# Flexi Quad Tan

As a quadrilateral Q is deformed (keeping the edge lengths constnt)
the diagonals and the angle X between them change. Prove that the
area of Q is proportional to tanX.

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.

Using the results of the two problems on quadrilaterals
Diagonals for Area and
Flexi Quads prove that the area of $Q$ is proportional to
$\tan\theta$.

Combine the result from the problem Flexi Quad Areas that the area $A(Q) = {\textstyle{1\over 2}}d_1d_2\sin\theta$ with the definition of the scalar product and use the result from the problem Flexi Quads that the scalar product of the diagonals is constant.

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 $.