RMO
Geometry
Level
2
Let ABC be a triangle. Let D, E be a 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 the area of the triangle APQ to that of the quadrilateral PDEQ.
1/3
2/11
9/11
5/3
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1 solution
Solution: If we can find [APQ]/[ADE], then we can get the required ratio as
[APQ] [PDEQ]
[APQ] [ADE]−[APQ] = 1 [ADE]/[APQ]−1.Now draw PM ⊥ AE and DL ⊥ AE. Ob-serve [APQ] = 1 2AQ·PM, [ADE] = 1 2AE ·DL. Further, since PM k DL, we also get PM/DL = AP/AD. Using these we obtain [APQ] [ADE] = AP AD · AQ AE . We have
AQ QE
[ABQ] [EBQ]
[ACQ] [ECQ]
[ABQ] + [ACQ] [BCQ]
[ABQ] [BCQ] +
[ACQ] [BCQ]
AF FC + AS SB . However
BS SA
[BQC] [AQC]
[BQC]/[AQB] [AQC]/[AQB]
CF/FA EC/BE
1 1/2 = 2. Besides AF/FC = 1. We obtain
AQ QE
AF FC + AS SB = 1 +
1 2
3 2 , AE QE = 1 +
3 2
5 2 ,
AQ AE
3 5 . Since EF k AD (since DE/EC = AF/FC = 1), we get AD = 2EF. Since EF k PD, we also have PD/EF = BD/DE = 1/2. Hence EF = 2PD. Thus AD = 4PD. This gives and AP/PD = 3 and AP/AD = 3/4. Thus
[APQ] [ADE]
AP AD ·
AQ AE
3 4 ·
3 5
9 20 . Finally,
[APQ] [PDEQ]
1 [ADE]/[APQ]−1 =
1 (20/9)−1
9 11 . (Note: BS/SA can also be obtained using Ceva’s theorem. Coordinate geometry solution can also be obtained.)