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 "2017 AMC 8 Problems/Problem 25"

(→‎Solution)
(→‎Solution 2 (tiny bit intuitional))
 
(47 intermediate revisions by 25 users not shown)
Line 1: Line 1:
−
==Problem 25==
+
==Problem==
  
 
In the figure shown, <math>\overline{US}</math> and <math>\overline{UT}</math> are line segments each of length 2, and <math>m\angle TUS = 60^\circ</math>. Arcs <math>\overarc{TR}</math> and <math>\overarc{SR}</math> are each one-sixth of a circle with radius 2. What is the area of the region shown?  
 
In the figure shown, <math>\overline{US}</math> and <math>\overline{UT}</math> are line segments each of length 2, and <math>m\angle TUS = 60^\circ</math>. Arcs <math>\overarc{TR}</math> and <math>\overarc{SR}</math> are each one-sixth of a circle with radius 2. What is the area of the region shown?  
Line 7: Line 7:
 
<math>\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}</math>
 
<math>\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}</math>
  
−
==See Also==
+
==Solution 1==
−
{{AMC8 box|year=2017|num-b=24|after=Last Problem}}
 
  
−
{{MAA Notice}}
+
<asy>draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label("$U$", (2,3.464), N); label("$S$", (1,1.732), W); label("$T$", (3,1.732), E); label("$R$", (2,0), S);</asy>
 +
 
 +
In addition to the given diagram, we can draw lines <math>\overline{SR}</math> and <math>\overline{RT}.</math> The area of rhombus <math>SRTU</math> is half the product of its diagonals, which is <math>\frac{2\sqrt3 \cdot 2}{2}=2\sqrt3</math>. However, we have to subtract off the circular segments. The area of those can be found by computing the area of the circle with radius 2, multiplying it by <math>\frac{1}{6}</math>, then finally subtracting the area of an equilateral triangle with a side length 2 from the sector. The sum of the areas of the circular segments is <math>2(\frac{4 \pi}{6}-\sqrt3).</math> The area of rhombus <math>SRTU</math> minus the circular segments is <math>2\sqrt3-\frac{4 \pi}{3}+2\sqrt3= \boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.</math>
 +
 
 +
~PEKKA
 +
 
 +
== Solution 2 (tiny bit intuitional) ==
 +
 
 +
We can extend <math>\overline{US}</math>, <math>\overline{UT}</math> to <math>X</math> and <math>Y</math>, respectively, such that <math>X</math> and <math>Y</math> are collinear to point <math>R</math>. Connect <math>\overline{XY}</math>. We can see points <math>X</math>, <math>Y</math> are probably circle centers of arc <math>SR</math>, <math>TR</math>, respectively. So, <math>\overline{XS} = 2 = \overline{TY}</math>. Thus, <math>\triangle{UXY}</math> is equilateral. The area of <math>\triangle{UXY}</math> is <math>\frac{\sqrt{3}}{4} \cdot 4^2</math>, or <math>4\sqrt{3}</math>, and both one sixth circles total up to <math>\frac{4\pi}{3}</math>. Finally, the answer is <math>\boxed{\textbf{(B)} 4\sqrt{3}-\frac{4\pi}{3}}</math>.
 +
 
 +
~ lovelearning999
 +
 
 +
==Video Solutions==
 +
 
 +
 
 +
https://youtu.be/sVclz6EmpEU
 +
 
 +
~savannahsolver

Latest revision as of 08:59, 14 December 2024

Problem

In the figure shown, $\overline{US}$ and $\overline{UT}$ are line segments each of length 2, and $m\angle TUS = 60^\circ$. Arcs $\overarc{TR}$ and $\overarc{SR}$ are each one-sixth of a circle with radius 2. What is the area of the region shown?

[asy]draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label("$U$", (2,3.464), N); label("$S$", (1,1.732), W); label("$T$", (3,1.732), E); label("$R$", (2,0), S);[/asy]

$\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}$

Solution 1

[asy]draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label("$U$", (2,3.464), N); label("$S$", (1,1.732), W); label("$T$", (3,1.732), E); label("$R$", (2,0), S);[/asy]

In addition to the given diagram, we can draw lines $\overline{SR}$ and $\overline{RT}.$ The area of rhombus $SRTU$ is half the product of its diagonals, which is $\frac{2\sqrt3 \cdot 2}{2}=2\sqrt3$. However, we have to subtract off the circular segments. The area of those can be found by computing the area of the circle with radius 2, multiplying it by $\frac{1}{6}$, then finally subtracting the area of an equilateral triangle with a side length 2 from the sector. The sum of the areas of the circular segments is $2(\frac{4 \pi}{6}-\sqrt3).$ The area of rhombus $SRTU$ minus the circular segments is $2\sqrt3-\frac{4 \pi}{3}+2\sqrt3= \boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.$

~PEKKA

Solution 2 (tiny bit intuitional)

We can extend $\overline{US}$, $\overline{UT}$ to $X$ and $Y$, respectively, such that $X$ and $Y$ are collinear to point $R$. Connect $\overline{XY}$. We can see points $X$, $Y$ are probably circle centers of arc $SR$, $TR$, respectively. So, $\overline{XS} = 2 = \overline{TY}$. Thus, $\triangle{UXY}$ is equilateral. The area of $\triangle{UXY}$ is $\frac{\sqrt{3}}{4} \cdot 4^2$, or $4\sqrt{3}$, and both one sixth circles total up to $\frac{4\pi}{3}$. Finally, the answer is $\boxed{\textbf{(B)} 4\sqrt{3}-\frac{4\pi}{3}}$.

~ lovelearning999

Video Solutions

https://youtu.be/sVclz6EmpEU

~savannahsolver

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_25&oldid=236963"