Difference between revisions of "2024 AMC 8 Problems/Problem 18"

Revision as of 17:53, 25 January 2024

Problem

Three concentric circles centered at $O$ have radii of $1$ , $2$ , and $3$ . Points $B$ and $C$ lie on the largest cirlce. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle $BOC$ , as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of $\angle{BOC}$ in degrees?

-figure-

$\textbf{(A) } 108\qquad\textbf{(B) } 120\qquad\textbf{(C) } 135\qquad\textbf{(D) } 144\qquad\textbf{(E) } 150$

Solution

Let $x=\angle{BOC}$ .

We see that the shaded region is the inner ring plus a sector $x^\circ$ of the outer ring. The area of this in terms of $x$ is $\left( 4 \pi - \pi \right)+\frac{x}{360} \left( 9 \pi - 4 \pi \right)$ . This simplifies to $3 \pi + \frac{x}{360}(5 \pi)$ .

Also, the unshaded portion is comprised of the smallest circle plus the sector $(360-x)^\circ$ of the outer ring. The area of this is $\pi + \frac{x}{360}(5 \pi)$ .

We are told these are equal, therefore $\pi + \frac{x}{360}(5 \pi) = 3 \pi + \frac{x}{360}(5 \pi)$ . Solving for $x$ reveals $x=\boxed{\textbf{(A) } 108}$ .

~MrThinker

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Solution 1

Video Solution 1 (super clear!) by Power Solve

https://youtu.be/TlTN7EQcFvE

@@ Line 1: / Line 1: @@
 ==Problem==
+Three concentric circles centered at <math>O</math> have radii of <math>1</math>, <math>2</math>, and <math>3</math>. Points <math>B</math> and <math>C</math> lie on the largest cirlce. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle <math>BOC</math>, as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of <math>\angle{BOC}</math> in degrees?
+-figure-
+<math>\textbf{(A) } 108\qquad\textbf{(B) } 120\qquad\textbf{(C) } 135\qquad\textbf{(D) } 144\qquad\textbf{(E) } 150</math>
+==Solution==
+Let <math>x=\angle{BOC}</math>.
+We see that the shaded region is the inner ring plus a sector <math>x^\circ</math> of the outer ring. The area of this in terms of <math>x</math> is <math>\left( 4 \pi - \pi \right)+\frac{x}{360} \left( 9 \pi - 4 \pi \right)</math>. This simplifies to <math>3 \pi + \frac{x}{360}(5 \pi)</math>.
+Also, the unshaded portion is comprised of the smallest circle plus the sector <math>(360-x)^\circ</math> of the outer ring. The area of this is <math>\pi + \frac{x}{360}(5 \pi)</math>.
+We are told these are equal, therefore <math>\pi + \frac{x}{360}(5 \pi) = 3 \pi + \frac{x}{360}(5 \pi)</math>. Solving for <math>x</math> reveals <math>x=\boxed{\textbf{(A) } 108}</math>.
+~MrThinker
+==See Also==
+{{AMC8 box|year=2024|num-b=17|num-a=19}}
+{{MAA Notice}}
 ==Solution 1==
 ==Video Solution 1 (super clear!) by Power Solve==
 https://youtu.be/TlTN7EQcFvE

Page

Toolbox

Search