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# Flexi Quad Tan

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 $.

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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 $.