Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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...")
 
(→‎Solution)
 
(3 intermediate revisions by 2 users not shown)
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>
−
 
 
−
[[1976 AHSME Problems/Problem 22|Solution]]
 
  
 
== Solution ==
 
== Solution ==
−
We can consider the locus of points <math>\mathit{P}</math> as the set of points satisfying the equation:
+
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.
−
<math>[(x_1-x_2)^2+(y_1-y_2)^2+(x_1-x_3)^2+(y_1-y_3)^2+(x_2-x_3)^2+(y_2-y_3)^2=a]</math>
 
−
where <math>(x_1,y_1)</math>, <math>(x_2,y_2)</math>, and <math>(x_3,y_3)</math> are the coordinates of the three vertices of the equilateral triangle.
 
  
−
If we simplify this equation, we get:
+
First, <math>\mathit{P}</math> must satisfy the equation:
−
<math>[2(x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_3^2)-2(x_1x_2+x_1x_3+x_2x_3+y_1y_2+y_1y_3+y_2y_3)=a]</math>
 
  
−
Since the vertices of the triangle are fixed, the left side of this equation is also fixed. Thus, the locus of points <math>\mathit{P}</math> is a circle centered at the centroid of the triangle with radius <math>\sqrt{a}</math>.
+
<math>|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</math>
  
−
We can now determine which of the answer choices is correct:
+
Second, the fact that the three sides of the triangle all have length <math>s</math> gives us:
 +
<math>|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</math>
  
−
<math>\textbf{(A) }\text{is a circle if }a>s^2</math> - This is correct, as the radius of the circle will be greater than zero when <math>a>s^2</math>.
+
Define the center of the triangle as <math>P_C = (x_C, y_C)</math>.
  
−
<math>\textbf{(B) }\text{contains only three points if }a=2s^2\text{ and is a circle if }a>2s^2</math> - This is incorrect, as the locus of points <math>\mathit{P}</math> is always a circle.
+
Note that <math>x_C = \frac{x_1+x_2+x_3}{3}</math> and <math>y_C = \frac{y_1+y_2+y_3}{3}</math>.
  
−
<math>\textbf{(C) }\text{is a circle with positive radius only if }s^2<a<2s^2</math> - This is incorrect, as the locus of points <math>\mathit{P}</math> is always a circle.
+
Also note that the distance between <math>P_C</math> and each of the triangle vertices is <math>\frac{\sqrt{3}s}{3}</math>, so:
  
−
<math>\textbf{(D) }\text{contains only a finite number of points for any value of }a</math> - This is incorrect, as the locus of points <math>\mathit{P}</math> is always a circle.
+
<math>|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</math>
  
−
<math>\textbf{(E) }\text{is none of these}</math> - This is incorrect, as the locus of points <math>\mathit{P}</math> is always a circle.
+
Rewrite <math>(x-x_1) = (x - x_C) + (x_C - x_1)</math>, and similarly for <math>(y-y_1), (x-x_2)</math>, etc.
 +
 
 +
Now expanding the first sum and using the second sum, we get:
 +
 
 +
<math>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)]</math>
 +
 
 +
Because <math>P_C</math> is centroid of the triangle, its coordinates are the averages of the coordinates of the three points.  Thus, <math>3x_C = x_1 + x_2 + x_3</math>, and the latter parts of the above equations are zero.
 +
 
 +
Thus, <math>a - s^2 = 3[(x-x_C)^2+(y-y_C)^2]</math>
 +
 
 +
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>.
 +
 +
This problem could be also solved with coordinate geometry, where we place <math>P_1</math> at the origin and <math>P_2</math> along the positive x-axis, or even let <math>P_C</math> be the origin and place <math>P_1</math> along the negative x-axis.
 +
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1976|num-b=21|num-a=23}}
 
{{AHSME box|year=1976|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:18, 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 above 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 $\boxed{\textbf{(A)}}$.

This problem could be also solved with coordinate geometry, where we place $P_1$ at the origin and $P_2$ along the positive x-axis, or even let $P_C$ be the origin and place $P_1$ along the negative x-axis.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1976_AHSME_Problems/Problem_22&oldid=254612"