Difference between revisions of "2017 AMC 8 Problems/Problem 25"
(almost everything) |
m (→Solution & Video) |
||
(32 intermediate revisions by 18 users not shown) | |||
Line 1: | Line 1: | ||
− | ==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? | ||
Line 6: | Line 6: | ||
<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== | |
+ | |||
+ | Let the centers of the circles containing arcs <math>\overarc{SR}</math> and <math>\overarc{TR}</math> be <math>X</math> and <math>Y</math>, respectively. Extend <math>\overline{US}</math> and <math>\overline{UT}</math> to <math>X</math> and <math>Y</math>, and connect point <math>X</math> with point <math>Y</math>. | ||
+ | <asy> | ||
+ | unitsize(1 cm); | ||
+ | pair U,S,T,R,X,Y; | ||
+ | U =(2,3.464); | ||
+ | S=(1,1.732); | ||
+ | T=(3,1.732); | ||
+ | R=(2,0); | ||
+ | X=(0,0); | ||
+ | Y=(4,0); | ||
+ | draw(U--S); | ||
+ | draw(S--U--T); | ||
+ | draw(S--X--Y--T,red); | ||
+ | draw(arc(X,R,S),red); | ||
+ | draw(arc(Y,T,R),red); | ||
+ | label("$U$",U, N); | ||
+ | label("$S$", S, W); | ||
+ | label("$T$", T, E); | ||
+ | label("$R$", R, S); | ||
+ | label("$X$",X, W); | ||
+ | label("$Y$", Y, E); | ||
+ | </asy> | ||
+ | We can clearly see that <math>\triangle UXY</math> is an equilateral triangle, because the problem states that <math>m\angle TUS = 60^\circ</math>. We can figure out that <math>m\angle SXR= 60^\circ</math> and <math>m\angle TYR = 60^\circ</math> because they are <math>\frac{1}{6}</math> of a circle. The area of the figure is equal to <math>[\triangle UXY]</math> minus the combined area of the <math>2</math> sectors of the circles (in red). Using the area formula for an equilateral triangle, <math>\frac{a^2\sqrt{3}}{4},</math> where <math>a</math> is the side length of the equilateral triangle, <math>[\triangle UXY]</math> is <math>\frac{\sqrt 3}{4} \cdot 4^2 = 4\sqrt 3.</math> The combined area of the <math>2</math> sectors is <math>2\cdot\frac16\cdot\pi r^2</math>, which is <math>\frac 13\pi \cdot 2^2 = \frac{4\pi}{3}.</math> Thus, our final answer is <math>\boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | <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 | ||
+ | |||
+ | ==Video Solutions== | ||
+ | https://youtu.be/LT4gyH--328 | ||
+ | |||
+ | https://youtu.be/wc5rGulTTR8 | ||
+ | |||
+ | - Happytwin | ||
+ | |||
+ | https://youtu.be/aE0oAq4Q_Ks | ||
+ | |||
+ | https://euclideanmathcircle.wixsite.com/emc1/videos?wix-vod-video-id=3a7970c3cd01453aa4263a8be7998588&wix-vod-comp-id=comp-kn8844mv | ||
+ | |||
+ | https://youtu.be/sVclz6EmpEU | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | =Solution 3 & Video= | ||
+ | |||
+ | First, we must make it like an equilateral triangle. If you add two of the same arcs, you will get an equilateral triangle. The goal is to find ONE of the arcs and multiply it by two. Then subtract it from the equilateral triangle. If you look at the triangle, T and S are midpoints of the equilateral triangle. So now we can find the arc.These arcs are <math>\frac{2}{6}\pi</math> so now you multiply by 2 so <math>\frac{4}{3}\pi</math>. And subtract. Equilateral's area is <math>4\sqrt{3}</math>. So now you do <math>4 \sqrt 3 - \frac{4}{3} \pi</math>. That is the answer and thank you. | ||
+ | |||
+ | ~ math.is.hard.to.understand | ||
+ | |||
+ | Here's a video to understand it (credits to ~ pi_is_3.14) | ||
+ | |||
+ | https://youtu.be/j3QSD5eDpzU?t=1350 | ||
==See Also== | ==See Also== |
Revision as of 21:00, 16 January 2024
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
Let the centers of the circles containing arcs and
be
and
, respectively. Extend
and
to
and
, and connect point
with point
.
We can clearly see that
is an equilateral triangle, because the problem states that
. We can figure out that
and
because they are
of a circle. The area of the figure is equal to
minus the combined area of the
sectors of the circles (in red). Using the area formula for an equilateral triangle,
where
is the side length of the equilateral triangle,
is
The combined area of the
sectors is
, which is
Thus, our final answer is
Solution 2
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
Video Solutions
- Happytwin
~savannahsolver
Solution 3 & Video
First, we must make it like an equilateral triangle. If you add two of the same arcs, you will get an equilateral triangle. The goal is to find ONE of the arcs and multiply it by two. Then subtract it from the equilateral triangle. If you look at the triangle, T and S are midpoints of the equilateral triangle. So now we can find the arc.These arcs are so now you multiply by 2 so
. And subtract. Equilateral's area is
. So now you do
. That is the answer and thank you.
~ math.is.hard.to.understand
Here's a video to understand it (credits to ~ pi_is_3.14)
https://youtu.be/j3QSD5eDpzU?t=1350
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.