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Difference between revisions of "2000 AMC 10 Problems/Problem 18"
 
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draw( (0,0)--(-1,0), dashed ); draw( (0,0)--(0,-1), dashed );
 
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( (5,0)--(6,0), dashed );  draw( (5,0)--(5,-1), dashed );
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draw( (5,5)--(6,5), dashed );  draw( (5,5)--(5,6), dashed );
+
draw( (5,5)--(6,5), dashed );  // This line is correct
−
rdxrdxcxts
+
draw( (5,5)--(5,6), dashed ); // Add semicolon here
−
dtgh );  draw( (5,5)--(5,6), dashed );
 
 
draw( (0,5)--(-1,5), dashed ); draw( (0,5)--(0,6), dashed );
 
draw( (0,5)--(-1,5), dashed ); draw( (0,5)--(0,6), dashed );
  

Latest revision as of 09:47, 8 July 2024

Problem

Charlyn walks completely around the boundary of a square whose sides are each $5$ km long. From any point on her path she can see exactly $1$ 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?

$\textbf{(A)} 24 \qquad\textbf{(B)}\ 27 \qquad\textbf{(C)}\ 39 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 42$

Video Solution

https://youtu.be/j7Hi5I8INII - Happytwin

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 $5\cdot 5 - 3\cdot 3 = 16$. The part outside the walk consists of four rectangles, and four arcs. Each of the rectangles has area $5\cdot 1=5$. The four arcs together form a circle with radius $1$.

Therefore the total area she can see is $16 + 4\cdot 5 + \pi\cdot 1^2 = 36+\pi \simeq 39.14$, which rounded to the nearest integer is $39$. $\boxed{C}$

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

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