Difference between revisions of "2002 AMC 10A Problems/Problem 19"

Latest revision as of 17:41, 15 October 2024

Problem

Spot's doghouse has a regular hexagonal base that measures one yard on each side. He is tethered to a vertex with a two-yard rope. What is the area, in square yards, of the region outside of the doghouse that Spot can reach?

$\text{(A)}\ 2\pi/3 \qquad \text{(B)}\ 2\pi \qquad \text{(C)}\ 5\pi/2 \qquad \text{(D)}\ 8\pi/3 \qquad \text{(E)}\ 3\pi$

Solution

$[asy] draw(polygon(6)); draw(Arc((1/2,sqrt(3)/2),2,-60,180)); draw(Arc((-1/2,sqrt(3)/2),1,180,240)); draw(Arc((1,0),1,240,300)); draw((-1/2,sqrt(3)/2)--(-3/2,sqrt(3)/2), dotted); draw((1,0)--(3/2,-sqrt(3)/2),dotted); [/asy]$

Part of what Spot can reach is $\frac{240}{360}=\frac{2}{3}$ of a circle with radius 2, which gives him $\frac{8\pi}{3}$ . He can also reach two $\frac{60}{360}$ parts of a unit circle, which combines to give $\frac{\pi}{3}$ . The total area is then $3\pi$ , which gives $\boxed{\text{(E)}}$ .

Note

We can clearly see that the area must be more than $\frac{8\pi}{3}$ , and the only such answer is $\boxed{\text{(E)}}$ .

Video Solution

https://www.youtube.com/watch?v=lkk3SAjsPaI ~David

Video Solution by Daily Dose of Math

https://youtu.be/C2pNSPfTxgA

~Thesmartgreekmathdude

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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 AMC 10 Problems and Solutions

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

@@ Line 5: / Line 5: @@
 ==Solution==
-Spot can reach a <math>2\pi-\frac{2\pi}{3}=\frac{4\pi}{3}</math> angle of a unit circle; hence he can reach a region with area <math>\boxed{\frac{2\pi}{3}\Rightarrow\text{(A)}}</math>.
+<center><asy>
+draw(polygon(6));
+draw(Arc((1/2,sqrt(3)/2),2,-60,180));
+draw(Arc((-1/2,sqrt(3)/2),1,180,240));
+draw(Arc((1,0),1,240,300));
+draw((-1/2,sqrt(3)/2)--(-3/2,sqrt(3)/2), dotted);
+draw((1,0)--(3/2,-sqrt(3)/2),dotted);
+</asy></center>
+Part of what Spot can reach is <math>\frac{240}{360}=\frac{2}{3}</math> of a circle with radius 2, which
+gives him <math>\frac{8\pi}{3}</math>. He can also reach two <math>\frac{60}{360}</math> parts of a unit circle, which combines to give <math>\frac{\pi}{3}</math>. The total area is then <math>3\pi</math>, which gives <math>\boxed{\text{(E)}}</math>.
+===Note===
+We can clearly see that the area must be more than <math>\frac{8\pi}{3}</math>, and the only such answer is <math>\boxed{\text{(E)}}</math>.
+==Video Solution==
+https://www.youtube.com/watch?v=lkk3SAjsPaI  ~David
+==Video Solution by Daily Dose of Math==
+https://youtu.be/C2pNSPfTxgA
+~Thesmartgreekmathdude
 ==See Also==
-{{AMC10 box|ab=A|year=2002|num-b=17|num-a=19}}
+{{AMC10 box|ab=A|year=2002|num-b=18|num-a=20}}
 [[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

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