# ratio between the area of square $wxyz$ and the area of square $abcd$ equal?

$ABCD$ is a square and $H$ is an interior point, which divides it for four triangles. If $W$, $X,$ $Y$ and $Z$ are the centroids of triangles $AHD$, $AHB$, $BHC$ and $CHD$ respectively , then what is the ratio between the area of the square $WXYZ$ and the area of the square $ABCD$ ?

Can anyone provide me a hint or a help to go ?

Thank you very much

• Think of this as a 3D problem, where $ABCDH$ is a square pyramid. – Akiva Weinberger Jul 28 '17 at 11:45

$XW=YZ=\frac{1}{3}BD$, $XY=WZ=\frac{1}{3}AC$ and $XYZW$ is square.

Thus, the ratio is $\frac{2}{9}$ because $S_{ABCD}=\frac{1}{2}AC\cdot BD$ and $$\frac{S_{XYZW}}{S_{ABCD}}=\frac{\frac{1}{3}AC\cdot\frac{1}{3}BD}{\frac{1}{2}AC\cdot BD}=\frac{2}{9}$$

For example, let $M$ is a midpoint of $BH$.

Hence, $$\frac{XY}{AC}=\frac{MY}{MC}=\frac{1}{3}$$

• but the answer is $2/9$ – user373141 Jul 27 '17 at 21:08
• @prayer smith I fixed. See now, please. – Michael Rozenberg Jul 27 '17 at 21:13
• please have another look , i add a graph of the problem – user373141 Jul 28 '17 at 11:37
• @prayer smith Yes it works. Just $\Delta MAC\sim\Delta XMY$. What is your question? – Michael Rozenberg Jul 28 '17 at 11:48
• @prayer smith You are welcome! – Michael Rozenberg Jul 28 '17 at 12:38

Draw the point on $AH$ that's closer to $A$ than $H$ in a 1:2 ratio. Call this $A'$. Repeat for $B$, $C$, and $D$. Show that $A'B'C'D'$ is another square, and that each vertex of $WXYZ$ is the midpoint of an edge of $A'B'C'D'$. Conclude that $WXYZ$ has half the area as $A'B'C'D'$.

Note, by the way, that there's no reason to assume that $H$ is coplanar with $ABCD$. In fact, it feels easier to visualize this problem if you assume it's not, making $ABCDH$ a square prism.