Solutions

Summary

Here are previously published solutions to the competition problems as they appear at www.edfinity.com. One can also find these solutions in the MAA's published texts on the AMC competitions (go to www.maa.org/publications/books and click on “Book Categories” to find their problem-solving books).

Warning: Each solution presented here is fast paced and to the point, simply working through the mathematics of the task at hand to get the job done. These solutions are written by a variety of authors.

For an account of the problem-solving practices behind each solution guiding themathematical steps presented—along with discussion on its connections to the Common Core State Standards and further, deeper, queries and possible explorations—see the Curriculum Bursts at www.maa.org/ci.

Comment. Sometimes multiple solutions to the same problem were published, with the alternate solutions using mathematical techniques different from those discussed in the sections of this text. We present here only the solutions featuring the tools of the Pythagorean Theorem and trigonometry. We've also made some minor editorial changes to these published solutions.

1. (A)

By the Pythagorean Theorem, (Or note that triangle AEB is similar to a 3-4-5 right triangle, so AE = 3 × 6 = 18.) It follows that CF = 24 and. The perimeter of the trapezoid is 50 + 30 + 18 + 50 + 7 + 25 = 180.

2. (A)

Assume that Alice starts at A in the figure and ends at B. In ΔABC, is right and. The Pythagorean Theorem shows that (AB)² = 13.

OR

In ΔABD in the second figure,

1) ∠ADB = 60°,

2) AD = 4, and

3) BD = 1.

By the law of cosines,

(AB)²= 1² + 4² – 2(1)(4)(1/ 2) =13.

Therefore

3. (B)

Because AB = 1, the smallest number of jumps is at least 2. The perpendicular bisector ofis the line with equation, which has no points with integer coordinates, so two jumps are not possible. A sequence of three jumps is possible; one such sequence is (0, 0) to (3, 4) to (6, 0) to (1, 0).