Difference between revisions of "1995 AIME Problems/Problem 9"
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== Solution 2 == | == Solution 2 == | ||
− | In a similar fashion, we encode the angles as complex numbers, so if <math>BM=x</math>, then <math>\angle BAD=\text{Arg}(11+xi)</math> and <math>\angle BDM=\text{Arg}(1+xi)</math>. So we need only find <math>x</math> such that <math>\text{Arg}((11+xi)^3)=\text{Arg}(1331-33x^2+(363x-x^3)i)=\text{Arg}(1+xi)</math>. This will happen when <math>\frac{363x-x^3}{1331-33x^2}=x</math>, which simplifies to <math>121x-4x^3=0</math>. Therefore, <math>x=\frac{11}{2}</math>. By the Pythagorean Theorem, <math>AB=\frac{11\sqrt{5}{2}</math>, so the perimeter is <math>11+11\sqrt{5}=11+\sqrt{605}</math>, giving us our answer, <math>\boxed{616}</math>. | + | In a similar fashion, we encode the angles as complex numbers, so if <math>BM=x</math>, then <math>\angle BAD=\text{Arg}(11+xi)</math> and <math>\angle BDM=\text{Arg}(1+xi)</math>. So we need only find <math>x</math> such that <math>\text{Arg}((11+xi)^3)=\text{Arg}(1331-33x^2+(363x-x^3)i)=\text{Arg}(1+xi)</math>. This will happen when <math>\frac{363x-x^3}{1331-33x^2}=x</math>, which simplifies to <math>121x-4x^3=0</math>. Therefore, <math>x=\frac{11}{2}</math>. By the Pythagorean Theorem, <math>AB=\frac{11\sqrt{5}}{2}</math>, so the perimeter is <math>11+11\sqrt{5}=11+\sqrt{605}</math>, giving us our answer, <math>\boxed{616}</math>. |
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== See also == | == See also == |
Revision as of 00:08, 11 June 2013
Contents
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, .
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |