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Assumed Knowledge
Motivation Is there a simple relationship between the length of the sides of a triangle? Apart from the fact that the sum of any two sides is greater than the third, there is, in general, no simple relationship between the three sides of a triangle. Among the set of all triangles, there is a special class, known as right-angled triangles or right triangles that contain a right angle. The longest side in a right-angled triangle is called the hypotenuse. The word is connected with a Greek word meaning to stretch because the ancient Egyptians discovered that if you take a piece of rope, mark off 3 units, then 4 units and then 5 units, this can be stretched to form a triangle that contains a right angle. This was very useful to the Egyptian builders. This raises all sorts of questions. What is so special about the lengths 3, 4 and 5? Are there other sets of numbers with this property? Is there a simple relationship between the lengths of the sides in a right-angled triangle? Given the lengths of the sides of a triangle, can we tell whether or not the triangle is right angled? Most adults remember the mathematical formula c2 = a2 + b2 or perhaps “the square on the hypotenuse is the sum of the squares on the other two sides.” The first version uses an implied standard notation, the second version uses archaic language but both are Pythagoras’ theorem. This theorem enables us to answer the questions raised in the previous paragraph. The discovery of Pythagoras’ theorem led the Greeks to prove the existence of numbers that could not be expressed as rational numbers. For example, taking the two shorter sides of a right triangle to be 1 and 1, we are led to a hypotenuse of length  Triples of integers such as (3, 4, 5) and (5, 12, 13) which occur as the side lengths of right-angled triangles are of great interest in both geometry and number theory − they are called Pythagorean triples. We find all of them in this module. Pythagoras’ theorem is used in determining the distance between two points in both Pythagoras’ theorem can be generalised to the cosine rule and used to establish Heron’s formula for the area of a triangle. Both of these are discussed in the Links Forward section. CONTENT Standard notation Let ABC be a triangle. We may write Then by convention, a is length of the interval BC. We also talk about angle A or So a is the length of the side opposite angle A. Using this notation we can succinctly state Pythagoras’ theorem and two of the most important theorems in trigonometry, the sine rule and the cosine rule. The sine rule and cosine rule are established in the Links Forward section. Right-angled triangles
We also talk about the shorter sides of a right-angled triangle.
If a and b are fixed then c is determined.
The theorem A triangle with sides 3 cm, 4 cm, 5 cm is a right-angled triangle. Similarly, if we draw a right-angled triangle with shorter sides 5 cm, 12 cm and measure the third side, we find that the hypotenuse has length ‘close to’ 13 cm. To understand the key idea behind Pythagoras’ theorem, we need to look at the squares of these numbers. You can see that in a 3, 4, 5 triangle, 9 + 16 = 25 or 32 + 42 = 52 and in the 5, 12, 13 triangle, 25 + 144 = 169 or 52 + 122 = 132. We state Pythagoras’ theorem:
EXAMPLE
Solution Let x be the length of the hypotenuse. Then by Pythagoras’ theorem, x2 = 122+ 162 = 400. So x = 20. Proof of the theorem A mathematical theorem is a logical statement, ‘If p then q’ where p and q are clauses involving mathematical ideas. The converse of ‘If p then q’ is the statement, ‘If q then p’. Converse of Pythagoras’ theorem: If c2 = a2 + b2 then There are many proofs of Pythagoras’ theorem. Proof 1 of Pythagoras’ theorem For ease of presentation let
From the left hand diagram (a + b)2 = a2 + b2 + 4 From the right hand diagram (a + b)2 = 4 Comparing the two equations we obtain c2 = a2 + b2 and the theorem is proved. Several other proofs of Pythagoras’ theorem are given in the Appendix.
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EXERCISE 1 Find the hypotenuse of the right angled triangles whose other sides are: a 5, 12 b 9, 12 c 35, 12 d 15, 8 e 15, 20 f 15, 112 Note: Clearly one can use a calculator and reduce each of the above calculations to half a dozen keystrokes. This leads to no insights at all. As a suggestion, if a perfect square is between 4900 and 6400 then the number is between 70 and 80. If the last digit of the square is 1 then the number ends in a 1 or a 9, etc. Applications of Pythagoras’ theorem
EXAMPLE
Solution Let b be the width, measured in cm. Then
The width of the rectangle is 15 cm.
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EXERCISE 2
The Converse theorem We now come to the question: Given the lengths of the sides of a triangle, can we tell whether or not the This is answered using the converse of Pythagoras’ theorem. The converse theorem says: If a2 + b2 = c2 then the triangle is right angled (with right angle at C). Thus, for example, a triangle with sides 20, 21, 29 is right angled since
The obvious question, which we shall answer later in this module, is can we find all such ‘Pythagorean triples of whole numbers’? We shall give two proofs of the converse − rather different in nature. However, both use the theorem itself in the proof! This does not often happen in elementary mathematics but is quite common in more advanced topics. First proof of the converse
Then, by Pythagoras’ theorem, But c2 = a2 + b2 so ED = c. Hence
So, Second proof of the converse
Clearly x < a and y < b so
This is a contradiction, so C = D or D is to the left of C on the line AC.
