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# 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 $(\frac {\triangle APQ}{\triangle ABF})$ = $(\frac {1.5x}{5x})$
Also $(\frac {\triangle ABF}{\triangle ABC} = \frac {1}{2})$
Thus $(\frac {\triangle APQ}{\triangle ABF}$ = $(\frac {\triangle ABF}{\triangle ABC})$  = $\frac {1.5x}{5x} \frac {1}{2}$ = $(\frac {\triangle APQ}{\triangle ABC})$ = $(\frac {1.5}{10})$ ...(1)

Again $\frac {\triangle ADE}{\triangle ABC}$ = $(\frac {1}{3})$ (as DE/BC = 1/3)
Thus $\frac {\triangle ADE}{\triangle ABC} - \frac {\triangle APQ}{\triangle ABC} = \frac {1}{3} - \frac {1.5}{10}$ $(\frac {PQED}{\triangle ABC} = \frac {5.5}{30})$ ...(2)
Using (1) and (2) we have $(\frac {\triangle APQ}{PQED} = \frac {4.5}{5.5} = \frac {9}{11})$

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