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 "2013 UNCO Math Contest II Problems/Problem 3"

m (moved 2013 UNC Math Contest II Problems/Problem 3 to 2013 UNCO Math Contest II Problems/Problem 3: disambiguation of University of Northern Colorado with University of North Carolina)
(→‎Solution)
 
(2 intermediate revisions by one other user not shown)
Line 14: Line 14:
  
 
== Solution ==
 
== Solution ==
 +
From the information that \( 9\pi \) is the area of small circle, we can infer that \( \frac{1}{2} AB = 3 \). 
 +
\(\because \text{area}_\text{small} = \pi (3)^{2} \implies \text{radius}_\text{small} = \frac{1}{2} AB = 3\)
  
 +
The line from \( C \) to the center of the small circle, i.e., the midpoint of \( AB \), will be perpendicular to \( AB \).
 +
 +
Let the midpoint of \( AB \) be \( D \).
 +
 +
<cmath>AD^{2} + CD^{2} = AC^{2}</cmath> 
 +
<cmath>3^{2} + 3^{2} = AC^{2}</cmath> 
 +
<cmath>AC^{2} = \text{radius}^{2}_\text{large} = 18</cmath> 
 +
 +
Now because \( A, B, C \) are part of the small triangle and \( AB \) is the diameter, the \( \angle ACB = 90^\circ \). 
 +
Thus, the area of the sector \( ABC \) in the large circle:
 +
 +
<cmath>\text{area}_{\text{sector}} = \frac{90}{360} \times \pi (\text{radius}_\text{large})^{2}</cmath> 
 +
<cmath>= \frac{1}{4} \times 18 \pi</cmath> 
 +
<cmath>= \boxed{ \frac{9}{2}\pi }</cmath> 
 +
 +
The area of the triangle \( ABC \):
 +
 +
<cmath>\text{area}_{\triangle} = \frac{1}{2} \times CD \times AB</cmath> 
 +
<cmath>= \frac{1}{2} \times 3 \times 6</cmath> 
 +
<cmath>= 9</cmath> 
 +
 +
Area of the segment \( ABC \):
 +
 +
<cmath>\text{area}_\text{segment} = \text{area}_\text{sector} - \text{area}_{\triangle}</cmath> 
 +
<cmath>= \frac{9}{2} \pi - 9</cmath> 
 +
<cmath>= 9\left( \frac{\pi}{2} -1 \right)</cmath> 
 +
 +
The area of the lune:
 +
 +
<cmath>\text{area}_\text{lune} = \frac{1}{2} \text{area}_\text{small} - \text{area}_\text{segment}</cmath> 
 +
<cmath>= \frac{1}{2} \cdot 9\pi - 9\left( \frac{\pi}{2} -1 \right)</cmath> 
 +
<cmath>= 9 \left( \frac{\pi}{2} -\frac{\pi}{2} + 1 \right)</cmath> 
 +
<cmath>= \boxed{ 9 }</cmath>
  
 
== See Also ==
 
== See Also ==
−
{{UNC Math Contest box|n=II|year=2013|num-b=2|num-a=4}}
+
{{UNCO Math Contest box|n=II|year=2013|num-b=2|num-a=4}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Latest revision as of 09:08, 7 August 2025

Problem

[asy] filldraw(circle((0,-sqrt(2)/2),sqrt(2)/2),grey); filldraw(circle((0,0),1),white); draw((-sqrt(2)/2,-sqrt(2)/2)--(sqrt(2)/2,-sqrt(2)/2)--(0,0)--cycle,black); MP("C",(0,0),N);MP("A",(-sqrt(2)/2,-sqrt(2)/2),W);MP("B",(sqrt(2)/2,-sqrt(2)/2),E); [/asy]

Point $C$ is the center of a large circle that passes through both $A$ and $B$, and $C$ lies on the small circle whose diameter is $AB$. The area of the small circle is $9\pi$. Find the area of the shaded $\textit{lune}$, the region inside the small circle and outside the large circle.

Solution

From the information that \( 9\pi \) is the area of small circle, we can infer that \( \frac{1}{2} AB = 3 \). \(\because \text{area}_\text{small} = \pi (3)^{2} \implies \text{radius}_\text{small} = \frac{1}{2} AB = 3\)

The line from \( C \) to the center of the small circle, i.e., the midpoint of \( AB \), will be perpendicular to \( AB \).

Let the midpoint of \( AB \) be \( D \).

\[AD^{2} + CD^{2} = AC^{2}\] \[3^{2} + 3^{2} = AC^{2}\] \[AC^{2} = \text{radius}^{2}_\text{large} = 18\]

Now because \( A, B, C \) are part of the small triangle and \( AB \) is the diameter, the \( \angle ACB = 90^\circ \). Thus, the area of the sector \( ABC \) in the large circle:

\[\text{area}_{\text{sector}} = \frac{90}{360} \times \pi (\text{radius}_\text{large})^{2}\] \[= \frac{1}{4} \times 18 \pi\] \[= \boxed{ \frac{9}{2}\pi }\]

The area of the triangle \( ABC \):

\[\text{area}_{\triangle} = \frac{1}{2} \times CD \times AB\] \[= \frac{1}{2} \times 3 \times 6\] \[= 9\]

Area of the segment \( ABC \):

\[\text{area}_\text{segment} = \text{area}_\text{sector} - \text{area}_{\triangle}\] \[= \frac{9}{2} \pi - 9\] \[= 9\left( \frac{\pi}{2} -1 \right)\]

The area of the lune:

\[\text{area}_\text{lune} = \frac{1}{2} \text{area}_\text{small} - \text{area}_\text{segment}\] \[= \frac{1}{2} \cdot 9\pi - 9\left( \frac{\pi}{2} -1 \right)\] \[= 9 \left( \frac{\pi}{2} -\frac{\pi}{2} + 1 \right)\] \[= \boxed{ 9 }\]

See Also

2013 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_UNCO_Math_Contest_II_Problems/Problem_3&oldid=256734"