Difference between revisions of "2023 AMC 10B Problems/Problem 24"
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==Solution 1== | ==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 <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 with , and ?
Solution 1
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 :
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