Difference between revisions of "2019 AMC 10B Problems/Problem 20"
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{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #20]] and [[2019 AMC 12B Problems|2019 AMC 12B #15]]}} | {{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #20]] and [[2019 AMC 12B Problems|2019 AMC 12B #15]]}} | ||
− | ==Problem== | + | |
+ | == 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> 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 | 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> 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> | <cmath>\frac{a}{b}\cdot\pi-\sqrt{c}+d,</cmath> | ||
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label("$D$", (3,-1), S); | label("$D$", (3,-1), S); | ||
</asy> | </asy> | ||
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<math>\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17</math> | <math>\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17</math> | ||
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− | + | == Solutions == | |
+ | === Solution 1 === | ||
+ | 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>. | |
− | <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>. | |
− | == | + | === Video Solution === |
+ | Video Solution from Youtube- https://www.youtube.com/watch?v=ZbWOZMfXtL8 | ||
+ | == See Also == | ||
{{AMC10 box|year=2019|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2019|ab=B|num-b=19|num-a=21}} | ||
{{AMC12 box|year=2019|ab=B|num-b=14|num-a=16}} | {{AMC12 box|year=2019|ab=B|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
− |
Latest revision as of 14:28, 3 January 2021
- 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 is trisected by points and so that Three semicircles of radius and have their diameters on and are tangent to line at and respectively. A circle of radius has its center on The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form where and are positive integers and and are relatively prime. What is ?
Solutions
Solution 1
Divide the circle into four parts: the top semicircle by connecting E, F, and G(); the bottom sector (), whose arc angle is because the large circle's radius is and the short length (the radius of the smaller semicircles) is , giving a triangle; the triangle formed by the radii of and the chord (); and the four parts which are the corners of a circle inscribed in a square (). Then the area is (in , we find the area of the bottom shaded region, and in we find the area of the shaded region above the semicircles but below the diameter).
The area of is .
The area of is .
For the area of , the radius of , and the distance of (the smaller semicircles' radius) to , creates two triangles, so 's area is .
The area of is .
Hence, finding , the desired area is , so the answer is .
Video Solution
Video Solution from Youtube- https://www.youtube.com/watch?v=ZbWOZMfXtL8
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.