2023 AMC 10B Problems/Problem 24
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
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 |
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