Difference between revisions of "2023 AMC 10B Problems/Problem 24"
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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>. When <math>w</math> is 0, we have the point <math>(0,0)</math>, and when <math>w</math> is 1, we have the point <math>(-3,4)</math>. We see that since this is a directly poportional function, we can just connect the dots like this | + | 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>. When <math>w</math> is 0, we have the point <math>(0,0)</math>, and when <math>w</math> is 1, we have the point <math>(-3,4)</math>. We see that since this is a directly poportional function, we can just connect the dots like this: |
<asy> | <asy> |
Revision as of 11:54, 21 November 2023
Problem
What is the perimeter of the boundary of the region consisting of all points which can be expressed as with
,
and
?
Solution
Notice that this we are given a parametric form of the region, and
is used in both
and
. We first fix
and
to
, and graph
from
. When
is 0, we have the point
, and when
is 1, we have the point
. We see that since this is a directly poportional function, we can just connect the dots like this:
Now, when we vary from
to
, this line is translated to the right
units:
We know that any points in the region between the line (or rather segment) and its translation satisfy and
, so we shade in the region:
We can also shift this quadrilateral one unit up, because of . Thus, this is our figure:
The length of the boundary is simply (
can be obtained by Pythagorean theorem, since we have side lengths
and
.). This equals
~Technodoggo ~ESAOPS
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.