PRMO 2017 question 24

24. Let P be an interior point of a triangle ABC whose sidelengths are 26, 65, 78. The line through P parallel to BC meets AB in K and AC in L. The line through P parallel to CA meets BC in M and BA in N. The line through P parallel to AB meets CA in S and CB in T. If KL, MN, ST are of equal lengths, find this common length.

Solution: 
Draw the triangle and see that PKBT, PMCL,PSAN are parallelograms.
Let:
PT = KB = x
PM = LC = y
PK = BT = z
KL = MN = ST = l (the common length)

Triangle PTM ~ ABC
y/65 = (26-l)/26

Similarly,
Triangle NKP ~ ABC
(l-y)/65 = (78-l)/78

Solving this you get l = 30.
But l should be less than each triangle side so it should be less than 26.
So no solution.

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