1995 AIME Problems/Problem 9
Triangle is isosceles, with and altitude Suppose that there is a point on with and Then the perimeter of may be written in the form where and are integers. Find
Let , so . Then, . Expanding using the angle sum identity gives Thus, . Solving, we get . Hence, and by the Pythagorean Theorem. The total perimeter is . The answer is thus .
In a similar fashion, we encode the angles as complex numbers, so if , then and . So we need only find such that . This will happen when , which simplifies to . Therefore, . By the Pythagorean Theorem, , so the perimeter is , giving us our answer, .
Let , so , , and thus We can then draw the angle bisector of , and let it intersect at Since , Let . Then we see by the Pythagorean Theorem, , , , and By the angle bisector theorem, Substituting in what we know for the lengths of those segments, we see that multiplying by both denominators and squaring both sides yields which simplifies to Substituting this in for x in the equations for and yields and Thus the perimeter is , and the answer is .
The triangle is symmetrical so we can split it in half ( and ).
Let and . By the Law of Sines on triangle , . Using we can get . We can use this information to relate to by using the Law of Sines on triangle .
(as is a right angle), so . Using the identity , we can turn the equation into::
Now that we've found , we can look at the side lengths of and (since they are symmetrical, the perimeter of is .
We note that and .
(Note it is positive since ).
The answer is .
Suppose , since , we have . Therefore, and . As a result, is isosceles, .
Let be a point on the extension of through such that and denote the intersection of and as . Then, , and by the Midpoint Theorem. So, and .
Consequently, , Assume and , then and . Since , and since , . Therefore, .
In , by the Pythagorean Theorem, . Similarly in , . So Since , we have and . Consequently, and . Thus, the perimeter of is , and the answer is .
|1995 AIME (Problems • Answer Key • Resources)|
|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.