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2023 AMC 10B Problems/Problem 24

Revision as of 20:55, 15 November 2023 by Eevee9406 (talk | contribs) (insert answers, see also)

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 $0\le w\le1$?

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

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 $0\le w\le1$:

[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

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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