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## 'Flexi Quad Tan' printed from http://nrich.maths.org/

Well done Shu Cao of the Oxford High School for girls for producing
this nice solution so promptly.

Consider any convex quadrilateral $Q$ made from four rigid rods
with flexible joints at the vertices so that the shape of $Q$ can
be changed while keeping the lengths of the sides constant. If the
diagonals of the quadrilateral cross at an angle $\theta$ in the
range $(0 \leq \theta < \pi/2)$, as we deform $Q$, the angle
$\theta$ and the lengths of the diagonals will change and we have
to prove that the area of of $Q$ is a constant multiple of $\tan
\theta $.

Notation: Let $|{\bf x}|$ mean the scalar quantity of vector ${\bf
x}$ and the area of $Q$ be represented by $S$.

In the problem Diagonals for Area it was shown that the area of a
quadrilateral is given by half the product of the lengths of the
diagonals multiplied by the sine of the angle between the
diagonals: $$S = {\textstyle{1\over 2}}|{\bf d_1}| \times |{\bf
d_2}|\sin \theta.$$ From the definition of the scalar product
$$|{\bf d_1}| \times |{\bf d_2}| = {{\bf d_1}\cdot {\bf d_2} \over
\cos \theta }.$$ So $$S = {\textstyle{1\over 2}}{{\bf d_1}\cdot
{\bf d_2}\sin \theta \over \cos \theta } = {\textstyle{1\over
2}}{\bf d_1}\cdot{\bf d_2}\tan \theta.$$ As shown in the problem
Flexi Quads, the scalar product of the diagonals is constant i.e.
$$2{\bf d}_1 \cdot{\bf d}_2 = {\bf a}_2^2+{\bf a}_4^2-{\bf
a}_1^2-{\bf a}_3^2.$$ As ${\bf a_1, a_2, a_3, a_4}$, the lengths of
the sides of the quadrilateral, all remain constant, hence ${\bf
d}_1 \cdot {\bf d}_2$ remains constant. Hence the area of the
quadrilateral $Q$ is a constant multiple of $\tan \theta$ and so it
is proportional to $\tan \theta $.