Difference between revisions of "1988 AIME Problems/Problem 12"
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[[Image:1988_AIME-12.png]] | [[Image:1988_AIME-12.png]] | ||
− | == Solution == | + | == Solution 1 == |
Call the [[cevian]]s AD, BE, and CF. Using area ratios (<math>\triangle PBC</math> and <math>\triangle ABC</math> have the same base), we have: | Call the [[cevian]]s AD, BE, and CF. Using area ratios (<math>\triangle PBC</math> and <math>\triangle ABC</math> have the same base), we have: | ||
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<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 = abc=\boxed{441}</cmath> | <cmath>9(a + b + c) + 54 = abc=\boxed{441}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let the labels <math>A,B,C</math> be the weights of the vertices. First off, replace <math>d</math> with <math>3</math>. 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 <cmath>\dfrac{a+b+c}{3}=\dfrac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B}{ABC}=\dfrac{43}{3}</cmath> | ||
+ | |||
+ | If we multiply them together, we get <cmath>\dfrac{abc}{27}=\dfrac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B+2ABC}{ABC}=\dfrac{43}{3}+2=\dfrac{49}{3}</cmath> | ||
+ | |||
+ | Multiplying both sides by <math>27</math>, we get that <math>abc=27\cdot \dfrac{49}{3}=\boxed{441}</math>. | ||
== See also == | == See also == |
Revision as of 20:54, 13 September 2014
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 the labels be the weights of the vertices. First off, replace with . 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
Multiplying both sides by , we get that .
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.