Download Convex quadrilaterals: Ptolemy, Bretschneider, Brahmagupta, et al

Convex quadrilaterals: Ptolemy, Bretschneider, Brahmagupta, et al
Let ABCD be a convex quadrilateral. Let sides AB, BC, CD, DA have
lengths a, b, c, d respectively, and let diagonals AC, BD have lengths m, n
respectively.
Ptolemy’s Theorem. We have mn ≤ ac + bd with equality iff ABCD is
cyclic.
I learned the following proof from Altshiller-Court’s book; he taught in
Oklahoma in the first half of last century and had a very positive impact on
many future school teachers. This pure proof is a gem.
Construct triangle AOB similar to triangle ACD but outside the quadrilateral. So angle OBA is angle D of the quadrilateral, and it abuts angle B
of the quadrilateral. So we have AO/AC = OB/CD = AB/AD. Since angle
OAC equals angle BAD and AO/AB = AC/AD, it follows that triangle
OAC is similar to triangle BAD. We now get
OC = mn/d,
OB = ac/d,
BC = bd/d
The triangle inequality gives mn ≤ ac + bd with equality iff O, B, C are
collinear iff angles B and D add to π iff ABCD is cyclic.
Let AC, BD intersect at E, and let angle BEC be θ. Apply the Cosine
Rule to the four triangles with vertex E to deduce the generalised Cosine
Rule
a2 − b2 + c2 − d2 = 2mn cos θ
Now let K be the area of quadrilateral ABCD. Apply the (1/2)bc sin A
formula to the four triangles to deduce that K = (1/2)mn sin θ and hence
that
16K 2 = 4m2 n2 − (a2 − b2 + c2 − d2 )2
which is Bretschneider’s formula.
Suppose now that ABCD is cyclic, so that mn = ac + bd. Some pretty
algebra with differences of squares now produces the Brahmagupta formula
for the area of a cyclic quadrilateral as
K 2 = (s − a)(s − b)(s − c)(s − d)
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where s is the semi-perimeter given by s = (a + b + c + d)/2. If ABCD is
also co-cyclic (there is an internal circle tangent to all four sides), then we
easily get a + c = b + d (angle and length are dual concepts!!); thus if ABCD
is cyclic and co-cyclic, then
K 2 = abcd
What a beautiful formula!
It remains to obtain a formula for K which does not involve the diagonal
lengths. The requisite formula is
K 2 = (s − a)(s − b)(s − c)(s − d) − abcd cos2 [(B + D)/2]
To deduce this from Bretschneider, go back to the diagram for Ptolemy’s
Theorem. We have
OC 2 = OB 2 + BC 2 − 2OB.BC cos(B + D)
and hence
m2 n2 = a2 c2 + b2 d2 − 2abcd cos(B + D) = (ac + bd)2 − 4abcd cos2 [(B + D)/2]
The result follows.
Colin’s problem was to fix a, b, c, d and show that K is maximal when
ABCD is cyclic. Of course this is immediate from the above formula. I
tried to think about the problem in terms of hinged rods. As an interesting
case, take a = 8, b = 2, c = 3, d = 7. The rods fold down to a degenerate
quadrilateral with zero area. An arbitrary quadrilateral is uniquely determined by angle A (which then fixes BD and so also triangle BDC). There
is a maximal angle allowable for A when B, C, D are collinear and we get a
final triangle. But I see no obvious intuitive reason why the area should be
maximal when angles B and D add to π (of course, equivalently for angles
A, C).
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