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

m (Fixed asymptote)
(Solution 1)
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==Solution 1==
 
==Solution 1==
  
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<asy>
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import geometry;
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pair A = (-3, 4);
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pair B = (-3, 5);
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pair C = (-1, 4);
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pair D = (-1, 5);
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pair AA = (0, 0);
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pair BB = (0, 1);
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pair CC = (2, 0);
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pair DD = (2, 1);
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//draw(A--B--D--C--cycle);
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draw(A--B);
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label("1",midpoint(A--B),W);
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label("2",midpoint(D--B),N);
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draw(A--C,dashed);
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draw(B--D);
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draw(C--D, dashed);
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draw(A--AA);
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label("5",midpoint(A--AA),W);
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draw(B--BB,dashed);
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draw(C--CC,dashed);
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draw(D--DD);
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label("5",midpoint(D--DD),E);
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label("1",midpoint(CC--DD),E);
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label("2",midpoint(AA--CC),S);
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// Dotted vertices
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dot(A); dot(B); dot(C); dot(D);
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dot(AA); dot(BB); dot(CC); dot(DD);
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draw(AA--BB,dashed);
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draw(AA--CC);
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draw(BB--DD,dashed);
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draw(CC--DD);
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label("(0,0)",AA,W);
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label("(-3,4)",A,SW);
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label("(-1,5)",D,E);
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label("(2,1)",DD,NE);
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</asy>
 
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>:  
 
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>:  
  

Revision as of 14:08, 15 November 2023

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 $0\le w\le1$?


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 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)--(-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);  	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