Difference between revisions of "1994 AIME Problems/Problem 8"
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Revision as of 23:42, 15 March 2020
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 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".
Solution Three
Plot this equilateral triangle on the complex plane. Translate the equilateral triangle so that its centroid is located at the origin. (The centroid can be found by taking the average of the three vertices of the triangle, which gives . The new coordinates of the equilateral triangle are . These three vertices are solutions of a cubic polynomial of form . By Vieta's Formulas, the sum of the paired roots of the cubic polynomial are zero. (Or for the three roots r1, r2, and r3, r1r2 + r2r3 + r3r1 = 0.) The vertices of the equilateral triangle represent the roots of a polynomial, so the vertices can be plugged into the above equation. Because both the real and complex components of the equation have to sum to zero, you really have two equations. Multiply out the equation given by Vieta's Formulas and isolate the ones with imaginary components. Simplify that equation, and that gives the equation Now use the equation with only real parts. This should give you a quadratic . Use your previously obtained equation to plug in for a and solve for b, which should yield . a is then . Multiplying a and b yields .
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 |
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