Difference between revisions of "2019 AMC 10B Problems/Problem 20"
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| − | For this solution to be a tad more clear, we are finding the area of the sector in B of 120 degrees because the large circle radius is 2, and the short length (the radius of the semicircle) is 1, and so the triangle is a 30-60-90 triangle. In A, we find the top semicircle part, in B minus C, we find the area of the shaded region above the semicircles but below the diameter, and in D we find the bottom shaded region. - edited by IronicNinja | + | For this solution to be a tad more clear, we are finding the area of the sector in B of 120 degrees because the large circle radius is 2, and the short length (the radius of the semicircle) is 1, and so the triangle is a 30-60-90 triangle. In A, we find the top semicircle part, in B minus C, we find the area of the shaded region above the semicircles but below the diameter, and in D we find the bottom shaded region. |
| + | - edited by IronicNinja | ||
==See Also== | ==See Also== | ||
Revision as of 21:32, 17 February 2019
- 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
![\[\frac{a}{b}\cdot\pi-\sqrt{c}+d,\]](http://latex.artofproblemsolving.com/d/8/b/d8bdd0e53bdd2b5b50aea9510be4bd6c45e2b867.png)
where 
and 
are positive integers and 
and 
are relatively prime. What is 
?
![[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]](http://latex.artofproblemsolving.com/a/c/9/ac9c68e87cdfb2dde115a063764bd2d48bf835e2.png)
Solution
Divide the circle into four parts: The top semicircle: (A), the bottom sector with arc length 120 degrees: (B), 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). The area is just (A) + (B) - (C) + (D).
Area of (A): 
Area of (B): 
Area of (C): Radius of 2, distance of 1 to BC, creates 2 30-60-90 triangles, so area of it is 
Area of (D): 
Total sum: 
For this solution to be a tad more clear, we are finding the area of the sector in B of 120 degrees because the large circle radius is 2, and the short length (the radius of the semicircle) is 1, and so the triangle is a 30-60-90 triangle. In A, we find the top semicircle part, in B minus C, we find the area of the shaded region above the semicircles but below the diameter, and in D we find the bottom shaded region. - edited by IronicNinja
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. 
