Difference between revisions of "1976 AHSME Problems/Problem 22"
(Created page with "== Problem 22 == Given an equilateral triangle with side of length <math>s</math>, consider the locus of all points <math>\mathit{P}</math> in the plane of the triangle such...") |
Agentdabber (talk | contribs) m (→Problem 22) |
||
Line 8: | Line 8: | ||
\textbf{(C) }\text{is a circle with positive radius only if }s^2<a<2s^2\qquad\\ | \textbf{(C) }\text{is a circle with positive radius only if }s^2<a<2s^2\qquad\\ | ||
\textbf{(D) }\text{contains only a finite number of points for any value of }a\qquad\\ | \textbf{(D) }\text{contains only a finite number of points for any value of }a\qquad\\ | ||
− | \textbf{(E) }\text{is none of these} </math> | + | \textbf{(E) }\text{is none of these} </math> |
− | |||
− | |||
== Solution == | == Solution == |
Latest revision as of 18:45, 10 November 2023
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.