2021 Fall AMC 12A Problems/Problem 24
Contents
Problem
Convex quadrilateral has and In some order, the lengths of the four sides form an arithmetic progression, and side is a side of maximum length. The length of another side is What is the sum of all possible values of ?
Solution 1
Let be a point on such that is a parallelogram. Suppose that and so as shown below. We apply the Law of Cosines to Let be the common difference of the arithmetic progression of the side-lengths. It follows that and are and in some order. It is clear that
If then is a rhombus with side-length which is valid.
If then we have six cases:
Note that becomes from which So, this case generates no valid solutions
Note that becomes from which So, this case generates
Note that becomes from which So, this case generates no valid solutions
Note that becomes from which So, this case generates
Note that becomes from which So, this case generates no valid solutions
Note that becomes from which So, this case generates no valid solutions
Together, the sum of all possible values of is
~MRENTHUSIASM
Solution 2
Let , and denote the sides , and respectively. Since , we get Using , we eliminate from above to get , which rearranges to , and, upon factoring, yields We divide into two cases, depending on whether is the smallest side.
If is not the smallest side then . If , we get a rhombus of side , so one possible value is . Otherwise, we can cancel the common factor from . After rearranging we get The first condition is false because ; the second condition is false because .
If is the smallest side, then . Assuming we can cancel common factors in to get The first condition yields the solution and the second condition yields the solution .
Together, the sum of all possible values of is
Solution 3
Denote , . Hence, , .
: .
This is a rhombus. So each side has length .
For the following cases, we consider four sides that have distinct lengths. To make their lengths an arithmetic sequence, we must have .
Therefore, in the subsequent analysis, we exclude the solution .
: .
Because the lengths of these sides form an arithmetic sequence, we have the following system of equations:
Hence,
By solving this system of equations, we get .
Thus, in this case, , , .
: .
Because the lengths of these sides form an arithmetic sequence, we have the following system of equations:
Hence,
By solving this system of equations, we get .
Thus, in this case, , , .
: .
By doing the similar analysis, we can show there is no solution in this case.
: .
By doing the similar analysis, we can show there is no solution in this case.
: .
By doing the similar analysis, we can show there is no solution in this case.
: .
By doing the similar analysis, we can show there is no solution in this case.
Therefore, the sum of all possible values of is
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Video Solution
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See Also
2021 Fall AMC 12A (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 12 Problems and Solutions |
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