1
$\begingroup$

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?

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

This problem deserves a proof without words:

enter image description here

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

see this picture.

$enter image description here

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!

enter image description here

$\endgroup$
0
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Anamaki May 10 '14 at 4:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.