Difference between revisions of "1988 AIME Problems/Problem 12"
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== Problem == | == Problem == | ||
+ | Let <math>P</math> be an interior point of triangle <math>ABC</math> and extend lines from the vertices through <math>P</math> to the opposite sides. Let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> denote the lengths of the segments indicated in the figure. Find the product <math>abc</math> if <math>a + b + c = 43</math> and <math>d = 3</math>. | ||
+ | |||
+ | {{image}} | ||
== Solution == | == Solution == | ||
+ | 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: | ||
+ | |||
+ | <math>\frac {d}{a + d} = \frac {[PBC]}{[ABC]}</math> | ||
+ | |||
+ | Similarily, <math>\frac {d}{b + d} = \frac {[PCA]}{[ABC]}</math> and <math>\frac {d}{c + d} = \frac {[PAB]}{[ABC]}</math>. | ||
+ | |||
+ | Then, | ||
+ | <math>\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = \frac {[PBC]}{[ABC]} + \frac {[PCA]}{[ABC]} + \frac {[PAB]}{[ABC]} = \frac {[ABC]}{[ABC]} = 1</math> | ||
+ | |||
+ | The [[identity]] <math>\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = 1</math> is a form of [[Ceva's Theorem]]. | ||
+ | |||
+ | Plugging in <math>d = 3</math>, we get | ||
+ | |||
+ | <cmath>\frac{3}{a + 3} + \frac{3}{b + 3} + \frac{3}{c+3} = 1</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>9(a + b + c) + 54 = abc</cmath> | ||
+ | <cmath>abc = 441</cmath> | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1988|num-b=11|num-a=13}} | |
− | + | [[Category:Intermediate Geometry Problems]] |
Revision as of 16:36, 28 September 2007
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 .
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
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 |