grayfalcon89 Posted June 12, 2006 Posted June 12, 2006 This is not that easy problem but give it a try. Suppose triangle ABC is equilateral, and AF = BD = CE = 1/3*AB (D,E, and F are on BC,AC, and AB respectively). Compute the ratio of the area of the triangle ABC to that of the area of triangle made by drawing liens CF,AD, and BE.
The Thing Posted July 3, 2006 Posted July 3, 2006 Compute the ratio of the area of the triangle ABC to that of the area of triangle made by drawing lines CF,AD, and BE. LINES CF, AD, and BE. Complicates the problem very dramatically. I did almost exactly what you did but just as I was about to post it I saw that. Lines CF, AD, and BE, NOT lines EF, FD and ED. EF, FD and ED just creates another equilaterial triangle and the Triangle FDB on your picture would be a 30-60-90 triangle (side of 1/3, side of 2/3, 60 triangle in between) and hence its area wouldn't be hard to figure out. Drawing lines CF, AD, and BE would create 3 lines that intersect to make some kind of triangle in the middle of the larger triangle.
alext87 Posted July 3, 2006 Posted July 3, 2006 i think I may have found the answer but it is quite tricky. I will scan in the solution tomo. But the ratio ended up being 7:1? Can you verify this or check the worked solution tomo once I have posted it.
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