# Areas in a rectangle

Suppose $P,Q, R$, and $S$ are the midpoints of the sides $AB, BC, CD$, and $DA$, respectively of rectangle $ABCD$. If the area of the rectangle is $\delta$, then prove that the area of the figure bounded by the straight lines $AQ, BR, CS$, and $DP$ is $\frac{\delta}{5}$.

I began by imposing a coordinate system but couldn't find a way to relate the given area with the area of the rectangle. How should I begin?

• You can also use vectors. – evil999man May 10 '14 at 3:52
• Yes, but I'm quite bad at vectors. Could you please give me a solution or a partial one? – Anamaki May 10 '14 at 4:01

Without loss of generality, make $ABCD$ a square with $A(0,0)\ B(0,2) \ C(2,2)\ D(2,0)$.
Using the lines $y = -2x + 2,y= -2x+4, y=x/2,y= x/2 + 1$. Two of these intersections are $(\frac{2}{5}, \frac{6}{5}), (\frac{4}{5},\frac{2}{5})$. The distance between them is a side of the small square and is $\frac{2}{\sqrt{5}}$. The area of the small square is $\frac{4}{5}$, divided by the total area of $4$ gives you a $\frac{1}{5}$ area. Transformations to a rectangle preserve the areas because everything is proportional.
$this is the picture of problem. we know the medians in a triangle are ... in$ G $!!!$ \dfrac{GM}{GA'}=\dfrac{1}{3} $( these points are not the points of our problem! ) so$ BM=MN=ND $also$ SA=SD $and$ SD' \parallel AA' $so$ D'D=D'A $thus$ S_{A'B'C'D'} = 2.S_{A'MND'} = 2*3.S_{DD'N} $so you need$ S_{DD'N} $if you need the compelte solution say please! Assume$A$at origin and vectors$\vec B$and$\vec C$. Now by writing a point on a line,say,$BC$, you can proceed by :$\vec B+\lambda(\vec{C}-\vec{B})$. Similarly, you can do it for other lines and find points of intersection in terms of$\vec B,\vec C$. Now you have all points, I assume know what to do... • Yes, the points will turn out to be$\frac{\vec B + \vec C}{2}\$ and so on. Now I take cross products to get areas? – Anamaki May 10 '14 at 4:10