The diagonals of a trapezoid $ ABCD $ whose bases are $ [AB] $ and $ [CD] $ intersect at $P.$ Prove that \[S_{PAB} + S_{PCD} > S_{PBC} + S_{PDA},\] Where $S_{XYZ} $ denotes the area of $\triangle XYZ $.
Problem
Source: Moroccan MO 11th grade 5th exam 2011
Tags: geometry, trapezoid, trigonometry, inequalities, geometry unsolved
Goutham
24.04.2011 12:09
Use the fact that $S_{XYZ}=\frac{1}{2}\cdot XY\cdot YZ\cdot \sin \angle XYZ$ to get that we have to prove \[PA\cdot PB+PC\cdot PD>PA\cdot PD+PB\cdot PC=2\cdot PA\cdot PD\] as $\frac{PB}{PD}=\frac{PA}{PC}=k$, say. Now, this implies that we have to prove \[k+\frac{1}{k}\ge 2\] which is true.
Dijkschneier
26.04.2011 18:42
Goutham wrote: the fact that $S_{XYZ}=XY\cdot YZ\cdot \sin \angle XYZ$ Really ?
Goutham
26.04.2011 18:44
Dijkschneier wrote: Goutham wrote: the fact that $S_{XYZ}=XY\cdot YZ\cdot \sin \angle XYZ$ Really ? Sorry about that. It should be okay now. LOL.
Dijkschneier
26.04.2011 19:02
Your solution is clean now !
quantumbyte
26.04.2011 20:55
Does this quadrilateral have to be a trapezoid? My solution is very similar but I called $AP=y, PC=yz, PB=x, PD=xz$. Then we have that $1+k^2>=2k$ which is obvious by the trivial inequality or AM-GM.