# 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

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