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 with these beastly numbers doesn't seem promising. How about properties of equilateral triangles? and perpendiculars inspires 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".
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