Difference between revisions of "2025 AMC 8 Problems/Problem 5"
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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)}\ 22 \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 == |
| + | |||
| + | 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 by Daily Dose of Math== | + | == 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 | 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 | + | == Video Solution 5 by Pi Academy == |
| − | |||
| − | |||
| − | |||
| − | + | https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK | |
==See Also== | ==See Also== | ||
| Line 103: | Line 98: | ||
{{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 21:27, 3 February 2025
Contents
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?
![[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]](http://latex.artofproblemsolving.com/a/3/6/a367bb91ccfd9ca190ffa1534b6f06436102ec08.png)
Solution 1
Each shortest possible path from to follows the edges of the rectangle. The following path outlines a path of 
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]](http://latex.artofproblemsolving.com/9/3/0/930173277b92d2c7a4f876d1f76eb42007e87e1a.png)
~ 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. 
