2023 AMC 10B Problems/Problem 24

Problem

What is the perimeter of the boundary of the region consisting of all points which can be expressed as $(2u-3w, v+4w)$ with $0\le u\le1$, $0\le v\le1,$ and $0\le w\le1$?

$\textbf{(A) } 10\sqrt{3} \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 16$

Solution

[asy] import geometry; pair A = (-3, 4); pair B = (-3, 5); pair C = (-1, 4); pair D = (-1, 5);   pair AA = (0, 0); pair BB = (0, 1); pair CC = (2, 0); pair DD = (2, 1);    //draw(A--B--D--C--cycle);    draw(A--B); label("1",midpoint(A--B),W); label("2",midpoint(D--B),N); draw(A--C,dashed); draw(B--D); draw(C--D, dashed);  draw(A--AA); label("5",midpoint(A--AA),W); draw(B--BB,dashed); draw(C--CC,dashed); draw(D--DD); label("5",midpoint(D--DD),E); label("1",midpoint(CC--DD),E); label("2",midpoint(AA--CC),S);  // Dotted vertices dot(A); dot(B); dot(C); dot(D);    dot(AA); dot(BB); dot(CC); dot(DD);  draw(AA--BB,dashed); draw(AA--CC); draw(BB--DD,dashed); draw(CC--DD);  label("(0,0)",AA,W); label("(-3,4)",A,SW); label("(-1,5)",D,E); label("(2,1)",DD,NE); [/asy] Notice that this we are given a parametric form of the region, and $w$ is used in both $x$ and $y$. We first fix $u$ and $v$ to $0$, and graph $(-3w,4w)$ from $0\le w\le1$. When $w$ is $0$, we have the point $(0,0)$, and when $w$ is $1$, we have the point $(-3,4)$. We see that since this is a directly proportional function, we can just connect the dots like this:

[asy] 	import graph; 	Label f; size(5cm); 	unitsize(0.7cm);  	xaxis(-5,5,Ticks(f, 5.0, 1.0)); 	yaxis(-5,5,Ticks(f, 5.0, 1.0));  	draw((0,0)--(-3,4)); 	[/asy]

Now, when we vary $u$ from $0$ to $2$, this line is translated to the right $2$ units:

[asy] 	import graph; 	Label f;  	unitsize(0.7cm); size(5cm); 	xaxis(-5,5,Ticks(f, 5.0, 1.0)); 	yaxis(-5,5,Ticks(f, 5.0, 1.0));  	draw((0,0)--(-3,4)); 	draw((2,0)--(-1,4)); 	[/asy]

We know that any points in the region between the line (or rather segment) and its translation satisfy $w$ and $u$, so we shade in the region:

[asy] 	import graph; 	Label f;  	unitsize(0.7cm); size(5cm); 	xaxis(-5,5,Ticks(f, 5.0, 1.0)); 	yaxis(-5,5,Ticks(f, 5.0, 1.0));  	draw((0,0)--(-3,4)); 	draw((2,0)--(-1,4));  	filldraw((0,0)--(-3,4)--(-1,4)--(2,0)--cycle, gray); 	[/asy]

We can also shift this quadrilateral one unit up, because of $v$. Thus, this is our figure:

[asy] 	import graph; 	Label f;  	unitsize(0.7cm); size(5cm); 	xaxis(-5,5,Ticks(f, 5.0, 1.0)); 	yaxis(-5,5,Ticks(f, 5.0, 1.0));  	draw((0,0)--(-3,4)); 	draw((2,0)--(-1,4));  	filldraw((0,0)--(-3,4)--(-1,4)--(2,0)--cycle, gray); 	filldraw((0,1)--(-3,5)--(-1,5)--(2,1)--cycle, gray);  draw((0,0)--(0,1),black+dashed); draw((2,0)--(2,1),black+dashed); draw((-3,4)--(-3,5),black+dashed); 	[/asy]

[asy] 	import graph; 	Label f;  	unitsize(0.7cm); size(5cm); 	xaxis(-5,5,Ticks(f, 5.0, 1.0)); 	yaxis(-5,5,Ticks(f, 5.0, 1.0));  	draw((0,0)--(-3,4)); 	draw((1,0)--(-2,4));  	filldraw((0,0)--(2,0)--(2,1)--(-1,5)--(-3,5)--(-3,4)--cycle, gray); 	[/asy]

The length of the boundary is simply $1+2+5+1+2+5$ ($5$ can be obtained by Pythagorean theorem, since we have side lengths $3$ and $4$.). This equals $\boxed{\textbf{(E) }16.}$

~Technodoggo ~ESAOPS

Video Solution 1 by OmegaLearn

https://youtu.be/KEV3ka5gWYU

Video Solution

https://youtu.be/bqIlsWTOL3k

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2023 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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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