1994 AIME Problems/Problem 8
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 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".
See also
1994 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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