2018 AMC 12B Problems/Problem 8: Difference between revisions
MRENTHUSIASM (talk | contribs) |
No edit summary |
||
| Line 5: | Line 5: | ||
<math>\textbf{(A) } 25 \qquad \textbf{(B) } 38 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 75 </math> | <math>\textbf{(A) } 25 \qquad \textbf{(B) } 38 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 75 </math> | ||
==Solution== | ==Solution 1== | ||
For each <math>\triangle ABC,</math> note that the length of one median is <math>OC=12.</math> Let <math>G</math> be the centroid of <math>\triangle ABC.</math> It follows that <math>OG=\frac13 OC=4.</math> | For each <math>\triangle ABC,</math> note that the length of one median is <math>OC=12.</math> Let <math>G</math> be the centroid of <math>\triangle ABC.</math> It follows that <math>OG=\frac13 OC=4.</math> | ||
| Line 41: | Line 41: | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
==Solution 2== | |||
We assign coordinates. Let <math>A = (-12,0)</math>, <math>B = (12,0)</math>, and <math>C = (x,y)</math> lie on the circle <math>x^2 +y^2 = 12^2</math>. Then, the centroid of <math>\triangle ABC</math> is <math>G = ((-12 + 12 + x)/3, (0 + 0 + y)/3) = (x/3,y/3)</math>. Thus, <math>G</math> traces out a circle with a radius <math>1/3</math> of the radius of the circle that point <math>C</math> travels on. Thus, <math>G</math> traces out a circle of radius <math>12/3 = 4</math>, which has area <math>16\pi\approx 50</math>, which is <math>\boxed{\textbf{(C) }50}.</math> | |||
==See Also== | ==See Also== | ||
Revision as of 18:01, 14 June 2022
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
For each 
note that the length of one median is 
Let 
be the centroid of 
It follows that 
As shown below, 
and 
are two shapes of with centroids 
and 
respectively:
![[asy] /* Made by MRENTHUSIASM */ size(200); pair O, A, B, C1, C2, G1, G2, M1, M2; O = (0,0); A = (-12,0); B = (12,0); C1 = (36/5,48/5); C2 = (-96/17,-180/17); G1 = O + 1/3 * C1; G2 = O + 1/3 * C2; M1 = (4,0); M2 = (-4,0); draw(Circle(O,12)); draw(Circle(O,4),red); dot("$O$", O, (3/5,-4/5), linewidth(4.5)); dot("$A$", A, W, linewidth(4.5)); dot("$B$", B, E, linewidth(4.5)); dot("$C_1$", C1, dir(C1), linewidth(4.5)); dot("$C_2$", C2, dir(C2), linewidth(4.5)); dot("$G_1$", G1, 1.5*E, linewidth(4.5)); dot("$G_2$", G2, 1.5*W, linewidth(4.5)); draw(A--B^^A--C1--B^^A--C2--B); draw(O--C1^^O--C2); dot(M1,red+linewidth(0.8),UnFill); dot(M2,red+linewidth(0.8),UnFill); [/asy]](http://latex.artofproblemsolving.com/f/7/d/f7d390ddcbaa38a41dd608a1c53e7396c707a2a3.png)
Therefore, point traces out a circle (missing two points) with the center 
and the radius 
as indicated in red. To the nearest positive integer, the area of the region bounded by the red curve is 
~MRENTHUSIASM
Solution 2
We assign coordinates. Let 
, 
, and 
lie on the circle 
. Then, the centroid of is 
. Thus, traces out a circle with a radius 
of the radius of the circle that point travels on. Thus, traces out a circle of radius 
, which has area 
, which is 
See Also
| 2018 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 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 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
