On the “quadrangular” inequalities

On the "quadratic" inequalities

For any four line segments we have

(the segments can be the edges of a 4-gcm) = (each of the segments is in length at most the sum of the lengths of the other three)

Presumably, this is well-known; if not, it is rather obvious. I would never have bothered to write it down, were it not for the fact that I could use all four inequalities to demonstrate that, given the lengths of the edges of a 4-gon, the latter can be bent in such a way that its 4 vertices lie on a circle.

The condition that the vertices lie on a circle is in the above terminology equivalent to $\varphi +\psi = \pi$ We shall demonstrate the possibility by computing

According to the cosine rule

$$ e ^ {2} = a ^ {2} + b ^ {2} - 2 \cdot a \cdot b \cdot \cos . \varphi $$

similarly , and because $\cos .\psi = - \cos .\varphi$ ,we have

$$ e ^ {2} = c ^ {2} + d ^ {2} + 2 \cdot c \cdot d \cdot \cos \varphi $$

Eliminating ${e}^{2}$ ,we find that $\cos \varphi$ is a root of the equation

$$ x: (a ^ {2} + b ^ {2} - (c ^ {2} + d ^ {2}) = (2 \cdot a \cdot b + 2 \cdot c \cdot d) \cdot x) $$

and, since $- 1 \leq \cos \varphi \leq + 1$ ,we have to show that the -unique- solution to the above equation is in absolute value at most 1 . Confining ourselves to the case that all four lengths are positive - if some lengths are zero, the theorem is trivial-, we have ${2.4} + {2.0d} > 0$ ,and our proof obligation boils down to demonstrating

(i) ${a}^{2} + {b}^{2} - \left( {{c}^{2} + {d}^{2}}\right) \leq 2 \cdot a \cdot b + 2 \cdot c \cdot d\;$ and (ii) ${a}^{2} + {b}^{2} - \left( {{c}^{2} + {d}^{2}}\right) \geq - 2 \cdot a \cdot b - 2 \cdot c \cdot d$

(i)

$$ = {\mathrm {a l g e b r a} } $$

$$ a ^ {2} - 2 \cdot a \cdot b + b ^ {2} \leq c ^ {2} + 2 \cdot c \cdot d + d ^ {2} $$

$$ = {\mathrm {a l g e b r a} } $$

$$ (a - b \max b - a) ^ {2} \leq (c + d) ^ {2} $$

$$ \neq {\mathrm {f o r n o n - n e g a t h e b a s e , t h e s q u a r e i s m o n o d u n i c} } $$

$$ (a - b \max b - a) \leq c + d $$

$$ = {\mathrm {a r i t h m e t i c} } $$

$$ a - b \leq c + d \land b - a \leq c + d $$

$$ = {\mathrm {a r i t h m e t i c} } $$

$$ a \leq b + c + d \land b \leq c + d + a $$

$$ s i m i l a r l y $$

(ii)

$$ \begin{array}{l} = {\mathrm {a l g e b r a} } \ a ^ {2} + 2 \cdot a \cdot b + b ^ {2} \geq c ^ {2} - 2 \cdot c \cdot d + d ^ {2} \ \Leftarrow {\mathrm {t h e a b o v e w i t h} (a, b) \mathrm {a n d} (c, d) \mathrm {i n t e r c h a n g e d} } \ c \leq d + a + b \land d \leq a + b + c \ \end{array} $$

Hence the demonstrandum follows from all four quadrangular inequalities.

I think it a nice proof because I had never ex- pected to use the quadrangular inequalities so explicitly. There is also something profoundly wrong with it because it cannot be readily generalized to polygons, for which the theorem is also valid. (It is, for given edge lengths, the polygon with maximum area; this has the property that its vertices lie on a circle.)

Los Alamos, 26 May 1986

prop. dr. Edsger W. Dgkstra Department of Computer Sciences The University of Texas at Austin Austin, TX 78712-1188 United States of America