Difference between revisions of "2017 AMC 8 Problems/Problem 25"
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− | ==Problem | + | ==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? | ||
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<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> | ||
− | ==Solution== | + | ==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> | |
− | <asy>draw((1,1.732)--(2,3.464)--(3,1.732)); | + | |
− | + | 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> | |
− | draw(arc((0,0),(2,0),(1,1.732))); | + | |
− | draw(arc((4,0),(3,1.732),(2,0))); | + | ~PEKKA |
− | label("$U$", (2,3.464), N); | + | |
− | label("$S$", (1,1.732), W); | + | == Solution 2 (tiny bit intuitional) == |
− | label("$T$", (3,1.732), E); | ||
− | label("$R$", (2,0), S); | ||
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− | |||
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− | ==See Also== | + | 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 | ||
+ | |||
+ | == See Also == | ||
{{AMC8 box|year=2017|num-b=24|after=Last Problem}} | {{AMC8 box|year=2017|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 17:23, 8 June 2025
Contents
[hide]Problem
In the figure shown, and
are line segments each of length 2, and
. Arcs
and
are each one-sixth of a circle with radius 2. What is the area of the region shown?
Solution 1
In addition to the given diagram, we can draw lines and
The area of rhombus
is half the product of its diagonals, which is
. 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
, 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
The area of rhombus
minus the circular segments is
~PEKKA
Solution 2 (tiny bit intuitional)
We can extend ,
to
and
, respectively, such that
and
are collinear to point
. Connect
. We can see points
,
are probably circle centers of arc
,
, respectively. So,
. Thus,
is equilateral. The area of
is
, or
, and both one sixth circles total up to
. Finally, the answer is
.
~ lovelearning999
Video Solutions
~savannahsolver
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.