I OBM

(i) $CIN$ is similar to $DCN$, so area $CIN={\left(\frac{CN}{DN}\right)}^2$ area $DCN={\left(\frac{\frac{1}{2}}{\sqrt{5/2}}\right)}^2\cdot \frac{1}{4}=$

$\frac{1}{20}$



(ii) If we stretch the plane parallel to one of sides of the square then all areas are increased by the same factor and hence the ratio $\frac{\text{area CIN}}{\text{area ABCD}}$ is unchanged. If we now shear parallel to one of the sides, areas are unchanged, so the ratio $\frac{\text{area CIN}}{\text{area ABCD}}$ remains $\frac{1}{20}$. Thus the result holds for parallelograms. The area outside the star is made up of 8 small triangles, each area $\frac{1}{20}$, so it is $\frac{2}{5}$.



(iii) Let the median be $AM$. Then $AGB$ and $ABC$ have the same base $AB$, so $\frac{\text{area }AGB}{\text{area }ABC}=\frac{GM}{AM}=\frac{1}{3}$. This is trivial, but this is a hint for part (iv).

(iv) If we take $A$ and $B$ close together, then we get the same figure as in (iii) and so the ratio tends to $\frac{1}{3}$. Hence the $\frac{2}{5}$ result is not true for all convex quadrilaterals.