1994 AIME Problems/Problem 8
The points , , and are the vertices of an equilateral triangle. Find the value of .
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.
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".
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