Difference between revisions of "2023 AMC 10B Problems/Problem 24"
m (→Solution 2) |
Roger8432v3 (talk | contribs) (→Solution 3 (Not rigorous)) |
||
(3 intermediate revisions by 2 users not shown) | |||
Line 72: | Line 72: | ||
</asy> | </asy> | ||
− | Now, when we vary <math> | + | Now, when we vary <math>2u</math> from <math>0</math> to <math>2</math>, this line is translated to the right <math>2</math> units: |
<asy> | <asy> | ||
Line 218: | Line 218: | ||
~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
+ | |||
+ | ==Solution 3 (Quick, Not rigorous)== | ||
+ | |||
+ | Let's assume that <math>w = 0</math>. If we plug <math>(0, 0), (0, 1), (1, 1), (1, 0)</math> into <math>u</math> and <math>v</math> respectively, we will form a rectangle that has side lengths <math>1</math> and <math>2</math>. If we change <math>w</math> to the maximum value (which is <math>1</math>), we will find out that w shifts the rectangle by <math>-3w</math> on the x-axis and <math>4w</math> on the y-axis. Since we need to find the perimeter of the shift and rectangle, we can calculate that the length that the corner has move during the shift using the [https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem?srsltid=AfmBOort48ZEr_91TCSYH_SXPuY9On1yzJxPuh5gqQ8d0P1dp_c8onDg Pythagorean Theorem]. Thus, the distance is <math>5</math>. Now, we can find the perimeter of the boundary which is <math>2(1 + 2 + 5) = 16</math>. This means that the answer is <math>\boxed{\textbf{(E)}~ 16}</math>. | ||
+ | |||
+ | ~ROGER8432V3 | ||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== |
Latest revision as of 10:31, 27 July 2025
Contents
Problem
What is the perimeter of the boundary of the region consisting of all points which can be expressed as with
,
and
?
Solution 1
Notice that 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
, we have the point
, and when
is
, we have the point
. We see that since this is a directly proportional 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
Solution 2
We can find the "boundary points" and work with our intuition to solve the problem. We set each of equal to
for a total of
combinations in
. We now test each one.
Case 1:
Case 2:
Case 3:
Case 4:
Case 5:
Case 6:
Case 7:
Case 8:
When graphed on a coordinate plane, the points appear as follows.
Notice how there are two distinct rectangles visible in the figure. This leads us to believe that the region tracks the motion of this region as it travels in space. To understand why this is true, we can imagine a fixed (as it is present in both the
and
coordinates). Then if we hold one of
or
fixed and let the other vary, we get a straight line parallel to the
or
axis respectively. If we let the other vary, we get the other type of straight line. Together, they form a rectangular region. In addition,
serves as a diagonal translation, so if we now let
vary, it traces out the motion of the rectangle. Keeping this in mind, we connect the dots.
Each of the diagonal sides have length by the distance formula on
and
(the other diagonal side is congruent), so our total area is
.
~ cxsmi
Solution 3 (Quick, Not rigorous)
Let's assume that . If we plug
into
and
respectively, we will form a rectangle that has side lengths
and
. If we change
to the maximum value (which is
), we will find out that w shifts the rectangle by
on the x-axis and
on the y-axis. Since we need to find the perimeter of the shift and rectangle, we can calculate that the length that the corner has move during the shift using the Pythagorean Theorem. Thus, the distance is
. Now, we can find the perimeter of the boundary which is
. This means that the answer is
.
~ROGER8432V3
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.