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2025 AMC 8 Problems/Problem 5: Difference between revisions

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==Problem==
== Problem ==
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?
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?


<asy>
<asy>
unitsize(20);
unitsize(20);


Line 34: Line 33:
draw((8.2,2)--(8.2,1), EndArrow(3));
draw((8.2,2)--(8.2,1), EndArrow(3));
draw(shift(8.88, 1.5) * scale(0.03) * texpath("1 block"));
draw(shift(8.88, 1.5) * scale(0.03) * texpath("1 block"));
</asy>
</asy>


<math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26\qquad \textbf{(E)}\ 28</math>
<math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26\qquad \textbf{(E)}\ 28</math>


==Solution 1==
== Solution 1 ==


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{24}</math> units:
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:


<asy>
<asy>
unitsize(20);
unitsize(20);


Line 67: Line 64:
draw(shift((6,5)) * w);
draw(shift((6,5)) * w);
label("$F$",(6,5));
label("$F$",(6,5));
</asy>
</asy>


~ [[zhenghua]]
~ [[zhenghua]]


==Solution 2==
== Solution 2 ==


Since it's a square grid, you can find the shortest distance using a line diagonally from one point to the other, creating a sort of slope, then find the rise and run of the slope, which also happens to be the shortest distance, repeat this process until you go back to Point <math>F</math>, and you should get this:
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>.
 
<math>2 + 1 + 3 + 7 + 4 + 2 + 1 + 4</math>, which is equal to <math>24</math>, so your answer will be <math>\boxed{\textbf{(C)}\ 24}</math>.
~Imhappy62789
~Imhappy62789


==Video Solution 1 (Detailed Explanation) 🚀⚡📊 ==
== Video Solution 1 (Detailed Explanation) 🚀⚡📊 ==
Youtube Link ⬇️


https://youtu.be/n6M3y_1dsOk
https://www.youtube.com/watch?v=n6M3y_1dsOk


~ ChillGuyDoesMath :)
~ ChillThingz :)


== Video Solution 2 by Daily Dose of Math ==


==Video Solution by Daily Dose of Math==
[//youtu.be/rjd0gigUsd0 ~Thesmartgreekmathdude]


https://youtu.be/rjd0gigUsd0
== Video Solution 3 ==


~Thesmartgreekmathdude
==Video Solution (A Clever Explanation You’ll Get Instantly)==
https://youtu.be/VP7g-s8akMY?si=2TfegPRg-_k1DEcz&t=257
https://youtu.be/VP7g-s8akMY?si=2TfegPRg-_k1DEcz&t=257
~hsnacademy
~hsnacademy


==Video Solution by Thinking Feet==
== Video Solution 4 by Thinking Feet ==
 
https://youtu.be/PKMpTS6b988
https://youtu.be/PKMpTS6b988
== Video Solution 5 by Pi Academy ==
https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
==Video Solution(Quick, fast, easy!)==
https://youtu.be/fdG7EDW_7xk
~MC


==See Also==
==See Also==
Line 104: Line 104:
{{AMC8 box|year=2025|num-b=4|num-a=6}}
{{AMC8 box|year=2025|num-b=4|num-a=6}}
{{MAA Notice}}
{{MAA Notice}}
[[Category:Introductory Geometry Problems]]

Latest revision as of 23:03, 2 November 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]

$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26\qquad \textbf{(E)}\ 28$

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 (Detailed Explanation) 🚀⚡📊

https://www.youtube.com/watch?v=n6M3y_1dsOk

~ ChillThingz :)

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

Video Solution(Quick, fast, easy!)

https://youtu.be/fdG7EDW_7xk

~MC

See Also

2025 AMC 8 (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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