A convex quadrilateral $ABCD$ is said to be dividable if for every internal point $P$, the area of $\triangle PAB$ plus the area of $\triangle PCD$ is equal to the area of $\triangle PBC$ plus the area of $\triangle PDA$. Characterize all quadrilaterals which are dividable.
Problem
Source: Canada Repêchage 2017/8
Tags: geometry
62861
13.04.2017 15:53
The only dividable quadrilaterals are parallelograms.
First, we show that a dividable quadrilateral is a parallelogram. Let $K = [ABCD]$. Taking the degenerate case $P = D$, we see that $[DAB] = [DBC]$, so $[BCD] = [DAB] = \tfrac{K}{2}$. Similarly, $[ABC] = [CDA] = \tfrac{K}{2}$. Thus, $[BCA] = [BCD]$, which forces $\overline{AD} \parallel \overline{BC}$ as $ABCD$ is convex. Similarly, $\overline{AB} \parallel \overline{CD}$, so $ABCD$ is a parallelogram.
To show that every parallelogram is dividable, take an affine transformation sending the parallelogram to a square; now, the conclusion is obvious.
GGPiku
13.04.2017 16:19
Take P the intersection of the diagonals. Using formula of area with sin we have $PA*PB+PC*PD=PB*PC+PA*PD$, equivalent with $(PA-PC)(PB-PD)=0$. So $PA=PC$. Now if we choose $Q=A$, we get $A_{ADC}=A_{ABC}$ and since $P$ is the midpoint of $AC$ we get that $P$ is the midpoint of $AC$ too. We conclude that $ABCD$ needs to be a paralellogram. To prove every paralellogram works, just note that $S_{ABCD}=d(A,BC)*BC=d(B,CD)*CD$ so $A_{PAB}+A_{PCD}=\frac{A_{ABCD}}{2}$.
NsisArman
17.07.2017 10:41
First we have: $ S_{PAD} + S_{PBC} + S_{PCD} + S_{PAB} = S_{ABCD}$ so we have $S_{PAB} + S_{PCD} = 1/2 * S_{ABCD}$ Suppose that $ AB \nparallel CD $ Let $ (AB , CD) = K $ Let $L$ on $KA$ and $ LK = AB $ and $F$ on $KB$ and $ FK=CD $ We have $ S_{PAB} = S_{PLK} and S_{PCD}=S_{PFK} $ so $S_{PLK}+S_{PFK} = 1/2 * S_{ABCD} $ We know that $S_{LFK}$ is constant. So $ S_{PLF} = 1/2 * S_{ABCD} - S_{LFK} $ is constant. So $ P $ is on a line parallel to$ LF $ which is Contradiction. So $ AB\parallel CD$ and $ AD \parallel BC$ and $ABCD$ is a parallelogram. And it's easy to show every parallelogram is dividable.
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