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EXERCISE 3 Work out the details of the proof when D is to the left of C on the line AC.
EXAMPLE Which of the triangles below are right-angled triangles?
Solution a Triangle ABC is not a right-angled triangle since 142 + 272 ≠ 322. b Triangle DEF is a right-angled triangle since 152 + 362 = 392. Irrational numbers Consider the sequence
This sequence of positive real numbers is strictly increasing and A right-angled triangle with equal side lengths is an isosceles triangle. Hence the angles are 45°, 45° and 90°. If the length of a side is 1 then the hypotenuse is of length Next we consider the right-angled triangle with shorter sides 1 and
Using the above constructions it follows that a length where n is a whole number greater than 1, can be constructed using just ruler and compass (see module, Constructions). Since 1 < When irrational numbers occur in problems involving Pythagoras’ theorem, we can either
EXAMPLE Find the length, correct to 2 decimal places, solution By Pythagoras’ theorem,
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EXERCISE 4 A cross-country runner runs 3km west, then 2km south and then 8km east. How far is she from her starting point? Give your answer in kilometres and correct to 2 decimal places.
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EXERCISE 5 Find the exact length of the long diagonal in a cube of side length 3 cm. Pythagorean triads Three whole numbers that are the lengths of the sides of a right-angled triangle are called a Pythagorean Triad or Pythagorean Triple. Thus, {3, 4, 5} is a Pythagorean Triad. The formula for how to generate such triples was known by about 2000BC. This is “proved” by a clay tablet (Plimpton 322) which contains fifteen different triples including (1679, 2400, 2929) The tablet is dated to 1800 BC. With your calculator check this is a Pythagorean triple. This was obviously not found by chance! Starting with (3, 4, 5) we can find or construct infinitely many such triples by taking integer multiples: (3, 4, 5), (6, 8, 10), (9, 12, 15), ….. Consider a triple (a, b, c) of positive whole numbers with a2 + b2 = c2. If a and b have a common factor then it also divides c. So a useful definition is that the Pythagorean triple (a, b, c) is primitive if, HCF (a, b) = HCF (b, c) = HCF (a, c) = 1 that is, the highest common factor of a and b is 1, etc. If we can find all primitive Pythagorean triples then we can find all triples by simply taking whole number multiples of the primitive triples. There are various families of examples. Consider the identity: (n + 1)2 − n2 = 2n + 1 So if 2n + 1 is a perfect square then we can construct a primitive triple ( (3, 4, 5); (5, 12, 13); (7, 24, 25); (9, 40, 41),…. It is possible to list all primitive triples. One form of this ‘classification’ is in the following theorem. We shall prove it using some elementary number theory including the use of the fundamental theorem of arithmetic and the use of the HCF. The symbol | is used for ‘divides exactly into’. The result gives a formula for all primitive Pythagorean Triads. Theorem If a2 + b2 = c2 and (a, b, c) is a primitive triad then a = p2 − q2, b = 2pq and c = p2 + q2 where the HCF of p and q is 1 and p and q are not both odd. Proof At least one of a, b and c is odd since the triad is primitive. The square of a whole number is either a multiple of 4 or one more than a multiple So we may assume a is odd, b is even and c is odd. c2 = a2 + b2 so b2 = c2 − a2 = (c − a)(c + a) Let d be the HCF of c − a and c + a, so d|c −a and d|c + a so d|2c and d|2a But a and c are coprime so d = 1 or 2 but c − a and c + a are even. So d = 2 Hence we have
Set Finally if p and q are odd then a and c are even which is not the case.
EXAMPLE 22 − 12 = 3, 2 × 2 × 1 = 4 and 22 + 12 = 5 so the triple (3, 4, 5) corresponds to (p, q) = (2, 1)
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EXERCISE 6 Investigate: a (p, q) = (2, 1), (3, 2), (4, 3), ….. b (p, q) = (4, 1), (5, 2), (6, 3), (7, 4), ….. c (p, q) = (2, 1), (4, 1), (6, 1), (8, 1), …..
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EXERCISE 7 Find the values of p and q corresponding to the triple (1679, 2400, 2929) from the clay tablet Plimpton 322. The largest triple on Plimpton 322 is (12 709, 13 500, 18 541) − find p and q in this case as well. LINKS FORWARD The distance formula in R2 (The coordinate plane)
Suppose that A(x1, y1) and B(x2, y2) are two points Consider the right-angled triangle AXB where
depending on the relative positions of A and B. By Pythagoras’ theorem
Therefore AB = BA = Distances in three-dimensional space The distance d from (0, 0) to (x, y) in the coordinate plane satisfies d2 = x2 + y2.
