Let $ABCD$ a generic convex quadrilateral. Let $P,Q,R,S$ the midpoints of $AB,BC,CD$ and $AD$ respectively. The segments $PR$ and $QS$ intersect each other in a point $O$, and divide $ABCD$ in 4 sub-quadrilaterals. If the area of $APOS$ is 25, the area of $PBQO$ is 16 and the area of $QCRO$ is 36, then what is the area of the remaining quadrilateral $SORD$ ?

I know that $PQRS$ is a parallelogram, so that $O$ is the midpoint of the segments $PR$ and $QS$, and its area is half the area of $ABCD$, but I can't move on from this. Any help?



Given that $AP=BP$, we have that triangles $APO$ and $BPO$ have same base and same height, and hence have the same area.

Likewise: $BQO$ and $CQO$ have the same area, and so on.

Now notice how the areas of the $4$ quadrilaterals are composed of those $8$ triangles.

  • $\begingroup$ @Möbius Yes, that's what you need to solve .. but adding the second and third you get: $p + q + r + s=61$, and subtracting the first from that we get $r+s=45$ $\endgroup$ – Bram28 Nov 7 '17 at 21:05
  • $\begingroup$ Thank you, I got 27 because of a wrong minus sign, I was solving by substitution (I got 36 - 9 instead of 36+9!) $\endgroup$ – Möbius Nov 7 '17 at 21:25
  • $\begingroup$ @Möbius Ah, I was wondering ... :) Glad I could help! $\endgroup$ – Bram28 Nov 7 '17 at 21:28

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