(Corollaries from)
Pythagoras' Theorem
Pythagoras' Theorem plays an important role in mathematics indeed. On this page I'll try to collect several statements some of whose proofs depend on the Pythagorean Theorem.
The Arithmetic Mean - Geometric Means Inequality
For positive a and b, (a + b)/2  (ab)
with equality iff a = b
For a proof, assume a>b and construct a right triangle with hypotenuse (a+b)/2 and one side equal to (a-b)/2. From the Pythagorean Theorem, the remaining side will equal (ab). Since, in a right triangle, the hypotenuse is the largest side, the inequality has been proven in the case a>b. It's obviously turns into equality when a=b.
One may argue that the proof followed from an algebraic identity
(a + b)2 - (a - b)2 = 4ab
In that case, the Pythagorean Theorem furnishes an intuitive geometric illustration. Just draw two touching circles with radii a and b as in the diagram.
As in the case Isoperimetric Inequality this too allows for two equivalent extremal problems:
- Among all pairs of numbers with a given product find two whose sum is minimal.
- Among all pairs of numbers with a given sum find two whose product is maximal.
In both cases, the extremal value is attained when the two numbers coincide. The latter fact has a nice geometric illustration which also suggests another proof for the Arithmetic Mean - Geometric Mean Inequality.
The former is often rewritten in a different form:
(1) |
For x > 0, x + 1/x 2
with equality iff x = 1
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The Arithmetic Mean - Geometric Mean Inequality for sequences of numbers was first proven when the length of the sequence was a power of 2 and from here for an arbitrary integer. (1) also extends for an arbitrary number of positive numbers:
Let xi > 0, i = 1,2,...,n. Then x1/x2 + x2/x3 + ... + xn/x1 n
Actually more is true. For positive xi's, let p be an arbitrary permutation of the set of indices {1,...,n}. Then
x1/xp(1) + x2/xp(2) + ... + xn/xp(n) n
Cosine Rule
The Cosine Rule is an obvious generalization of the Pythagorean Theorem. However, its variant that does not use trigonometric functions is a direct consequence of the latter.
Lemma
The difference of squares of two sides of a triangle equals the difference of squares of their projections on the third side:
(2) |
AB2 - BC2 = AH2 - CH2
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For a proof, use Pythagoras' Theorem twice: AB2=AH2+BH2 and BC2=CH2+BH2. Subtract one equation from the other.
From (2), BC2 = AB2 + CH2 - AH2. If ABC is an acute triangle then CH = AC - AH which gives
(3.1) |
BC2 = AB2 + AC2 - 2*AC*AH
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If angle A is obtuse, then CH = AC + AH and (2) yields
(3.2) |
BC2 = AB2 + AC2 + 2*AC*AH
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Stewart's Theorem
Coxeter and Greitzer remark that the theorem below was named after M.Stewart, who stated it in 1746, but was probably discovered by Archimedes about 300 B.C. However, the first known proof is by R.Simson, 1751.
Let point D lie between the vertices A and C of ABC. Then
AB2*DC + BC2*AD - BD2*AC = AC*DC*AD
For the definiteness sake, let H (the foot of the altitude BH) lie between D and C, as in the diagram. Apply (3.1) to BCD and (3.2) to ABD:
BC2 = BD2 + DC2 - 2*DC*DH
AB2 = BD2 + AD2 + 2*AD*DH
Multiply the first identity by AD, the second by DC, and sum them up
BC2*AD + AB2*DC =
BD2*(AD + DC) + DC2*AD + AD2*DC =
BD2*AC + AD*DC*AC
Medians
Let D be the midpoint between A and C. As usual, let a = |BC|, b = |AC|, c = |AB|, and mb = |BD|. After division by a, Stewart's Theorem then takes the form
mb2 = (a2 + c2)/2 - b2/4
Altitudes and Heron's Formula
Note that (3.1) applies to an acute while (3.2) to an obtuse angle. Since, in a triangle, only one angle may be obtuse, (3.1) is always applicable. Let, for example, angle A be acute. Then from (3.1)
a2 = b2 + c2 - 2b*AH
On the other hand, in the right triangle AHB, BH2 = c2 - AH2 which, combined with the previous identity, yields
BH2 | = c2 - (b2 + c2 - a2)2/(4b2) |
| =
(c - (b2 + c2 - a2)/(2b))(c + (b2 + c2 - a2)/(2b)) |
| =
(2bc - b2 - c2 + a2)(2bc + b2 + c2 - a2)/(4b2) |
| =
(a2 - (b - c)2)((b + c)2 - a2)/(4b2) |
| =
(b - a + c)(b + a - c)(a + c - b)(a + b + c)/(4b2) |
Introduce the semiperimeter p = (a + b + c)/2. Then
BH2 = 4p(p - a)(p - b)(p - c)/b2
Which is just another way of writing Heron's formula. Thus, the two facts: Pythagoras' Theorem and Heron's formula each have an independent proof. But, in addition, each may be derived from the other.
Remark
Dr. S.Brodie kindly prepared a demonstration of how Pythagoras' Theorem is used to construct a regular pentagon.
References
- H.S.M.Coxeter and S.L.Greitzer, Geometry Revisited, MAA, 1967
- R.B.Nelsen, Proofs Without Words, MAA, 1993
Copyright © 1996-1998 Alexander Bogomolny
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