Difference between revisions of "2025 AMC 8 Problems/Problem 5"
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− | + | == 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? | ||
+ | |||
+ | <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> | ||
+ | |||
+ | <math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26\qquad \textbf{(E)}\ 28</math> | ||
+ | |||
+ | == 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{\textbf{(C)}\ 24}</math> 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 <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>. | ||
+ | ~Imhappy62789 | ||
+ | |||
+ | == Video Solution 1 == | ||
+ | |||
+ | [//youtu.be/n6M3y_1dsOk ~ ChillGuyDoesMath] | ||
+ | |||
+ | == Video Solution 2 by Daily Dose of Math == | ||
+ | |||
+ | [//youtu.be/rjd0gigUsd0 ~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== | ||
+ | |||
+ | {{AMC8 box|year=2025|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 21:27, 3 February 2025
Contents
[hide]Problem
Betty drives a truck to deliver packages in a neighborhood whose street map is shown below. Betty starts at the factory (labled ) and drives to location
, then
, then
, before returning to
. What is the shortest distance, in blocks, she can drive to complete the route?
Solution 1
Each shortest possible path from to
follows the edges of the rectangle. The following path outlines a path of
units:
~ 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 , and you should get
, which is equal to
.
~Imhappy62789
Video Solution 1
Video Solution 2 by Daily Dose of Math
Video Solution 3
https://youtu.be/VP7g-s8akMY?si=2TfegPRg-_k1DEcz&t=257 ~hsnacademy
Video Solution 4 by Thinking Feet
Video Solution 5 by Pi Academy
https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.