Difference between revisions of "2003 AIME I Problems/Problem 7"
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== See also == | == See also == |
Revision as of 11:16, 21 March 2020
Contents
[hide]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 .
Solution 2
Using Stewart's Theorem, letting the side length be c, and the cevian be d, then we have . Dividing both sides by thirty leaves . The solution follows as above.
Solution 3
drop an altitude from angle D to side AC. let the intersection point be E. since the triangle is isosceles, AE is half of AC, which is 15. Then, label side AD as x. since AED is a right triangle, you can figure out cos(A) with adjacent divided by hypotenuse, which in this case is AE divided by x, which is just 15/x. now law of cosines. label bd as y. so, by law of cosines, Since cosine A is equal to 15/x, , which can be simplified to then continue with the first solution
u guys can edit the formatting if youd like to
-intelligence_20
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.