Difference between revisions of "2018 AMC 10B Problems/Problem 12"
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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 | ||
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| + | ==Solution 2 (no coordinates)== | ||
| + | We know the centroid of a triangle splits the medians into segments of ratio 2:1 and the median of the triangle that goes to the center of the circle is the radius (<math>12</math>), so the length from the centroid of the triangle to the center of the circle is always <math>\dfrac{1}{3} \cdot 12 = 4</math>. The area of a circle with radius <math>4</math> is <math>16\pi</math>, or around <math>\boxed{\textbf{(C)} \text{ 50}}</math>. | ||
| + | -That_Crazy_Book_Nerd | ||
==See Also== | ==See Also== | ||
Revision as of 16:01, 16 February 2018
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
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 2:1 and the median of the triangle that goes to the center of the circle is the radius (
), 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. 
