Ptolemaeus and Brahmagupta (or: Baffled by Symmetry)

Ptolemaeus and Brahmagupta (or: Baffled by Symmetry)

f. g. o

Fig.1

We call a convex 4-gon whose vertices lie on a circle a "circular quadrangle", and consider circular quadrangles with sides $a, b, c$ and $d$ . There are essentially - i.e. apart from mirror images - 3 such circular quadrangles, with sides in order $(a, b, c, d)$ - figo-, (a, b, d, c) - fig1- and (a, c, b, d) - no figure-. Their diagonals are (e, f), (e, g) and (f, g); they have circumscribed circles with the same diameter D and have the same area A.

Furthermore we have with positive angles $\alpha ,\beta ,\gamma$ , and $\delta$ ,such that $\alpha + \beta + \gamma + \delta = \pi$

$a = D \cdot \sin .\alpha \;b = D \cdot \sin .\beta \;c = D \cdot \sin .\gamma \;d = D \cdot \sin .\delta$

$e = D \cdot \sin .\left( {\alpha + \beta }\right) \;f = D \cdot \sin .\left( {\beta + \gamma }\right) \;g = D \cdot \sin .\left( {\beta + \delta }\right)$

(Note that $\sin \left( {\alpha + \beta }\right) = \sin \left( {\gamma + \delta }\right)$ etc.)

Because twice the area of a convex 4-gon equals the product of the diagonals times the sine of the angle between them, we derive from Fig.0

$$ 2 \cdot A = e \cdot f \cdot \sin . (\beta + \delta) \quad (0) $$

Because twice the area of a triangle equals the product of two sides times the sine of the angle between them, we derive from Fig.1

$$ 2 \cdot A = (a \cdot c + b \cdot d) \cdot \sin . (\beta + \delta) \quad (1) $$

Hence

$$ e \cdot f = a \cdot c + b \cdot d $$

of symmetry we also have

$$ e \cdot g = a \cdot d + b \cdot c \quad a n d \quad f \cdot g = a \cdot b + c \cdot d. ] $$

From (0) and $g = {D \cdot \sin \left( {\beta + \delta }\right) }$ we derive

$$ 2 \cdot A \cdot D = e \cdot f \cdot g \tag {2} $$

(somethiing I was not aware of).

The above proof of the theorem of Ptolemaeus is much nicer than the tricky one I was shown in my schooldays (which was the same one my parents had grown up with: even ugly arguments have a long life, once they have penetrated the school curriculum!). Yet I am annoyed by it because I compared fig.0 with fig.1, from which f has disappeared, whereas I could also have com

pared it with the omitted figure in which diagonal e has disappeared. I have been overspecific in a way I could not avoid.

An alternative proof of the theorem of Ptolemaeus shows that for

$$ \alpha + \beta + \gamma + \delta = \pi \tag {3} $$

we have

$$ \sin . (\alpha + \beta) \cdot \sin . (\beta + \gamma) = \sin . \alpha \cdot \sin . \gamma + \sin . \beta \cdot \sin . \delta . (4) $$

Here the formulation of (the left-hand side of) (4) is arbitrary; the simplest computation I can think of makes it even worse. We can use (3) by eliminating $\delta$ from (4), i.e. for arbitrary $\alpha, \beta$ , and $\gamma$ we have to show

$$ \sin . (\alpha + \beta) \cdot \sin (\beta + \gamma) = \sin . \alpha \cdot \sin . \gamma + \sin . \beta \cdot \sin . (\alpha + \beta + \gamma). (5) $$

Relation (5) holds for $\beta = 0$ ; furthermore, the derivatives of both sides with respect to $\beta$ are equal, viz. to $\sin (\alpha + 2\beta + \gamma)$ . Hence, (5) holds for any value of $\alpha, \beta$ , and $\gamma$ . But as far as destruction of symmetry is concerned, this is even worse.

With the theorem of Brahmagupta I have had similar experiences; it states

$$ \begin{array}{l} A ^ {2} = (s - a) \cdot (s - b) \cdot (s - c) \cdot (s - d) \tag {6} \ w h e r e s = (a + b + c + d) / 2 \ \end{array} $$

The least cumbersome argument I know of

(i) establishes that ${A}^{2}$ is a homogeneous polynomial of degree 4 in $a,b,c$ ,and $d$ ; (ii) establishes that the right-hand side is a factor of ${A}^{2}$ ,i.e. ${A}^{2} =$ a constant times that right-hand side; (iii) establishes that the constant equals 1 (e.g. by inspection of a square).

Step (ii) Follows From

$$ s \cdot a = 0 \equiv a = b + c + d $$

since the right-hand side implies ${A}^{2} = 0$ ,s-a is a factor, and so for the others. Step (i) is the problem.

With the cosine rule we derive From Fig.

$$ \begin{array}{l} g ^ {2} = a ^ {2} + c ^ {2} - 2 \cdot a \cdot c \cdot \cos (\beta + \delta) \ g ^ {2} = b ^ {2} + d ^ {2} + 2 \cdot b \cdot d \cdot \cos (\beta + \delta) \ \end{array} $$

hence, eliminating ${g}^{2}$ ,we find

(a.c + b.d) $\cdot \cos .\left( {\beta + \delta }\right) = {2}^{\text{nd }}$ degree polynomial Eliminating $\beta + \delta$ from the above and (1) by squaring both sides of both,one establishes the conclusion of (i).

Finally, one establishes (iii) by looking at the unit square.

Needless to say, the above proof of Brahmagupta's Theorem - though the least cumbersome I have ever

seen - still amoys me. It is a little bit less ugly than the proof of the Theorem of Ptolemaeus in that the demonstration of (ii) uses only fig.1 . But, firstly, that choice is unattractively arbitrary, and, secondly, we then made the arbitrary choice of focussing our attention on diagonal $g$ .

My shumbling block is probably the algebraic identity

$$ \begin{array}{l} (a \cdot b + c \cdot d) \cdot (a \cdot c + b \cdot d) \cdot (a \cdot d + b \cdot c) = \ a ^ {3} \cdot b \cdot c \cdot d + b ^ {3} \cdot c \cdot d \cdot a + c ^ {3} \cdot d \cdot a \cdot b + d ^ {3} \cdot a \cdot b \cdot c + \ a ^ {2} \cdot b ^ {2} \cdot c ^ {2} + b ^ {2} \cdot c ^ {2} \cdot d ^ {2} + c ^ {2} \cdot d ^ {2} \cdot a ^ {2} + d ^ {2} \cdot a ^ {2} \cdot b ^ {2} \ \end{array} $$

the one side showing a 3-fold symmetry, the other one

Austin 24 August 1986

prof. d. Edsger W. Dgksta

Department of Computer Sciences

The University of Texas at Austin

Austin, TX 78712-1188

United States of America