Difference between revisions of "2018 AMC 10B Problems/Problem 12"
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− | + | {{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #8]] and [[2018 AMC 10B Problems|2018 AMC 10B #12]]}} | |
− | + | ==Problem == | |
− | <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? |
− | + | <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> | |
− | + | ||
+ | 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 | ||
+ | |||
+ | ==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 14:19, 5 June 2023
- The following problem is from both the 2018 AMC 12B #8 and 2018 AMC 10B #12, so both problems redirect to this page.
Contents
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:
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
.
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 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
. Now we know that all triangles
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
-
-
triangle in conjunction with the properties of a centroid, we can quickly see that the length of the centroid is
now we can plug it in to the area formula where we get
.
Video Solution (HOW TO THINK CRITICALLY!!!)
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.