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 "2003 AMC 12A Problems/Problem 15"
 
(8 intermediate revisions by 6 users not shown)
Line 36: Line 36:
 
</asy>
 
</asy>
  
−
Let <math>[X]</math> denote the area of region <math>X</math> in the figure above.
 
  
 
The shaded area <math>[A]</math> is equal to the area of the smaller semicircle <math>[A+B]</math> minus the area of a [[sector]] of the larger circle <math>[B+C]</math> plus the area of a [[triangle]] formed by two [[radius | radii]] of the larger semicircle and the diameter of the smaller semicircle <math>[C]</math>.
 
The shaded area <math>[A]</math> is equal to the area of the smaller semicircle <math>[A+B]</math> minus the area of a [[sector]] of the larger circle <math>[B+C]</math> plus the area of a [[triangle]] formed by two [[radius | radii]] of the larger semicircle and the diameter of the smaller semicircle <math>[C]</math>.
  
−
The area of the smaller semicircle is <math>[A+B] = \frac{1}{2}\pi\cdot(\frac{1}{2})^{2}=\frac{1}{8}\pi</math>.  
+
The area of the smaller semicircle is <math>[A+B] = \frac{1}{2}\pi\cdot\left(\frac{1}{2}\right)^{2}=\frac{1}{8}\pi</math>.  
  
−
Since the radius of the larger semicircle is equal to the diameter of the smaller semicircle, the triangle is an equilateral triangle and the sector measures <math>60^\circ</math>.  
+
Since the radius of the larger semicircle is equal to the diameter of the smaller half circle, the triangle is an equilateral triangle and the sector measures <math>60^\circ</math>.  
  
−
The area of the <math>60^\circ</math> sector of the larger semicircle is <math>[B+C] = \frac{60}{360}\pi\cdot(\frac{2}{2})^{2}=\frac{1}{6}\pi</math>.  
+
The area of the <math>60^\circ</math> sector of the larger semicircle is <math>[B+C] = \frac{60}{360}\pi\cdot\left(\frac{2}{2}\right)^{2}=\frac{1}{6}\pi</math>.  
  
 
The area of the triangle is <math>[C] = \frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}</math>.
 
The area of the triangle is <math>[C] = \frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}</math>.
  
−
So the shaded area is <math>[A] = [A+B]-[B+C]+[C] = \left(\frac{1}{8}\pi\right)-\left(\frac{1}{6}\pi\right)+\left(\frac{\sqrt{3}}{4}\right)=\boxed{\mathrm{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}</math>.
+
So the shaded area is <math>[A] = [A+B]-[B+C]+[C] = \left(\frac{1}{8}\pi\right)-\left(\frac{1}{6}\pi\right)+\left(\frac{\sqrt{3}}{4}\right)=\boxed{\mathrm{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}</math>. We have thus solved the problem.
 +
 
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=-4qnOdE0Dyk
  
 
== See Also ==
 
== See Also ==
Line 55: Line 57:
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 +
[[Category:Area Problems]]
 +
{{MAA Notice}}

Latest revision as of 18:06, 3 June 2024

The following problem is from both the 2003 AMC 12A #15 and 2003 AMC 10A #19, so both problems redirect to this page.

Problem

A semicircle of diameter $1$ sits at the top of a semicircle of diameter $2$, as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.

[asy] import graph; size(150); defaultpen(fontsize(8)); pair A=(-2,0), B=(2,0); filldraw(Arc((0,sqrt(3)),1,0,180)--cycle,mediumgray); filldraw(Arc((0,0),2,0,180)--cycle,white); draw(2*expi(2*pi/6)--2*expi(4*pi/6)); label("1",(0,sqrt(3)),(0,-1)); label("2",(0,0),(0,-1)); [/asy]

$\mathrm{(A) \ } \frac{1}{6}\pi-\frac{\sqrt{3}}{4}\qquad \mathrm{(B) \ } \frac{\sqrt{3}}{4}-\frac{1}{12}\pi\qquad \mathrm{(C) \ } \frac{\sqrt{3}}{4}-\frac{1}{24}\pi\qquad \mathrm{(D) \ } \frac{\sqrt{3}}{4}+\frac{1}{24}\pi\qquad \mathrm{(E) \ } \frac{\sqrt{3}}{4}+\frac{1}{12}\pi$

Solution

[asy] import graph; size(150); defaultpen(fontsize(8)); pair A=(-2,0), B=(2,0); filldraw(Arc((0,sqrt(3)),1,0,180)--cycle,mediumgray); fill(Arc((0,0),2,0,180)--cycle,white); draw(Arc((0,0),2,0,180)--cycle); draw((0,0)--2*expi(2*pi/6)--2*expi(2*pi/6*2)--(0,0)); label("A",(0,2),(0,4)); label("B",(0,2),(0,-1)); label("C",(0,sqrt(3)/2),(0,2)); label("1",(-0.5,sqrt(3)/2),(-1,0)); label("1",(0.5,sqrt(3)/2),(1,0)); [/asy]


The shaded area $[A]$ is equal to the area of the smaller semicircle $[A+B]$ minus the area of a sector of the larger circle $[B+C]$ plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle $[C]$.

The area of the smaller semicircle is $[A+B] = \frac{1}{2}\pi\cdot\left(\frac{1}{2}\right)^{2}=\frac{1}{8}\pi$.

Since the radius of the larger semicircle is equal to the diameter of the smaller half circle, the triangle is an equilateral triangle and the sector measures $60^\circ$.

The area of the $60^\circ$ sector of the larger semicircle is $[B+C] = \frac{60}{360}\pi\cdot\left(\frac{2}{2}\right)^{2}=\frac{1}{6}\pi$.

The area of the triangle is $[C] = \frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}$.

So the shaded area is $[A] = [A+B]-[B+C]+[C] = \left(\frac{1}{8}\pi\right)-\left(\frac{1}{6}\pi\right)+\left(\frac{\sqrt{3}}{4}\right)=\boxed{\mathrm{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}$. We have thus solved the problem.

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=-4qnOdE0Dyk

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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
2003 AMC 12A (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=2003_AMC_12A_Problems/Problem_15&oldid=220031"