Difference between revisions of "1988 AIME Problems/Problem 12"
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− | Let | + | Let <math>A,B,C</math> be the weights of the respective vertices. We see that the weights of the feet of the cevians are <math>A+B,B+C,C+A</math>. By [[mass points]], we have that: <cmath>\dfrac{a}{3}=\dfrac{B+C}{A}</cmath> <cmath>\dfrac{b}{3}=\dfrac{C+A}{B}</cmath> <cmath>\dfrac{c}{3}=\dfrac{A+B}{C}</cmath> |
If we add the equations together, we get <math>\frac{a+b+c}{3}=\frac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B}{ABC}=\frac{43}{3}</math> | If we add the equations together, we get <math>\frac{a+b+c}{3}=\frac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B}{ABC}=\frac{43}{3}</math> |
Revision as of 19:32, 16 March 2017
Contents
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 1
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
Solution 2
Let be the weights of the respective vertices. We see that the weights of the feet of the cevians are . By mass points, we have that:
If we add the equations together, we get
If we multiply them together, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.