Question

Let $a,$ $b,$ $c$ and $d$ be the lengths of sides $MN,$ $NP,$ $PQ$ and $QM,$ respectively, of quadrilateral $MNPQ.$ If $A$ is the area of $MNPQ,$ then

$A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$ if and only if $MNPQ$ is convex

$A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$ if and only if $MNPQ$ is a rectangle

$A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$ if and only if $MNPQ$ is a rectangle

$A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$ if and only if $MNPQ$ is a parallelogram

$A\ge\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$ if and only if $MNPQ$ is a parallelogram

Answer: B

Concepts:area inequality rectangle

Difficulty rating: 2320

Small Hint:

Split the quadrilateral along each diagonal and bound every sine by $1$

Big Hint:

Combine the two resulting area bounds and analyze when every bound is an equality

Solution:

Splitting along diagonal $MP$ and using $\sin\theta\le1$ gives $A\le\frac{ab+cd}{2}.$ Splitting along $NQ$ similarly gives $A\le\frac{ad+bc}{2}.$ These bounds remain valid for a nonconvex quadrilateral by taking the appropriate difference of triangle areas. Adding them yields $\begin{aligned} 2A&\le\frac{ab+ad+bc+cd}{2}\\ &=\frac{(a+c)(b+d)}2, \end{aligned}$ or $A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right).$ Equality requires equality in all four sine bounds, so all four angles are right angles; conversely a rectangle gives equality. Thus the equality holds exactly for rectangles.

Therefore, the correct answer is B.

1977 AMC 12 Problem 26

Answer: B

Difficulty rating: 2320

Small Hint:

Big Hint:

Solution:

Problem 26 in Other Years