Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AMC 10B Problems/Problem 20"

(Created page with "I got 14")
 
(→‎Solution 2)
 
(78 intermediate revisions by 34 users not shown)
Line 1: Line 1:
−
I got 14
+
{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #20]] and [[2019 AMC 12B Problems|2019 AMC 12B #15]]}}
 +
 
 +
== Problem ==
 +
As shown in the figure, line segment <math>\overline{AD}</math> is trisected by points <math>B</math> and <math>C</math> so that <math>AB=BC=CD=2.</math> Three semicircles of radius <math>1,</math> <math>\overarc{AEB},\overarc{BFC},</math> and <math>\overarc{CGD},</math> have their diameters on <math>\overline{AD},</math> lie in the same halfplane determined by line <math>AD</math>, and are tangent to line <math>EG</math> at <math>E,F,</math> and <math>G,</math> respectively. A circle of radius <math>2</math> has its center on <math>F. </math> The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form
 +
<cmath>\frac{a}{b}\cdot\pi-\sqrt{c}+d,</cmath>
 +
where <math>a,b,c,</math> and <math>d</math> are positive integers and <math>a</math> and <math>b</math> are relatively prime. What is <math>a+b+c+d</math>?
 +
 
 +
<asy>
 +
size(6cm);
 +
filldraw(circle((0,0),2), grey);
 +
filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0));
 +
filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0));
 +
filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0));
 +
dot((-3,-1));
 +
label("$A$",(-3,-1),S);
 +
dot((-2,0));
 +
label("$E$",(-2,0),NW);
 +
dot((-1,-1));
 +
label("$B$",(-1,-1),S);
 +
dot((0,0));
 +
label("$F$",(0,0),N);
 +
dot((1,-1));
 +
label("$C$",(1,-1), S);
 +
dot((2,0));
 +
label("$G$", (2,0),NE);
 +
dot((3,-1));
 +
label("$D$", (3,-1), S);
 +
</asy>
 +
 
 +
<math>\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17</math>
 +
 
 +
==Solution 1==
 +
 
 +
[[File:10b20.png]]
 +
 
 +
~mathboy282
 +
 
 +
== Solution 2==
 +
Divide the circle into four parts: the top semicircle by connecting E, F, and G(<math>A</math>); the bottom sector (<math>B</math>), whose arc angle is <math>120^{\circ}</math> because the large circle's radius is <math>2</math> and the short length (the radius of the smaller semicircles) is <math>1</math>, giving a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle; the triangle formed by the radii of <math>A</math> and the chord (<math>C</math>); and the four parts which are the corners of a circle inscribed in a square (<math>D</math>). Then the area is <math>A + B - C + D</math> (in <math>B-C</math>, we find the area of the bottom shaded region, and in <math>D</math> we find the area of the shaded region above the semicircles but below the diameter).
 +
 
 +
The area of <math>A</math> is <math>\frac{1}{2} \pi \cdot 2^2 = 2\pi</math>.
 +
 
 +
The area of <math>B</math> is <math>\frac{120^{\circ}}{360^{\circ}} \pi \cdot 2^2 = \frac{4\pi}{3}</math>.
 +
 
 +
For the area of <math>C</math>, the radius of <math>2</math>, and the distance of <math>1</math> (the smaller semicircles' radius) to <math>BC</math>, creates two <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangles, so <math>C</math>'s area is <math>2 \cdot \frac{1}{2} \cdot 1 \cdot \sqrt{3} = \sqrt{3}</math>.
 +
 
 +
The area of <math>D</math> is <math>4 \cdot 1-\frac{1}{4}\pi \cdot 2^2=4-\pi</math>.
 +
 
 +
Hence, finding <math>A+B-C+D</math>, the desired area is <math>\frac{7\pi}{3}-\sqrt{3}+4</math>, so the answer is <math>7+3+3+4=\boxed{\textbf{(E) } 17}</math>.
 +
 
 +
----------
 +
 
 +
Note from ~milquetoast: I found this solution incredibly unspecific and difficult to understand, especially in defining C,  because of the wording. I think what this solution is trying to say is the same as the first video solution down below.
 +
 
 +
Note: Wow that is harsh
 +
 
 +
Note from ~<B+: I also believe this solution is worded inefficiently and is not very comprehensible. I hope that someone can make this solution a little bit more understandably good as I'm not very good at explaining things so I cannot. :)
 +
 
