1988 AIME Problems/Problem 12
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
Solution 3
You can use mass points to derive Plugging it in yields We proceed as we did in Solution 1 - however, to make the equation look less messy, we do the substitution
Then we have Clearing fractions gives us Factoring yields and the left hand side looks suspiciously like what we want to find. (It is.)
Substituting yields our answer as
Solution 4 (Ceva Identity)
A cool identity derived from Ceva's Theorem is that:
To see this, we use another Ceva's Theorem identity (sometimes attributed to Gergonne): , and similarly for cevians and . And then:
Inserting into our identity gives:
This problem is quite similar to the 1992 AIME problem 14, which also featured concurrent cevians and is trivialized by the identity used in this solution. For reference, below is a link. https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_14
See also
1988 AIME (Problems • Answer Key • Resources) | ||
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Followed by Problem 13 | |
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