \(BC\) and \(AD\).
Since \(EF\) is the middle perpendicular to \(BC\) the triangles \(BEF\) and \(CEF\) are equal which implies the equality of segments \(BF\) and \(CF\) and of angles \(\angle EBF = \angle ECF\), the same about the segments \(AF=FD\). By condition we have \(AB=CD\), thus the triangles \(ABF\) and \(CDF\) are equal, thus \(\angle ABF = \angle DCF\). But then we have \[\angle ABE = \angle ABF + \angle FBE = \angle DCF + \angle FCE = \angle DCE.\]

A equilateral triangle made of paper bends in a straight line so that one of the vertices falls on the opposite side as shown on the picture. Prove that the corresponding angles of the two white triangles are equal.

A square has been divided into \(4\) rectangles and a square. If the rectangle in the bottom left corner has dimensions \(1 \times 4\) and the one in the top right is \(2 \times 5\), what is the area of the small square in the middle?

In a convex quadrilateral \(ABCD\), all the triangles \(\triangle ABC\), \(\triangle BCD\), \(\triangle CDA\) and \(\triangle DAB\) have equal perimeters. Show that \(ABCD\) is a rectangle.

Consider two congruent triangles \(ABC\) and \(A_1B_1C_1\). We draw a point \(M\) on the side \(BC\) and a point \(M_1\) on the side \(B_1C_1\) such that the ratio of lengths \(BM:MC\) is equal to the ratio of lengths \(B_1M_1:M_1C_1\). Prove that \(AM = A_1M_1\).

We call a median the segment from the vertex of a triangle to the midpoint of the opposite side. Prove that in two congruent triangles, the corresponding medians are of equal length.