 +
== Solution 3 ==
 +
 
 +
First we have to solve the area of the non-shaded area(the semicircles) that are in Circle <math>F</math>.The middle semicircle has area <math>\frac12\pi</math> and the other two have about half of their are inside the circle = <math>\frac14\pi\ + \frac14\pi\ + \frac12\pi\ = \pi</math>. Then we subtract the part of the quartercircle that isn't in Circle <math>F</math>. This is an area equal to that of a triangle minus an minor segment. The height of the triangle is the radius of the semicircles, which is <math>1</math>. The length is the radius of Circle <math>F</math> minus the length from the center of the middle semicircle up to until it is on the edge of the circle. Using the Pythagorean Theorem we can figure out that the length is: <cmath>\sqrt{2^2 - 1^2} = \sqrt{3}.</cmath> This means that the length of the triangle is <math>2 - \sqrt{3}</math> and so the area of the triangle is <math>\frac{2 - \sqrt{3}}{2}</math>. For the area of the segment, it's the area of the sector minus the area of the triangle. The triangle's length is the radius of <math>F</math>: <math>2</math>, while its height is the radius of the semicircles: <math>1</math>, so the area is 1. The  angle is <math>30^{\circ}</math> as the hypotenuse is the radius of <math>F</math> and the opposite side is the radius of the semicircles, which means the area is <math>\frac{1}{12}</math> of the whole area, which is <math>4\pi</math> so the area of the sector is <math>\frac{\pi}{3}</math> and the area of the segment is <math>\frac{\pi}{3} - 1</math> and so the area of the part of the quartercircles that stick out of Circle <math>F</math> is: <cmath>(\frac{2 - \sqrt{3}}{2})-(\frac{\pi}{3} - 1) = \frac{2 - \sqrt{3}}{2} + 1 - \frac{\pi}{3} = \frac{4 - \sqrt{3}}{2} - \frac{\pi}{3}.</cmath>
 +
 
 +
Since there are two, one for each side, we have to multiply it by 2, so we have <math>{4 - \sqrt{3}} - \frac{2\pi}{3}</math>, which we subtract from <math>\pi</math> which gets us <math>\frac{5\pi}{3} - 4 + \sqrt{3}</math> which we subtract from <math>4\pi</math> <math>=</math> <math>\frac{12\pi}{3}</math>, which is <math>\frac{7\pi}{3} + 4 - \sqrt{3}</math> so we get <math>7+3+4+3=\boxed{\textbf{(E) } 17}</math>.
 +
 
 +
==Video Solution by Education, the Study of Everything==
 +
https://youtu.be/nhDYNevwH1g
 +
 
 +
~Education, the Study of Everything
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/t3EWtMnJu2Y?t=516
 +
 
 +
~pi_is_3.14
 +
 
 +
== See Also ==
 +
{{AMC10 box|year=2019|ab=B|num-b=19|num-a=21}}
 +
{{AMC12 box|year=2019|ab=B|num-b=14|num-a=16}}
 +
{{MAA Notice}}

Latest revision as of 19:15, 27 October 2024

The following problem is from both the 2019 AMC 10B #20 and 2019 AMC 12B #15, so both problems redirect to this page.

Problem

As shown in the figure, line segment $\overline{AD}$ is trisected by points $B$ and $C$ so that $AB=BC=CD=2.$ Three semicircles of radius $1,$ $\overarc{AEB},\overarc{BFC},$ and $\overarc{CGD},$ have their diameters on $\overline{AD},$ lie in the same halfplane determined by line $AD$, and are tangent to line $EG$ at $E,F,$ and $G,$ respectively. A circle of radius $2$ has its center on $F.$ The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form \[\frac{a}{b}\cdot\pi-\sqrt{c}+d,\] where $a,b,c,$ and $d$ are positive integers and $a$ and $b$ are relatively prime. What is $a+b+c+d$?

