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1988 AIME Problems/Problem 12

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Problem

Let P be an interior point of triangle ABC and extend lines from the vertices through P to the opposite sides. Let a, b, c, and d denote the lengths of the segments indicated in the figure. Find the product abc if a + b + c = 43 and d = 3.

Image:1988_AIME-12.png

Solution

Call the cevians AD, BE, and CF. Using area ratios (\triangle PBC and \triangle ABC have the same base), we have:

\frac {d}{a + d} = \frac {[PBC]}{[ABC]}

Similarily, \frac {d}{b + d} = \frac {[PCA]}{[ABC]} and \frac {d}{c + d} = \frac {[PAB]}{[ABC]}.

Then, \frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = \frac {[PBC]}{[ABC]} + \frac {[PCA]}{[ABC]} + \frac {[PAB]}{[ABC]} =...

The identity \frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = 1 is a form of Ceva's Theorem.

Plugging in d = 3, we get

\frac{3}{a + 3} + \frac{3}{b + 3} + \frac{3}{c+3} = 1 3[(a + 3)(b + 3) + (b + 3)(c + 3) + (c + 3)(a + 3)] = (a+3)(b+3)(c+3) 3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27 9(a + b + c) + 54 = abc=\boxed{441}

See also

1988 AIME (ProblemsResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
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