1998 AIME Problems/Problem 6
Contents
[hide]Problem
Let be a parallelogram. Extend through to a point and let meet at and at Given that and find
Solution
Solution 1
There are several similar triangles. , so we can write the proportion:
Also, , so:
Substituting,
Thus, .
Solution 2
We have so . We also have so . Equating the two results gives and so which solves to
See also
1998 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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