Problem
GEO-B2-M07-P017 Division of the Second Cevian
#17
★★★★☆ Level 4 of 5
In the same type of configuration: \(D\in BC\), \(E\in CA\), \(BD:DC=3:4\), \(CE:EA=2:5\), and \(AD\cap BE=P\). Find \(BP:PE\).
Find the areas around \(P\), then compare \([PCA]\) with the total area.
Let \([PAB]=x\), \([PBC]=y\), \([PCA]=z\). From \(BD:DC=3:4\), \(x:z=3:4\). From \(CE:EA=2:5\), \(y:x=2:5\). Take \(x=15\); then \(z=20\), \(y=6\), total \(41\). Since \(P\in BE\), \(\frac{PE}{BE}=\frac{[PCA]}{[ABC]}=\frac{20}{41}\). Thus \(BP:PE=21:20\).
The problem requires choosing the correct small area carefully.