Let ABC be a triangle with AB=7,BC=9, and CA=4. Let D be the point such that AB∥CD and CA∥BD. Let R be a point within triangle BCD. Lines ℓ and m going through R are parallel to CA and AB respectively. Line ℓ meets AB and BC at P and P′ respectively, and m meets CA and BC at Q and Q′ respectively. If S denotes the largest possible sum of the areas of triangles BPP′,RP′Q′, and CQQ′, determine the value of S2.
A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.
Solution
Let R′ denote the intersection of the lines through Q′ and P′ parallel to ℓ and m respectively. Then [RP′Q′]=[R′P′Q′]. Triangles BPP′,R′P′Q′, and CQQ′ lie in ABC without overlap, so that on the one hand, S≤ABC. On the other, this bound is realizable by taking R to be a vertex of triangle BCD. We compute the square of the area of ABC to be 10⋅(10−9)⋅(10−7)⋅(10−4)= 180.
Want a route through all this instead of an archive? The track
puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.
Source: Omni-MATH,
licensed Apache-2.0.
Statement and solution reproduced as published; topic and difficulty added by this site.