1995 AIME Problems/Problem 9
Contents
[hide]Problem
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
Solution 1
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 .
Solution 2
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, .
Solution 3
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 moth denominators and squaring both sides yields which simplifies to Substituting this in for x yields and Thus the perimeter is , and the answer is .
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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