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We can obtain a two-fold symmetric figure by folding a square sheet of paper in the middle and then cutting along some curves drawn on the paper. By making two perpendicular folds through the centre of the paper and then cutting, we can obtain a four-fold symmetric figure. We can also get an eight-fold symmetric figure by making a fold bisecting an angle made by the two perpendicular folds before cutting. But it is not possible to obtain a three-fold, five-fold or six-fold symmetric figure in this way; we need to make more folds before cutting. Making a three-fold (respectively five-fold and six-fold) figure involves the division of the angle at the centre (360°) of a square sheet of a paper into six (respectively ten and twelve) equal parts. In other words, we need to construct the angles 60°, 36° and 30°. But these angles cannot be obtained by repeated bisections of 180° by simple folding as in the making of two-fold, four-fold and eight-fold figures. In [1], we see that each of the constructions of 60° and 30° applies the fact that sin 30° = ½ and takes only a few simple folding steps. The construction of 36° is more tedious (see, for example, [2] and [3]) as sin 36° is not a simple fraction but an irrational number. In this Article, we show how to make, by paper-folding and cutting a regular pentagon, a five-pointed star and create any five-fold figure as we want. The construction obtained by dividing the angle at the centre of a square paper into ten equal parts is called a pentagon base. We gained much insight from [2] and [3] when developing the method for making the pentagon base to be presented below.

Example 1Take a cardboard equilateral triangle. It is well known that this has six symmetries. To display these symmetries we colour the vertices of the triangle on both sides of the card with three different colours.

Pythagorean triples are triples of integers (a, b, c) satisfying the equation a2 + b2 = c2. For the purpose of this paper, we will take a, b and c to be positive, unless otherwise stated. Then, of course, it follows that a triple represents the lengths of sides of a right triangle. Also, for the purpose of this paper, we will consider the triples (a, b, c) and (b, a, c) to be distinct, even though they represent the same right triangle. A primitive Pythagorean triple is one for which a, b and c are relatively prime.

In a previous paper [1], I described a method for computing sums of powers via a matrix composed from Pascal's triangle. (The motivation for this came from a mathematical puzzle posed by Paul Maslanka in the Guardian.) Since then I have read or viewed a number of papers on sums of powers. Some were nearly four hundred years old and the most recent one was published this year. One rather obscure work I stumbled across was ‘On the sums of powers of natural numbers’ by the Japanese mathematician Mr T. Kariya, published in Tokyo in 1907. If any readers know more of Mr Kariya I would be glad to hear about him.

The main goal of this paper is to give a deeper understanding of the geometrical inequality proposed by Martin Lukarevski in [1]. In order to formulate our results we shall introduce and use the following notation throughout this paper. Let A1A2A3 be a triangle a1, a2, a3, the lengths of the sides opposite to A1, A2, A3 respectively, P an arbitrary inner point of it xi, the distance of P from the side of length ai. Let r, R be the inradius and circumradius of the triangle hi, the altitude belonging to side ai, Δ the area and finally let α be a real parameter. We adopt also the use of Σui to refer the sum taken over the suffices i = 1, 2, 3. Now we are in the position to reformulate the original problem into a more general form namely: find bounds for Σxαi in terms of r and R. The main results of our investigation are summarised in the following theorem.

Tangential quadrilaterals are defined to be quadrilaterals in which a circle can be inscribed that is tangent to all four sides. It is well known and easy to prove that a convex quadrilateral is tangential if, and only if, the angle bisectors of all four vertex angles are concurrent at a point, which is the centre of the inscribed circle (incircle). The most well-known and in problem solving useful characterisation of tangential quadrilaterals is Pitot's theorem, which states that a convex quadrilateral is tangential if and only if its consecutive sides a, b, c, d satisfy the relation a + c = b + d [1, pp. 64-67]. If you want to have more background information about characterisations of tangential quadrilaterals, then we recommend you to check out the lovely papers [2, 3, 4], as well as our previous contributions on the subject [5, 6, 7]. These six papers together include about 30 characterisations that are either proved or reviewed there with references.

The general class of quadrilaterals where one diagonal is bisected by the other diagonal has appeared very rarely in the geometrical literature, but they have been named several times in connection with quadrilateral classifications. Günter Graumann strangely gave these objects two different names in [1, pp. 192, 194]: sloping-kite and sliding-kite. A. Ramachandran called them slant kites in [2, p. 54] and Michael de Villiers called them bisecting quadrilaterals in [3, pp. 19, 206]. The latter is a pretty good name, although a bit confusing: what exactly is bisected?We have found no papers and only two books where any theorems on such quadrilaterals are studied. In each of the books, one necessary and sufficient condition for such quadrilaterals is proved (see Theorem 1 and 2 in the next section). The purpose of this paper is to investigate basic properties of convex bisecting quadrilaterals, but we have chosen to give them a slightly different name. Let us first remind the reader that a quadrilateral whose diagonals have equal lengths is called an equidiagonal quadrilateral and one whose diagonals are perpendicular is called an orthodiagonal quadrilateral.