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Difference between revisions of "2023 AMC 10B Problems/Problem 24"

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==Solution 2==
 
==Solution 2==
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We can find the "boundary points" and work with our intuition to solve the problem. We set each of <math>u, v, w</math> equal to <math>0, 1</math> for a total of <math>8</math> combinations in <math>u, v, w</math> (these points will all be on the boundary of the region). We now test each one.
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We can find the "boundary points" and work with our intuition to solve the problem. We set each of <math>u, v, w</math> equal to <math>0, 1</math> for a total of <math>8</math> combinations in <math>u, v, w</math>. We now test each one.
  
 
Case 1: <math>u = 0, v = 0, w = 0 \implies (0, 0)</math>   
 
Case 1: <math>u = 0, v = 0, w = 0 \implies (0, 0)</math>   

Revision as of 12:30, 4 April 2024

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 1

[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 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

Solution 2

We can find the "boundary points" and work with our intuition to solve the problem. We set each of $u, v, w$ equal to $0, 1$ for a total of $8$ combinations in $u, v, w$. We now test each one.

Case 1: $u = 0, v = 0, w = 0 \implies (0, 0)$

Case 2: $u = 0, v = 0, w = 1 \implies (-3, 4)$

Case 3: $u = 0, v = 1, w = 0 \implies (0, 1)$

Case 4: $u = 0, v = 1, w = 1 \implies (-3, 5)$

Case 5: $u = 1, v = 0, w = 0 \implies (2, 0)$

Case 6: $u = 1, v = 0, w = 1 \implies (-1, 4)$

Case 7: $u = 1, v = 1, w = 0 \implies (2, 1)$

Case 8: $u = 1, v = 1, w = 1 \implies (-1, 5)$

When graphed on a coordinate plane, the points appear as follows.

[asy] import graph; import geometry; 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)); pair A = (0, 0); dot (A); pair B = (-3, 4); dot (B); pair C = (0, 1); dot (C); pair D = (-3, 5); dot (D); pair E = (2, 0); dot (E); pair F = (-1, 4); dot (F); pair G = (2, 1); dot (G); pair H = (-1, 5); dot (H); [/asy]

Notice how there are two distinct rectangles visible in the figure. This leads us to believe that the region tracks the motion of this region as it travels in space. To understand why this is true, we can imagine a fixed $w$ (as it is present in both the $x$ and $y$ coordinates). Then if we hold one of $u$ or $v$ fixed and let the other vary, we get a straight line parallel to the $x$ or $y$ axis respectively. If we let the other vary, we get the other type of straight line. Together, they form a rectangular region. In addition, $w$ serves as a diagonal translation, so if we now let $w$ vary, it traces out the motion of the rectangle. Keeping this in mind, we connect the dots.

[asy] import graph; import geometry; 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)--(-3,5)--(-1,5)--(2,1)--(2,0)--cycle, gray); [/asy]

Each of the diagonal sides have length $5$ by the distance formula on $(0,0)$ and $(-3,4)$ (the other diagonal side is congruent), so our total area is $2 + 1 + 5 + 2 + 1 + 5 = \boxed{\textbf{(E)}~ 16}$.

~ cxsmi

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 (Problems • Answer Key • Resources)
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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