Difference between revisions of "2013 UNCO Math Contest II Problems/Problem 3"
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== 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 == | ||
− | {{ | + | {{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
Point is the center of a large circle that passes through both
and
, and
lies on the small circle whose diameter is
. The area of the
small circle is
. Find the area of the shaded
, 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 \( D \).
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:
The area of the triangle \( ABC \):
Area of the segment \( ABC \):
The area of the lune:
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 |