[asy] size(6cm); filldraw(circle((0,0),2), grey); filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0)); dot((-3,-1)); label("$A$",(-3,-1),S); dot((-2,0)); label("$E$",(-2,0),NW); dot((-1,-1)); label("$B$",(-1,-1),S); dot((0,0)); label("$F$",(0,0),N); dot((1,-1)); label("$C$",(1,-1), S); dot((2,0)); label("$G$", (2,0),NE); dot((3,-1)); label("$D$", (3,-1), S); [/asy]

$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17$

Solution 1

10b20.png

~mathboy282

Solution 2

Divide the circle into four parts: the top semicircle by connecting E, F, and G($A$); the bottom sector ($B$), whose arc angle is $120^{\circ}$ because the large circle's radius is $2$ and the short length (the radius of the smaller semicircles) is $1$, giving a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle; the triangle formed by the radii of $A$ and the chord ($C$); and the four parts which are the corners of a circle inscribed in a square ($D$). Then the area is $A + B - C + D$ (in $B-C$, we find the area of the bottom shaded region, and in $D$ we find the area of the shaded region above the semicircles but below the diameter).

The area of $A$ is $\frac{1}{2} \pi \cdot 2^2 = 2\pi$.

The area of $B$ is $\frac{120^{\circ}}{360^{\circ}} \pi \cdot 2^2 = \frac{4\pi}{3}$.

For the area of $C$, the radius of $2$, and the distance of $1$ (the smaller semicircles' radius) to $BC$, creates two $30^{\circ}-60^{\circ}-90^{\circ}$ triangles, so $C$'s area is $2 \cdot \frac{1}{2} \cdot 1 \cdot \sqrt{3} = \sqrt{3}$.

The area of $D$ is $4 \cdot 1-\frac{1}{4}\pi \cdot 2^2=4-\pi$.

Hence, finding $A+B-C+D$, the desired area is $\frac{7\pi}{3}-\sqrt{3}+4$, so the answer is $7+3+3+4=\boxed{\textbf{(E) } 17}$.

Note from ~milquetoast: I found this solution incredibly unspecific and difficult to understand, especially in defining C, because of the wording. I think what this solution is trying to say is the same as the first video solution down below.

Note: Wow that is harsh

Note from ~<B+: I also believe this solution is worded inefficiently and is not very comprehensible. I hope that someone can make this solution a little bit more understandably good as I'm not very good at explaining things so I cannot. :)

Solution 3

First we have to solve the area of the non-shaded area(the semicircles) that are in Circle $F$.The middle semicircle has area $\frac12\pi$ and the other two have about half of their are inside the circle = $\frac14\pi\ + \frac14\pi\ + \frac12\pi\ = \pi$. Then we subtract the part of the quartercircle that isn't in Circle $F$. This is an area equal to that of a triangle minus an minor segment. The height of the triangle is the radius of the semicircles, which is $1$. The length is the radius of Circle $F$ minus the length from the center of the middle semicircle up to until it is on the edge of the circle. Using the Pythagorean Theorem we can figure out that the length is: \[\sqrt{2^2 - 1^2} = \sqrt{3}.\] This means that the length of the triangle is $2 - \sqrt{3}$ and so the area of the triangle is $\frac{2 - \sqrt{3}}{2}$. For the area of the segment, it's the area of the sector minus the area of the triangle. The triangle's length is the radius of $F$: $2$, while its height is the radius of the semicircles: $1$, so the area is 1. The angle is $30^{\circ}$ as the hypotenuse is the radius of $F$ and the opposite side is the radius of the semicircles, which means the area is $\frac{1}{12}$ of the whole area, which is $4\pi$ so the area of the sector is $\frac{\pi}{3}$ and the area of the segment is $\frac{\pi}{3} - 1$ and so the area of the part of the quartercircles that stick out of Circle $F$ is: \[(\frac{2 - \sqrt{3}}{2})-(\frac{\pi}{3} - 1) = \frac{2 - \sqrt{3}}{2} + 1 - \frac{\pi}{3} = \frac{4 - \sqrt{3}}{2} - \frac{\pi}{3}.\]

Since there are two, one for each side, we have to multiply it by 2, so we have ${4 - \sqrt{3}} - \frac{2\pi}{3}$, which we subtract from $\pi$ which gets us $\frac{5\pi}{3} - 4 + \sqrt{3}$ which we subtract from $4\pi$ $=$ $\frac{12\pi}{3}$, which is $\frac{7\pi}{3} + 4 - \sqrt{3}$ so we get $7+3+4+3=\boxed{\textbf{(E) } 17}$.

Video Solution by Education, the Study of Everything

https://youtu.be/nhDYNevwH1g

~Education, the Study of Everything

Video Solution by OmegaLearn

https://youtu.be/t3EWtMnJu2Y?t=516

~pi_is_3.14

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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
2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_20&oldid=230671"