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Difference between revisions of "1976 AHSME Problems/Problem 22"

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If we define point <math>\mathit{P} = (x, y)</math>, and the three points of the triangle as <math>P_1 = (x_1,y_1)</math>, <math>P_2 = (x_2,y_2)</math>, and <math>P_3 = (x_3,y_3)</math>, then we have two equations to work with.   
 
If we define point <math>\mathit{P} = (x, y)</math>, and the three points of the triangle as <math>P_1 = (x_1,y_1)</math>, <math>P_2 = (x_2,y_2)</math>, and <math>P_3 = (x_3,y_3)</math>, then we have two equations to work with.   
  
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First, <math>\mathit{P}</math> is the locus of points satisfying the equation:
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First, <math>\mathit{P}</math> must satisfy the equation:
  
 
<math>|P - P_1|^2 + |P - P_2|^2 + |P - P_3|^2 =\newline
 
<math>|P - P_1|^2 + |P - P_2|^2 + |P - P_3|^2 =\newline
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Thus, <math>a - s^2 = 3[(x-x_C)^2+(y-y_C)^2]</math>
 
Thus, <math>a - s^2 = 3[(x-x_C)^2+(y-y_C)^2]</math>
  
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This means that if <math>a > s^2</math>, we will get a circle centered at <math>P_C</math>.
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This means that when <math>a > s^2</math>, the locus of points <math>P</math> will be a circle centered at <math>P_C</math>.
  
 
Therefore, the correct answer is <math>\boxed{\textbf{(A)}}</math>.
 
Therefore, the correct answer is <math>\boxed{\textbf{(A)}}</math>.

Revision as of 16:11, 18 July 2025

Problem 22

Given an equilateral triangle with side of length $s$, consider the locus of all points $\mathit{P}$ in the plane of the triangle such that the sum of the squares of the distances from $\mathit{P}$ to the vertices of the triangle is a fixed number $a$. This locus

$\textbf{(A) }\text{is a circle if }a>s^2\qquad\\ \textbf{(B) }\text{contains only three points if }a=2s^2\text{ and is a circle if }a>2s^2\qquad\\ \textbf{(C) }\text{is a circle with positive radius only if }s^2<a<2s^2\qquad\\ \textbf{(D) }\text{contains only a finite number of points for any value of }a\qquad\\ \textbf{(E) }\text{is none of these}$

Solution

If we define point $\mathit{P} = (x, y)$, and the three points of the triangle as $P_1 = (x_1,y_1)$, $P_2 = (x_2,y_2)$, and $P_3 = (x_3,y_3)$, then we have two equations to work with.

First, $\mathit{P}$ must satisfy the equation:

$|P - P_1|^2 + |P - P_2|^2 + |P - P_3|^2 =\newline [(x-x_1)^2+(y-y_1)^2]+[(x-x_2)^2+(y-y_2)^2]+[(x-x_3)^2+(y-y_3)^2] = a$

Second, the fact that the three sides of the triangle all have length $s$ gives us: $|P_2 - P_1|^2 + |P_3 - P_2|^2 + |P_1 - P_3|^2 =\newline [(x_2-x_1)^2+(y_2-y_1)^2]+[(x_3-x_2)^2+(y_3-y_2)^2]+[(x_1-x_3)^2+(y_1-y_3)^2] = 3s^2$

Define the center of the triangle as $P_C = (x_C, y_C)$.

Note that $x_C = \frac{x_1+x_2+x_3}{3}$ and $y_C = \frac{y_1+y_2+y_3}{3}$.

Also note that the distance between $P_C$ and each of the triangle vertices is $\frac{\sqrt{3}s}{3}$, so:

$|P_C - P_1|^2 + |P_C - P_2|^2 + |P_C - P_3|^2 =\newline [(x_C-x_1)^2+(y_C-y_1)^2]+[(x_C-x_2)^2+(y_C-y_2)^2]+[(x_C-x_3)^2+(y_C-y_3)^2] = s^2$

Rewrite $(x-x_1) = (x - x_C) + (x_C - x_1)$, and similarly for $(y-y_1), (x-x_2)$, etc.

Now expanding the first sum and using the second sum, we get:

$a = 3[(x-x_C)^2+(y-y_C)^2] + s^2\newline + 2(x - x_C)[(x_C - x_1) + (x_C - x_2) + (x_C - x_3)] + 2(y - y_C)[(y_C - y_1) + (y_C - y_2) + (y_C - y_3)]$

Because $P_C$ is centroid of the triangle, its coordinates are the averages of the coordinates of the three points. Thus, $3x_C = x_1 + x_2 + x_3$, and the latter parts of the equations are zero.

Thus, $a - s^2 = 3[(x-x_C)^2+(y-y_C)^2]$

This means that when $a > s^2$, the locus of points $P$ will be a circle centered at $P_C$.

Therefore, the correct answer is $\boxed{\textbf{(A)}}$.

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

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