2003 AIME I Problems/Problem 7
Contents
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 1 (Pythagorean Theorem)
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 .
Solution 2 (Stewart's Theorem)
Let and , then by Stewart's Theorem we have:
. After simplifying:
.
The solution follows as above.
Solution 3 (Law of Cosines)
Drop an altitude from point to side . Let the intersection point be . Since triangle is isosceles, AE is half of , or . Then, label side AD as . Since is a right triangle, you can figure out with adjacent divided by hypotenuse, which in this case is divided by , or . Now we apply law of cosines. Label as . Applying law of cosines, . Since is equal to , , which can be simplified to . The solution proceeds as the first solution does.
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See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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