Olympiad Maths Prep

Track / Stage 3 / 209 of 260 #209 of 2000

Problem 209

AMC 10/12, early questions
Geometry Difficulty 3.8 Find the answer

What is the perimeter of the boundary of the region consisting of all points which can be expressed as (2u3w,v+4w)(2u-3w, v+4w) with 0u10\le u\le1, 0v1,0\le v\le1, and 0w10\le w\le1?
(A) 103(B) 13(C) 12(D) 18(E) 16\textbf{(A) } 10\sqrt{3} \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 16

Official solution

Notice that this we are given a parametric form of the region, and ww is used in both xx and yy. We first fix uu and vv to 00, and graph (3w,4w)(-3w,4w) from 0w10\le w\le1. When ww is 00, we have the point (0,0)(0,0), and when ww is 11, we have the point (3,4)(-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 uu from 00 to 22, this line is translated to the right 22 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 ww and uu, 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 vv. 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+51+2+5+1+2+5 (55 can be obtained by Pythagorean theorem, since we have side lengths 33 and 44.). This equals (E) 16.\boxed{\textbf{(E) }16.}
~Technodoggo ~ESAOPS

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.