The Theorem of Pythagoras

Pythagorean Theorem

If we make four copies of a right-angled triangle with shorter sides x and y and longer side r, we can arrange them in the corners of a square of side x + y, leaving an area r² uncovered, as in the first diagram here. Alternatively we can arrange them in pairs forming rectangles x by y, leaving square areas of sizes x² and y² uncovered, as in the second diagram. It is evident that the areas uncovered in the two diagrams must be equal, that is r² = x² + y².
This is probably the simplest proof of the most well known proposition in geometry, namely the Theorem of Pythagoras, that in a right-angled triangle the square of the side (known unnecessarily as the hypotenuse) opposite the right angle is equal to the sum of the squares of the other two sides.

Rectangular Coordinates

Since a right-angled triangle with sides x, y and r is half a rectangle with sides x and y the Pythagorean formula also enables one to calculate the length of the diagonal of a rectangle from the lengths of its sides.

This is the basis for rectangular coordinates in which the position of a point P is specified by the sides x and y of a rectangle whose diagonal is OP = r and whose sides are parallel to a fixed pair of unit perpendicular lines OX and OY. The point O (called the origin of coordinates) has coordinates (0, 0) and P is (x, y). The length r = √(x² + y²) being the distance of P from O. The coordinates x and y can also be described as the distances of P from the extended lines OY and OX, as shown by the perpendiculars PN and PM respectively, or as the distances OM and ON from O to the projections M and N of P onto the lines OX and OY.

Another way of proving the Pythagorean theorem is to draw the perpendicular MK from the right-angled vertex to the long side, and note that the two triangles that result are similar to the whole triangle. By similarity of OMP and OKM we have: OM/OP = OK/OM whence OK.OP = OM². By similarity of OMP and MKP we have MP/OP = KP/MP whence KP.OP = MP². And since OP = OK + KP we have OP² = OK.OP + KP.OP = OM² + MP².

The Pythagorean formula also leads us to a more general formula to calculate the distance between two points P = (x, y) and P' = (x', y') in the form
PP' = √[(x' – x)² + (y' – y)²].

Pythagorean Dissection

This pattern shows the most symmetrical way to dissect two squares into five pieces that will fit together to form a single square. The pattern is formed by superimposing a tessellation of alternating squares of two sizes with a tessellation of the larger squares.
The larger squares are arranged so their sides are parallel to the third side of the right angled triangle formed by the sides of the smaller squares. The relative positions of the two grids are arranged so that the smallest square is centred in the largest square, and the points where four of the largest squares meet are at the centres of the intermediate sized squares. The points where the sides of the intermediate squares cross the sides of the largest squares bisect the sides of the larger squares.

This constructional proof of the Pythagorean theorem is due to Henry Perigal 1830 [Messenger of Mathematics, new series, vol.ii, pp.103-106]

Pythagorean Triples

The earliest knowledge of the theorem of Pythagoras probably came from the observation that a triangle whose sides are 3, 4 and 5 units in length forms a right-angled triangle, together with the observation that 3² + 4² = 5².

Any three whole numbers a, b, c which satisfy the rule a² + b² = c² are called Pythagorean triplets. They form two endless sequences: (a) 3, 4, 5; 5, 12, 13; 7, 24, 25; 9, 40, 41; ... (b) 8, 15, 17; 12, 35, 37; 16, 63, 65;... Babylonian tablets containing list of Pythagorean triples have been discovered which date from 1600bc or earlier.

A Square Root Construction

Beginning with a unit line P0-P1 draw another unit line at right angles to P2 then the distance P0-P2 will be √2.

Now draw a unit line from P2 at right angles to P0-P2 to reach P3. Then P0-P3 = √3.

Continuing this process from P(n−1) we draw a unit line at right angles to P0-P(n−1) and reach Pn and by Pythagoras P0-Pn = √(1² + (√(n − 1))²) = √(1 + n − 1) = √n. In a plane the unit lines form a spiral pattern.

Alternatively, the perpendiculars can be regarded as being in an extra dimension. This we get √2 as the diagonal of a unit square, √3 as the diagonal of a unit cube, √4 = 2 the diagonal of a hypercube, and so on.