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December 3, 2012

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.


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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