It is possible to go from O to any point P by ‘moving’ We say the coordinates of the point P are (a, b, c). Again a basic question is ‘What is the distance OP?’ The answer is OP2 = a2 + b2 + c2. To see this, let A = (a, 0, 0) and B = (a, b, 0). The triangle OAB is right-angled at A. Hence OB2 = OA2 + AB2 = a2 + b2. Next OBP is right−angled at B so OP2 = OB2 + BP2 = a2 + b2 + c2. Circles in the plane, centre the origin A circle is the path traced out by a point moving a fixed distance from a fixed point First suppose we draw a circle in the Cartesian plane centre the origin and radius 1
Then by Pythagoras’ theorem (or the distance formula in R2) x2 + y2 = 12 Conversely, if x2 + y2 = 1 then the point (x, y) lies on the circle of radius 1. Similarly, if x2 + y2 = r2 then (x, y) lies on the circle, centre the origin, of radius r and conversely all points on this circle satisfy the equation. Pythagoras’ theorem in trigonometry
Now by definition cosθ = By Pythagoras’ theorem, a2 + b2 = c2. Therefore Hence, one of the most fundamental identities in trigonometry. cos2 θ + sin2θ = 1 There are 2 important formulas linking the side lengths Theorem − The cosine rule
Then a2 = b2 + c2 − 2bc cos A Proof
By Pythagoras’ theorem b2 = h2 + x2 and a2 = h2 + y2 Also c = x + y so, eliminating h2
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EXERCISE 8 Show that the cosine rule is still true when Theorem − The sine rule
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EXERCISE 9 Write down expressions for sin A and sin B and hence prove the sine rule.
Apollonius’ theorem and Heron’s formula Euclid’s Elements was written about 300BC. As discussed elsewhere in these modules this amazing set of thirteen books collected together most of the geometry and number theory known at that time. During the next century Apollonius and Archimedes developed mathematics considerably. Apollonius is best remembered for his study of ellipses, parabolas and hyperbolas. Archimedes is often ranked as one of the most important mathematicians of all time. He carried out a number of calculations, which anticipated ideas from integral calculus. In this section we discuss Heron’s formula that scholars believe was discovered by Pythagoras. We shall prove Apollonius’ theorem and Heron’s formula which both follow from Pythagoras’ theorem using algebra. Apollonius’ theorem Suppose b2 + c2 = 2x2 + 2m2 Proof Let length of the altitude AE be h. Also let CE = t, ED = s.
Clearly s + t = x and there are three right-angled triangles so m2 = h2 + s2 c2 = h2 + (s + x)2 b2 = h2 + t2 Taking into account the formula to be proved we consider
In the above diagram we asumed then
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EXERCISE 10 Use the cosine rule to write m2 in Deduce Apollonius’ theorem in a couple of steps. Heron’s formula This is an amazing formula expressing the area of a triangle in terms of its side lengths. To write this in its standard form consider a
We give a proof that uses only Pythagoras’ theorem, the formula for the area of a triangle and some algebra. Proof
Then x + y = c. The area of a triangle is half base times height so
Pythagoras’ theorem gives a2 = h2 + x2 (2) b2 = h2 + y2 = h2 + (c − x)2 (3) We must eliminate h and x from equations (1), (2) and (3). This is non-trivial!
