Difference between revisions of "2000 AMC 10 Problems/Problem 18"
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==Problem== | ==Problem== | ||
+ | |||
+ | Charlyn walks completely around the boundary of a square whose sides are each <math>5</math> km long. From any point on her path she can see exactly <math>1</math> km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number? | ||
+ | |||
+ | <math>\textbf{(A)} 24 \qquad\textbf{(B)}\ 27 \qquad\textbf{(C)}\ 39 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 42</math> | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/j7Hi5I8INII - Happytwin | ||
==Solution== | ==Solution== | ||
+ | |||
+ | The area she sees looks at follows: | ||
+ | |||
+ | <asy> | ||
+ | unitsize(0.8cm); | ||
+ | |||
+ | path p1 = (0,0)--(5,0)--(5,5)--(0,5)--cycle; | ||
+ | path p2 = (1,1)--(4,1)--(4,4)--(1,4)--cycle; | ||
+ | path p3 = arc((0,0),1,180,270) -- arc((5,0),1,270,360) -- arc((5,5),1,0,90) -- arc((0,5),1,90,180) -- cycle; | ||
+ | fill(p3,lightgray); | ||
+ | unfill(p2); | ||
+ | |||
+ | draw(p1,linewidth(bp)); draw(p2); draw(p3); | ||
+ | draw( (0,0)--(-1,0), dashed ); draw( (0,0)--(0,-1), dashed ); | ||
+ | draw( (5,0)--(6,0), dashed ); draw( (5,0)--(5,-1), dashed ); | ||
+ | draw( (5,5)--(6,5), dashed ); // This line is correct | ||
+ | draw( (5,5)--(5,6), dashed ); // Add semicolon here | ||
+ | draw( (0,5)--(-1,5), dashed ); draw( (0,5)--(0,6), dashed ); | ||
+ | |||
+ | draw( (0,-1)--(5,-1), Arrows ); label( "$5$", (2.5,-1), S ); | ||
+ | draw( (1,1)--(4,1), Arrows ); label( "$3$", (2.5,1), N ); | ||
+ | draw( (4,3.5)--(5,3.5), Arrows ); label( "$1$", (4.5,3.5), N ); | ||
+ | draw( (5,3.5)--(6,3.5), Arrows ); label( "$1$", (5.5,3.5), N ); | ||
+ | </asy> | ||
+ | |||
+ | The part inside the walk has area <math>5\cdot 5 - 3\cdot 3 = 16</math>. The part outside the walk consists of four rectangles, and four arcs. Each of the rectangles has area <math>5\cdot 1=5</math>. The four arcs together form a circle with radius <math>1</math>. | ||
+ | |||
+ | Therefore the total area she can see is <math>16 + 4\cdot 5 + \pi\cdot 1^2 = 36+\pi \simeq 39.14</math>, which rounded to the nearest integer is <math>39</math>. <math>\boxed{C}</math> | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2000|num-b=17|num-a=19}} | {{AMC10 box|year=2000|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 10:47, 8 July 2024
Contents
Problem
Charlyn walks completely around the boundary of a square whose sides are each km long. From any point on her path she can see exactly
km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?
Video Solution
https://youtu.be/j7Hi5I8INII - Happytwin
Solution
The area she sees looks at follows:
The part inside the walk has area . The part outside the walk consists of four rectangles, and four arcs. Each of the rectangles has area
. The four arcs together form a circle with radius
.
Therefore the total area she can see is , which rounded to the nearest integer is
.
See Also
2000 AMC 10 (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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.