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

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You ain't getting this one
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==Solution 1==
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Notice that this we are given a parametric form of the region, and <math>w</math> is used in both <math>x</math> and <math>y</math>. We first fix <math>u</math> and <math>v</math> to <math>0</math>, and graph <math>(-3w,4w)</math> from <math>0\le w\le1</math>:
 +
 
 +
<asy>
 +
import graph;
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Label f;
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 +
unitsize(0.7cm);
 +
 
 +
xaxis(-5,5,Ticks(f, 5.0, 1.0));
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yaxis(-5,5,Ticks(f, 5.0, 1.0));
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draw((0,0)--(-3,4));
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</asy>
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Now, when we vary <math>u</math> from <math>0</math> to <math>2</math>, this line is translated to the right <math>2</math> units:
 +
 
 +
<asy>
 +
import graph;
 +
Label f;
 +
 
 +
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));
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draw((2,0)--(-1,4));
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</asy>
 +
 
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We know that any points in the region between the line (or rather segment) and its translation satisfy <math>w</math> and <math>u</math>, so we shade in the region:
 +
 
 +
<asy>
 +
import graph;
 +
Label f;
 +
 
 +
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));
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draw((2,0)--(-1,4));
 +
 
 +
filldraw((0,0)--(-3,4)--(-1,4)--(2,0)--cycle, gray);
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</asy>
 +
 
 +
We can also shift this quadrilateral one unit up, because of <math>v</math>. Thus, this is our figure:
 +
 
 +
<asy>
 +
import graph;
 +
Label f;
 +
 
 +
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));
 +
draw((2,0)--(-1,4));
 +
 
 +
filldraw((0,0)--(-3,4)--(-1,4)--(2,0)--cycle, gray);
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filldraw((0,1)--(-3,5)--(-2,5)--(1,1)--cycle, gray);
 +
 
 +
draw((0,0)--(0,1),black+dashed);
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draw((2,0)--(2,1),black+dashed);
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draw((-3,4)--(-3,5),black+dashed);
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draw((-1,4)--(-1,5),black+dashed);
 +
</asy>
 +
 
 +
<asy>
 +
import graph;
 +
Label f;
 +
 
 +
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));
 +
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 <math>1+1+5+1+1+5</math> (<math>5</math> can be obtained by Pythagorean theorem, since we have side lengths <math>3</math> and <math>4</math>.). This equals <math>\boxed{\textbf{(E) }16.}</math>
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 +
~Technodoggo

Revision as of 13:20, 15 November 2023

Solution 1

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; 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); 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); 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); 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)--(-2,5)--(1,1)--cycle, gray); draw((0,0)--(0,1),black+dashed); draw((2,0)--(2,1),black+dashed); draw((-3,4)--(-3,5),black+dashed); draw((-1,4)--(-1,5),black+dashed); [/asy]

[asy] import graph; Label f; 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)); 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 $1+1+5+1+1+5$ ($5$ can be obtained by Pythagorean theorem, since we have side lengths $3$ and $4$.). This equals $\boxed{\textbf{(E) }16.}$

~Technodoggo

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