2003 AIME I Problems/Problem 7
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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
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 
.
The pairs of divisors of 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 • Resources) | ||
| Preceded by Problem 6 | Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
