Difference between revisions of "1994 AIME Problems/Problem 8"
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'''Note''': There is another solution where the point <math>b+37i</math> is a rotation of <math>-60</math> degrees of <math>a+11i</math>; however, this triangle is just a reflection of the first triangle by the <math>y</math>-axis, and the signs of <math>a</math> and <math>b</math> are flipped. However, the product <math>ab</math> is unchanged. | '''Note''': There is another solution where the point <math>b+37i</math> is a rotation of <math>-60</math> degrees of <math>a+11i</math>; however, this triangle is just a reflection of the first triangle by the <math>y</math>-axis, and the signs of <math>a</math> and <math>b</math> are flipped. However, the product <math>ab</math> 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 <math>\sqrt{3}</math> can lead you down the path of using simple vectors. | ||
+ | |||
+ | Let's begin: we drop a perpendicular from <math>O</math> to <math>AB</math>. Call the point <math>M</math>, as it is the midpoint of <math>AB</math> as well. Thus, <math>M=(\frac{a+b}{2}, 24)</math>. Now we can use perpendicularity and slope to find that of <math>OM</math> first: we get <math>\frac{48}{a+b}</math>. Its direction 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> directions to get a displacement of <math>[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}]</math>. (Do you see why we had to switch <math>x</math> and <math>y</math> due to the rotation?) | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | '''Note''': This solution was also present in Titu Andreescu and Zuming Feng's "103 Trigonometry Problems". | ||
== See also == | == See also == | ||
{{AIME box|year=1994|num-b=7|num-a=9}} | {{AIME box|year=1994|num-b=7|num-a=9}} |
Revision as of 16:08, 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 can lead you down the path of using simple vectors.
Let's begin: we drop a perpendicular from to . Call the point , as it is the midpoint of as well. Thus, . Now we can use perpendicularity and slope to find that of first: we get . Its direction is . Meanwhile from point we can use a vector with the distance; we have to switch the and directions to get a displacement of . (Do you see why we had to switch and due to the rotation?)
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.