2
$\begingroup$

$ABCD$ 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 $ABCD$ ?

enter image description here

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

Thank you very much

$\endgroup$
  • $\begingroup$ Think of this as a 3D problem, where $ABCDH$ is a square pyramid. $\endgroup$ – Akiva Weinberger Jul 28 '17 at 11:45
1
$\begingroup$

$XW=YZ=\frac{1}{3}BD$, $XY=WZ=\frac{1}{3}AC$ 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, $$\frac{XY}{AC}=\frac{MY}{MC}=\frac{1}{3}$$

$\endgroup$
  • $\begingroup$ but the answer is $2/9$ $\endgroup$ – user373141 Jul 27 '17 at 21:08
  • $\begingroup$ @prayer smith I fixed. See now, please. $\endgroup$ – Michael Rozenberg Jul 27 '17 at 21:13
  • $\begingroup$ please have another look , i add a graph of the problem $\endgroup$ – user373141 Jul 28 '17 at 11:37
  • $\begingroup$ @prayer smith Yes it works. Just $\Delta MAC\sim\Delta XMY$. What is your question? $\endgroup$ – Michael Rozenberg Jul 28 '17 at 11:48
  • 1
    $\begingroup$ @prayer smith You are welcome! $\endgroup$ – Michael Rozenberg Jul 28 '17 at 12:38
0
$\begingroup$

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.

$\endgroup$

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.