Difference between revisions of "2025 AMC 8 Problems/Problem 5"

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== Problem ==
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Betty drives a truck to deliver packages in a neighborhood whose street map is shown below. Betty starts at the factory (labled <math>F</math>) and drives to location <math>A</math>, then <math>B</math>, then <math>C</math>, before returning to <math>F</math>. What is the shortest distance, in blocks, she can drive to complete the route?
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<asy>
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unitsize(20);
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add(grid(8,6));
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path w = circle((0,0),0.4);
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fill(w, white);
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draw(w);
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label("$B$",(0,0));
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fill(shift((2,4)) * w, white);
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draw(shift((2,4)) * w);
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label("$C$",(2,4));
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fill(shift((7,3)) * w, white);
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draw(shift((7,3)) * w);
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label("$A$",(7,3));
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fill(shift((6,5)) * w, white);
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draw(shift((6,5)) * w);
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label("$F$",(6,5));
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draw((6,-0.2)--(7,-0.2), EndArrow(3));
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draw((7,-0.2)--(6,-0.2), EndArrow(3));
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draw(shift(6.5, -0.48) * scale(0.03) * texpath("1 block"));
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draw((8.2,1)--(8.2,2), EndArrow(3));
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draw((8.2,2)--(8.2,1), EndArrow(3));
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draw(shift(8.88, 1.5) * scale(0.03) * texpath("1 block"));
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</asy>
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<math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26\qquad \textbf{(E)}\ 28</math>
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== Solution 1 ==
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Each shortest possible path from <math>A</math> to <math>B</math> follows the edges of the rectangle. The following path outlines a path of <math>\boxed{\textbf{(C)}\ 24}</math> units:
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<asy>
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unitsize(20);
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add(grid(8,6));
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draw((6,5)--(7,5)--(7,0)--(0,0)--(0,4)--(2,4)--(2,5)--cycle,green);
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path w = circle((0,0),0.4);
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fill(w, white);
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draw(w);
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label("$B$",(0,0));
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fill(shift((2,4)) * w, white);
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draw(shift((2,4)) * w);
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label("$C$",(2,4));
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fill(shift((7,3)) * w, white);
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draw(shift((7,3)) * w);
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label("$A$",(7,3));
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fill(shift((6,5)) * w, white);
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draw(shift((6,5)) * w);
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label("$F$",(6,5));
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</asy>
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~ [[zhenghua]]
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== Solution 2 ==
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We can find the shortest distance using a line diagonally from one point to the other, creating a sort of slope, then find the sum of rise and run of the slope, which happens to be the shortest distance, repeat this process until you get back to Point <math>F</math>, and you should get <math>2 + 1 + 3 + 7 + 4 + 2 + 1 + 4</math>, which is equal to <math>\boxed{\textbf{(C)}\ 24}</math>.
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~Imhappy62789
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== Video Solution 1 ==
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[//youtu.be/n6M3y_1dsOk ~ ChillGuyDoesMath]
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== Video Solution 2 by Daily Dose of Math ==
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[//youtu.be/rjd0gigUsd0 ~Thesmartgreekmathdude]
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== Video Solution 3 ==
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https://youtu.be/VP7g-s8akMY?si=2TfegPRg-_k1DEcz&t=257
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~hsnacademy
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== Video Solution 4 by Thinking Feet ==
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https://youtu.be/PKMpTS6b988
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== Video Solution 5 by Pi Academy ==
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https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
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==See Also==
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{{AMC8 box|year=2025|num-b=4|num-a=6}}
 +
{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 21:27, 3 February 2025

Problem

Betty drives a truck to deliver packages in a neighborhood whose street map is shown below. Betty starts at the factory (labled $F$) and drives to location $A$, then $B$, then $C$, before returning to $F$. What is the shortest distance, in blocks, she can drive to complete the route?

[asy] unitsize(20);  add(grid(8,6));  path w = circle((0,0),0.4);  fill(w, white); draw(w); label("$B$",(0,0));  fill(shift((2,4)) * w, white); draw(shift((2,4)) * w); label("$C$",(2,4));  fill(shift((7,3)) * w, white); draw(shift((7,3)) * w); label("$A$",(7,3));  fill(shift((6,5)) * w, white); draw(shift((6,5)) * w); label("$F$",(6,5));  draw((6,-0.2)--(7,-0.2), EndArrow(3)); draw((7,-0.2)--(6,-0.2), EndArrow(3)); draw(shift(6.5, -0.48) * scale(0.03) * texpath("1 block"));  draw((8.2,1)--(8.2,2), EndArrow(3)); draw((8.2,2)--(8.2,1), EndArrow(3)); draw(shift(8.88, 1.5) * scale(0.03) * texpath("1 block")); [/asy]

$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26\qquad \textbf{(E)}\ 28$

Solution 1

Each shortest possible path from $A$ to $B$ follows the edges of the rectangle. The following path outlines a path of $\boxed{\textbf{(C)}\ 24}$ units:

[asy] unitsize(20);  add(grid(8,6)); draw((6,5)--(7,5)--(7,0)--(0,0)--(0,4)--(2,4)--(2,5)--cycle,green);  path w = circle((0,0),0.4);  fill(w, white); draw(w); label("$B$",(0,0));  fill(shift((2,4)) * w, white); draw(shift((2,4)) * w); label("$C$",(2,4));  fill(shift((7,3)) * w, white); draw(shift((7,3)) * w); label("$A$",(7,3));  fill(shift((6,5)) * w, white); draw(shift((6,5)) * w); label("$F$",(6,5)); [/asy]

~ zhenghua

Solution 2

We can find the shortest distance using a line diagonally from one point to the other, creating a sort of slope, then find the sum of rise and run of the slope, which happens to be the shortest distance, repeat this process until you get back to Point $F$, and you should get $2 + 1 + 3 + 7 + 4 + 2 + 1 + 4$, which is equal to $\boxed{\textbf{(C)}\ 24}$. ~Imhappy62789

Video Solution 1

~ ChillGuyDoesMath

Video Solution 2 by Daily Dose of Math

~Thesmartgreekmathdude

Video Solution 3

https://youtu.be/VP7g-s8akMY?si=2TfegPRg-_k1DEcz&t=257 ~hsnacademy

Video Solution 4 by Thinking Feet

https://youtu.be/PKMpTS6b988

Video Solution 5 by Pi Academy

https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK

See Also

2025 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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. AMC logo.png