2016/17 AHSMC Part 1 Problem 16

1. Assume that the sides of the quadrilateral are of lengths $3,4,3,4$, in this order. The quadrilateral is a parallelogram (so it is convex). Let $\alpha \in \bigl(0, \tfrac{\pi}{2}\bigr]$ be the measure in radians of an angle of the parallelogram and $S$ its area. Then we have $$S = 12\sin\alpha \iff \sin\alpha = \frac{S}{12}.$$ Since $0 < \sin\alpha \le 1$ there are twelve convenient integral values of $S$, namely $1,2,\ldots,12$, and thus twelve distinct values of $\alpha$. Therefore we get twelve non-congruent parallelograms with area a positive integer.
2. Assume that the sides of the quadrilateral are of lengths $4,4,3,3$, in this order. Let $\alpha$ be the measure in radians of the angle between two sides of lengths $4$ and $3$. The quadrilateral is convex if $\alpha \in (\beta,\pi)$ with $\sin\beta = \dfrac{\sqrt{7}}{4}$. Note that the lower limit $\beta$ for the angle $\alpha$ occurs when the sides of length $3$ are both on the same line and hence, the quadrilateral degenerates to an isosceles triangle of sides $4,4,6$.

   As above, one obtains $\sin\alpha = \dfrac{S}{12}$. If $\alpha \in \bigl[\tfrac{\pi}{2}, \pi\bigr)$ there are $12$ integer values for $S$ for which we get twelve distinct values for $\alpha$ and hence one obtains twelve non-congruent quadrilaterals. If $\alpha \in \bigl(\beta, \tfrac{\pi}{2}\bigr)$ then $$\sin\beta < \sin\alpha < 1 \iff \frac{\sqrt{7}}{4} < \sin\alpha < 1 \iff \sqrt{63} < S < 12.$$ There are four integer values for $S$ in $(\sqrt{63}, 12)$ and thus four distinct values for $\alpha \in \bigl(\beta, \tfrac{\pi}{2}\bigr)$ such that $\sin\alpha = \dfrac{S}{12}$ for which one obtains four non-congruent convex quadrilaterals. The number of requested convex quadrilaterals is $12+12+4=28$.

Alternative Approach: The largest parallelogram of sides in the order $3,4,3,4$ is clearly the rectangle (as it has largest altitude) of area $12$, and so other parallelograms in this family can have areas $1$ to $11$. Similarly the quadrilaterals with sides in the order $4,4,3,3$ are all composed of two congruent triangles with two of the sides being $4$ and $3$, with area at most $6$, so the largest such quadrilateral will also have area $12$. As the angle between the sides $4$ and $3$ becomes greater than $90^\circ$, we get $11$ more convex quadrilaterals of area $1$ to $11$. When this angle is less than $90^\circ$, we stay convex as long as the two sides of length $3$ do not align, which happens when the quadrilateral becomes an isosceles triangle of sides $4,4,6$ of area $\sqrt{63} < 8$. Thus we get four more convex quadrilaterals of areas $8,9,10$ and $11$ as well, for a total of $28$. The answer is (C).