Difference between revisions of "1988 AIME Problems/Problem 12"
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<cmath>3[(a + 3)(b + 3) + (b + 3)(c + 3) + (c + 3)(a + 3)] = (a+3)(b+3)(c+3)</cmath> | <cmath>3[(a + 3)(b + 3) + (b + 3)(c + 3) + (c + 3)(a + 3)] = (a+3)(b+3)(c+3)</cmath> | ||
<cmath>3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27</cmath> | <cmath>3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27</cmath> | ||
− | <cmath>9(a + b + c) + 54 = | + | <cmath>9(a + b + c) + 54 = abc=\boxed{441}</cmath> |
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== See also == | == See also == |
Revision as of 18:17, 10 April 2008
Problem
Let be an interior point of triangle and extend lines from the vertices through to the opposite sides. Let , , , and denote the lengths of the segments indicated in the figure. Find the product if and .
Solution
Call the cevians AD, BE, and CF. Using area ratios ( and have the same base), we have:
Similarily, and .
Then,
The identity is a form of Ceva's Theorem.
Plugging in , we get
See also
1988 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |