Difference between revisions of "2003 AIME I Problems/Problem 7"
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Revision as of 17:14, 8 March 2007
Problem
Point is on
with
and
Point
is not on
so that
and
and
are integers. Let
be the sum of all possible perimeters of
. Find
Solution
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Denote the height of as
,
, and
. Using the Pythagorean theorem, we find that
and
. Thus,
. The LHS is difference of squares, so
. As both
are integers,
must be integral divisors of 189.
The divisors of 189 are . This yields the four potential sets for
as
. The last is not a possibility since it simply degenerates into a line. The sum of the three possible perimeters of
is equal to
.
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |