1976 AHSME Problems/Problem 22
Problem 22
Given an equilateral triangle with side of length , consider the locus of all points in the plane of the triangle such that the sum of the squares of the distances from to the vertices of the triangle is a fixed number . This locus
Solution
We can consider the locus of points as the set of points satisfying the equation: where , , and are the coordinates of the three vertices of the equilateral triangle.
If we simplify this equation, we get:
Since the vertices of the triangle are fixed, the left side of this equation is also fixed. Thus, the locus of points is a circle centered at the centroid of the triangle with radius .
We can now determine which of the answer choices is correct:
- This is correct, as the radius of the circle will be greater than zero when .
- This is incorrect, as the locus of points is always a circle.
- This is incorrect, as the locus of points is always a circle.
- This is incorrect, as the locus of points is always a circle.
- This is incorrect, as the locus of points is always a circle.
Therefore, the correct answer is .
See Also
1976 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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