RMO 2012 solution to Question No. 5

5. Let ABC be a triangle. Let D, E be points on the segment BC such that BD = DE = EC. Let F be the mid point of AC. Let BF intersect AD in P and AE in Q respectively. Determine the ratio of triangle APQ to that of the quadrilateral PDEQ.

Solution:

Applying Menelaus' theorem to ΔBCF with AD as the transversal, we have
 \(\frac {BD}{DC} \frac{CA}{AF} \frac {FP}{PB} = 1 \)
But BD/DC = 1/2 (as BD = DE = EC) and CA/AF = 2/1 (as CF = FA).
Hence we have BP = PF.
Again applying Menelaus' Theorem to ΔBCF with AE as the transversal we have
\(\frac {BE}{EC} \frac{CA}{AF} \frac {FQ}{QB} = 1 \)
But BE/EC = 2/1 and CA/AF = 2/1
Hence 4FQ = QB.

Suppose FQ= x unit. The QB = 4x unit. That is BF = 5x unit. Since BP = PF hence each is 2.5x unit.
Thus PQ = 2.5x - x = 1.5x unit

Hence  \(\frac{ΔAPQ}{ΔABF} = \frac{1.5x}{5x}\)
Also \(\frac{ΔABF}{ΔABC} =\frac{1}{2}\)

Thus  \(\frac{ΔAPQ}{ΔABF} \frac{ΔABF}{ΔABC}= \frac{1.5x}{5x} \frac{1}{2}\)
=> \(\frac{ΔAPQ}{ΔABC} = \frac{1.5}{10}\) ...(1)

Again \(\frac{ΔADE}{ΔABC} = \frac{1}{3}\) (as DE/BC = 1/3)
Thus \(\frac{ΔADE}{ΔABC} - \frac{ΔAPQ}{ΔABC}  = \frac{1}{3} -  \frac{1.5}{10}\)
=> \(\frac{PQED}{ΔABC}  = \frac{5.5}{30}\) ...(2)

Using (1) and (2) we have 

 \(\frac{ΔAPQ}{PQED} = \frac{4.5}{5.5} = \frac{9}{11}\)