Difference between revisions of "2018 AMC 10B Problems/Problem 12"
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<math>\textbf{(A)} \text{ 25} \qquad \textbf{(B)} \text{ 38} \qquad \textbf{(C)} \text{ 50} \qquad \textbf{(D)} \text{ 63} \qquad \textbf{(E)} \text{ 75}</math> | <math>\textbf{(A)} \text{ 25} \qquad \textbf{(B)} \text{ 38} \qquad \textbf{(C)} \text{ 50} \qquad \textbf{(D)} \text{ 63} \qquad \textbf{(E)} \text{ 75}</math> | ||
− | ==Solution (Coordinate Bash)== | + | ==Solution 1 (Coordinate Bash)== |
Let <math>A=(-12,0),B=(12,0)</math>. Therefore, <math>C</math> lies on the circle with equation <math>x^2+y^2=144</math>. Let it have coordinates <math>(x,y)</math>. Since we know the centroid of a triangle with vertices with coordinates of <math>(x_1,y_1),(x_2,y_2),(x_3,y_3)</math> is <math>\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)</math>, the centroid of <math>\triangle ABC</math> is <math>\left(\frac{x}{3},\frac{y}{3}\right)</math>. Because <math>x^2+y^2=144</math>, we know that <math>\left(\frac{x}{3}\right)^2+\left(\frac{y}{3}\right)^2=16</math>, so the curve is a circle centered at the origin. Therefore, its area is <math>16\pi\approx \boxed{\text{(C) }50}</math>. | Let <math>A=(-12,0),B=(12,0)</math>. Therefore, <math>C</math> lies on the circle with equation <math>x^2+y^2=144</math>. Let it have coordinates <math>(x,y)</math>. Since we know the centroid of a triangle with vertices with coordinates of <math>(x_1,y_1),(x_2,y_2),(x_3,y_3)</math> is <math>\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)</math>, the centroid of <math>\triangle ABC</math> is <math>\left(\frac{x}{3},\frac{y}{3}\right)</math>. Because <math>x^2+y^2=144</math>, we know that <math>\left(\frac{x}{3}\right)^2+\left(\frac{y}{3}\right)^2=16</math>, so the curve is a circle centered at the origin. Therefore, its area is <math>16\pi\approx \boxed{\text{(C) }50}</math>. | ||
-tdeng | -tdeng |
Latest revision as of 16:35, 27 January 2021
Problem
Line segment is a diameter of a circle with . Point , not equal to or , lies on the circle. As point moves around the circle, the centroid (center of mass) of traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?
Solution 1 (Coordinate Bash)
Let . Therefore, lies on the circle with equation . Let it have coordinates . Since we know the centroid of a triangle with vertices with coordinates of is , the centroid of is . Because , we know that , so the curve is a circle centered at the origin. Therefore, its area is . -tdeng
Solution 2 (No Coordinates)
We know the centroid of a triangle splits the medians into segments of ratio , and the median of the triangle that goes to the center of the circle is the radius (length ), so the length from the centroid of the triangle to the center of the circle is always . The area of a circle with radius is , or around . -That_Crazy_Book_Nerd
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.