Next (2) gives substitute into (1)
Square (4) and substitute into (5)
This is the difference of two squares, so
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EXERCISE 11 Find the areas of the triangles with side lengths: a 13, 14, 15 b 13, 20, 21 c 10, 17, 21 d 51, 52, 53 HISTORY As outlined above, the theorem, named after the sixth century BC Greek philosopher and mathematician Pythagoras, is arguably the most important elementary theorem in mathematics, since its consequences and generalisations have wide ranging applications. It is often difficult to determine via historical sources how long certain facts have been known. However, in the case of Pythagoras’ theorem there is a Babylonian tablet, known as Plimpton 322, that dates from about 1700BC. This tablet lists fifteen Pythagorean triples including (3, 4, 5), (28, 45, 53) and (65, 72, 97). It does not include (5, 12, 13) or (8, 15, 17) but it does include (12 709, 13 500, 18 541)! The fifteen triples correspond (very roughly) to angles between 30° and 45° in the right-angled triangle. The Babylonian number system is base 60 and all of the even sides are of the form 2a 3b 5c presumably to facilitate calculations in base 60. Most historical documents are found as fragments and one could call this the Rosetta Stone of mathematics. Whichever interpretation of the purpose of Plimpton 322 is correct, and there are several, it is clear that both Pythagoras’ theorem and how to construct Pythagorean Triples was known well before 1700BC. The nature of mathematics began to change about 600 BC. This was closely linked to the rise of the Greek city states. There was constant trade and hence ideas spread freely from the earlier civilisations of Egypt and Babylonia. Most of the history is lost forever, but tradition has it the Thales, Pythagoras and their students, were responsible for developing many of the key ideas − in particular the need to prove theorems! What we do know is what was known at about 300BC. This is because Euclid of Alexandria wrote his thirteen volume book the Elements. Contrary to popular belief, this book is by no means solely about geometry. Book 1 of the Elements is on geometry and attempts to set geometry on a sound logical basis by giving some twenty-three definitions and lists five postulates and five common notions. This axiomatic approach, although flawed and incomplete gave a logical approach to the study of geometry which was a central part of a classical education right up to the twentieth century. The imperfections of Euclid were not fixed until 1900 when David Hilbert gave a modern correct system of axioms. In Book 1 of Euclid a number of theorems are proved such as the well-known result that in an isosceles triangle the base angles are equal. The final theorem, Proposition 1-47, is Pythagoras’ theorem. The proof given is not the easiest known at the time, but uses only congruence and other results proved in Book 1. Euclid’s Elements are very sophisticated. APPENDIX Other proofs of Pythagoras’ theorem There are hundreds of proofs of Pythagoras’ theorem − We shall present a few more including Euclid’s proof. Second proof
The large square is a + b by a + b so (a + b)2 = 4 a2 + 2ab + b2 = 4 or c2 = a2 + b2. Third proof
The side length of the inner square is b − a Hence (b − a)2 + 4 a2 − 2ab + b2 + 2ab = c2 a2 + b2 = c2 and we have another proof of Pythagoras’ theorem. The first three proofs are essentially based on congruence of triangles, partially disguised as sums of areas. Some proofs use similarity. One of the nicest or perhaps minimalist proofs comes from considering, a simple diagram which contains three triangles all similar to each other. Fourth proof We take an arbitrary right-angled triangle ABC with CD is perpendicular to AB and x + y = c
That is,
That is,
That is, From the second group of equations we obtain (by cross multiplication) b2 = y(x + y) Similarly, from the third group we obtain a2 = x(x + y) Hence
and the proof is complete. We draw squares of areas a2, b2 and c2 adjacent to the sides of the triangle ABC.
The area of the shaded rectangle is cx = and the other rectangle is cy = So we have divided the square, area c2 , into two rectangles of area a2 and b2. This is the key idea in Euclid’s proof. Fifth proof: Euclid’s proof
of half the area of the two regions. Here are the details:
since
Hence area area ACDE = 2 × area Similarly, area AGHF = 2 × area area AGFH = area ACDE = b2. In a similar way it can be shown that area BHFI = a2 and the theorem is proved. Answers to exercises EXERCISE 1 a 13 b 15 c 37 d 17 e 25 f 113 EXERCISE 2 90 cm EXERCISE 3 Let DC = y where D is the point on AC produced so that BD is perpendicular to AC produced. Let BD = x. Assume c2 = a2 + b2 Using Pythagoras’ theorem in triangle BDC: a2 = x2 + y2. Use the three equations to show 2by = 0 which is a contradiction. EXERCISE 4 5.39 km EXERCISE 5 3 EXERCISE 6 a
b
c
EXERCISE 7 p = 48 and q = 25 and p = 125 and q = 54. EXERCISE 8 Let D be the point on BA produced so that CD is perpendicular to BA produced. Use Pythagoras’ theorem twice: b2 = h2 + x2 and a2 = h2 + (x + c)2. Eliminate h and substitute x = b cos(180 − A) = −b cos A to obtain the result. EXERCISE 9 Use h = b sin A = a sin B so EXERCISE 10 m2 = b2 + x2 − 2bx cos C and c2 = b2 + 4x2 − 4bx cos C. Multiply the first equation by 2 and subtract the second equation to obtain the result. EXERCISE 11 a 84 b 126 c 84 d 1170 The Improving Mathematics Education in Schools (TIMES) Project 2009-2011 was funded by the Australian Government Department of Education, Employment and Workplace Relations. The views expressed here are those of the author and do not necessarily represent the views of the Australian Government Department of Education, Employment and Workplace Relations. © The University of Melbourne on behalf of the International Centre of Excellence for Education in Mathematics (ICE-EM), the education division of the Australian Mathematical Sciences Institute (AMSI), 2010 (except where otherwise indicated). This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License. |
What theorem states that in any triangle the lengths of any two sides is greater than the length of its third side?
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