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2018 AMC 10B Problems/Problem 12: Difference between revisions

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==Problem==
{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #8]] and [[2018 AMC 10B Problems|2018 AMC 10B #12]]}}


Line segment <math>\overline{AB}</math> is a diameter of a circle with <math>AB=24</math>. Point <math>C</math>, not equal to <math>A</math> or <math>B</math>, lies on the circle. As point <math>C</math> moves around the circle, the centroid (center of mass) of <math>\triangle{ABC}</math> traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?
==Problem ==


<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>
Line segment <math>\overline{AB}</math> is a diameter of a circle with <math>AB = 24</math>. Point <math>C</math>, not equal to <math>A</math> or <math>B</math>, lies on the circle. As point <math>C</math> moves around the circle, the centroid (center of mass) of <math>\triangle ABC</math> 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 (Coordinate Bash)==
<math>\textbf{(A) } 25 \qquad \textbf{(B) } 38 \qquad \textbf{(C) } 50  \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 75  </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


==Solution 2 (no coordinates)==
==Solution 1==
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 (length <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
By the Inscribed Angle Theorem, <math>\triangle ABC</math> is a right triangle with <math>\angle C=90^{\circ}.</math> So, its circumcenter is the midpoint of <math>\overline{AB},</math> and its median from <math>C</math> is half as long as <math>\overline{AB}.</math> For each <math>\triangle ABC,</math> let <math>O</math> and <math>G</math> be its circumcenter and centroid, respectively. It follows that <math>OA=OB=OC=12.</math> In any triangle, since the centroid divides each median into parts in the ratio <math>2:1,</math> with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex, we have <math>OG=\frac13 OC=4.</math>
 
As shown below, <math>\triangle ABC_1</math> and <math>\triangle ABC_2</math> are two shapes of <math>\triangle ABC</math> with centroids <math>G_1</math> and <math>G_2,</math> 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>
Therefore, point <math>G</math> traces out a circle (missing two points) with the center <math>O</math> and the radius <math>\overline{OG},</math> as indicated in red. To the nearest positive integer, the area of the region bounded by the red curve is <math>\pi\cdot OG^2=16\pi\approx\boxed{\textbf{(C) } 50}.</math>
 
~MRENTHUSIASM ~megacleverstarfish15
 
==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 = \left(\frac{-12 + 12 + x}{3}, \frac{0 + 0 + y}{3}\right) = \left(\frac x3,\frac y3\right)</math>. Thus, <math>G</math> traces out a circle with a radius <math>\frac13</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>\frac{12}{3} = 4</math>, which has area <math>16\pi\approx \boxed{\textbf{(C) } 50}</math>.
 
==Solution 3==
First we can draw a few conclusions from the given information. Firstly we can see clearly that the distance from the centroid to the center of the circle will remain the same no matter <math>C</math> is on the circle. Also we can see that because the two legs of every triangles will always originate on the diameter, using inscribed angle rules, we know that <math>\angle C = \frac{180^\circ}{2} = 90^\circ</math>. Now we know that all triangles <math>ABC</math> will be a right triangle. We also know that the closed curve will simply be a circle with radius equal to the centroid of each triangle. We can now pick any arbitrary triangle, calculate its centroid, and the plug it in to the area formula. Using a <math>45^\circ</math>-<math>45^\circ</math>-<math>90^\circ</math> triangle in conjunction with the properties of a centroid, we can quickly see that the length of the centroid is <math>4</math> now we can plug it in to the area formula where we get <math>16\pi\approx\boxed{\textbf{(C) } 50}</math>.
 
==Video Solution (HOW TO THINK CRITICALLY!!!)==
https://youtu.be/CXOOhQVsOo8
 
~Education, the Study of Everything


==See Also==
==See Also==
{{AMC10 box|year=2018|ab=B|num-b=11|num-a=13}}
{{AMC10 box|year=2018|ab=B|num-b=11|num-a=13}}
{{AMC12 box|year=2018|ab=B|num-a=9|num-b=7}}
{{MAA Notice}}
{{MAA Notice}}
[[Category:Intermediate Geometry Problems]]

Latest revision as of 23:49, 27 October 2025

The following problem is from both the 2018 AMC 12B #8 and 2018 AMC 10B #12, so both problems redirect to this page.

Problem

Line segment $\overline{AB}$ is a diameter of a circle with $AB = 24$. Point $C$, not equal to $A$ or $B$, lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?

$\textbf{(A) } 25 \qquad \textbf{(B) } 38 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 75$

Solution 1

By the Inscribed Angle Theorem, $\triangle ABC$ is a right triangle with $\angle C=90^{\circ}.$ So, its circumcenter is the midpoint of $\overline{AB},$ and its median from $C$ is half as long as $\overline{AB}.$ For each $\triangle ABC,$ let $O$ and $G$ be its circumcenter and centroid, respectively. It follows that $OA=OB=OC=12.$ In any triangle, since the centroid divides each median into parts in the ratio $2:1,$ with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex, we have $OG=\frac13 OC=4.$

As shown below, $\triangle ABC_1$ and $\triangle ABC_2$ are two shapes of $\triangle ABC$ with centroids $G_1$ and $G_2,$ 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] Therefore, point $G$ traces out a circle (missing two points) with the center $O$ and the radius $\overline{OG},$ as indicated in red. To the nearest positive integer, the area of the region bounded by the red curve is $\pi\cdot OG^2=16\pi\approx\boxed{\textbf{(C) } 50}.$

~MRENTHUSIASM ~megacleverstarfish15

Solution 2

We assign coordinates. Let $A = (-12,0)$, $B = (12,0)$, and $C = (x,y)$ lie on the circle $x^2 +y^2 = 12^2$. Then, the centroid of $\triangle ABC$ is $G = \left(\frac{-12 + 12 + x}{3}, \frac{0 + 0 + y}{3}\right) = \left(\frac x3,\frac y3\right)$. Thus, $G$ traces out a circle with a radius $\frac13$ of the radius of the circle that point $C$ travels on. Thus, $G$ traces out a circle of radius $\frac{12}{3} = 4$, which has area $16\pi\approx \boxed{\textbf{(C) } 50}$.

Solution 3

First we can draw a few conclusions from the given information. Firstly we can see clearly that the distance from the centroid to the center of the circle will remain the same no matter $C$ is on the circle. Also we can see that because the two legs of every triangles will always originate on the diameter, using inscribed angle rules, we know that $\angle C = \frac{180^\circ}{2} = 90^\circ$. Now we know that all triangles $ABC$ will be a right triangle. We also know that the closed curve will simply be a circle with radius equal to the centroid of each triangle. We can now pick any arbitrary triangle, calculate its centroid, and the plug it in to the area formula. Using a $45^\circ$-$45^\circ$-$90^\circ$ triangle in conjunction with the properties of a centroid, we can quickly see that the length of the centroid is $4$ now we can plug it in to the area formula where we get $16\pi\approx\boxed{\textbf{(C) } 50}$.

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/CXOOhQVsOo8

~Education, the Study of Everything

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
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

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