Difference between revisions of "1994 AIME Problems/Problem 8"
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== Solution Two == | == Solution Two == | ||
− | Using the Pythagorean theorem doesn't seem promising (you can look at the beastly numbers). It's better to use some properties of equilateral triangles. Thinking about the number <math>\sqrt{3}</math> can | + | Using the Pythagorean theorem doesn't seem promising (you can look at the beastly numbers). It's better to use some properties of equilateral triangles. Thinking about the number <math>\sqrt{3}</math> and looking for perpendiculars gives this solution: |
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+ | First, drop a perpendicular from <math>O</math> to <math>AB</math>. Call this midpoint of <math>AB M</math>. Thus, <math>M=(\frac{a+b}{2}, 24)</math>. The vector from <math>O</math> to <math>M</math> is <math>[\frac{a+b}{2}, 24]</math>. Meanwhile from point <math>M</math> we can use a vector with <math>\frac{\sqrt{3}}{3}</math> the distance; we have to switch the <math>x</math> and <math>y</math> and our displacement is <math>[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}]</math>. (Do you see why we switched <math>x</math> and <math>y</math> due to the rotation of 90 degrees?) | ||
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We see this displacement from <math>M</math> to <math>A</math> is <math>[\frac{a-b}{2}, 13]</math> as well. Equating the two vectors, we get <math>a+b=26\sqrt{3}</math> and <math>a-b=16\sqrt{3}</math>. Therefore, <math>a=21\sqrt{3}</math> and <math>b=5\sqrt{3}</math>. And the answer is <math>\boxed{315}</math>. | We see this displacement from <math>M</math> to <math>A</math> is <math>[\frac{a-b}{2}, 13]</math> as well. Equating the two vectors, we get <math>a+b=26\sqrt{3}</math> and <math>a-b=16\sqrt{3}</math>. Therefore, <math>a=21\sqrt{3}</math> and <math>b=5\sqrt{3}</math>. And the answer is <math>\boxed{315}</math>. |
Revision as of 17:13, 3 September 2017
Contents
Problem
The points , , and are the vertices of an equilateral triangle. Find the value of .
Solution
Consider the points on the complex plane. The point is then a rotation of degrees of about the origin, so:
Equating the real and imaginary parts, we have:
Solving this system, we find that . Thus, the answer is .
Note: There is another solution where the point is a rotation of degrees of ; however, this triangle is just a reflection of the first triangle by the -axis, and the signs of and are flipped. However, the product is unchanged.
Solution Two
Using the Pythagorean theorem doesn't seem promising (you can look at the beastly numbers). It's better to use some properties of equilateral triangles. Thinking about the number and looking for perpendiculars gives this solution:
First, drop a perpendicular from to . Call this midpoint of . Thus, . The vector from to is . Meanwhile from point we can use a vector with the distance; we have to switch the and and our displacement is . (Do you see why we switched and due to the rotation of 90 degrees?)
We see this displacement from to is as well. Equating the two vectors, we get and . Therefore, and . And the answer is .
Note: This solution was also present in Titu Andreescu and Zuming Feng's "103 Trigonometry Problems".
See also
1994 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.