Difference between revisions of "2005 AMC 12A Problems/Problem 13"
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== Problem == | == Problem == | ||
In the five-sided star shown, the letters <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> and <math>E</math> are replaced by the | In the five-sided star shown, the letters <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> and <math>E</math> are replaced by the | ||
− | numbers 3, 5, 6, 7 and 9, although not necessarily in that order. The sums of the | + | numbers <math>3, 5, 6, 7</math> and <math>9,</math> although not necessarily in that order. The sums of the |
numbers at the ends of the line segments <math>\overline{AB}</math>, <math>\overline{BC}</math>, <math>\overline{CD}</math>, <math>\overline{DE}</math>, and <math>\overline{EA}</math> form an | numbers at the ends of the line segments <math>\overline{AB}</math>, <math>\overline{BC}</math>, <math>\overline{CD}</math>, <math>\overline{DE}</math>, and <math>\overline{EA}</math> form an | ||
arithmetic sequence, although not necessarily in that order. What is the middle | arithmetic sequence, although not necessarily in that order. What is the middle |
Latest revision as of 10:01, 22 August 2024
Problem
In the five-sided star shown, the letters , , , and are replaced by the numbers and although not necessarily in that order. The sums of the numbers at the ends of the line segments , , , , and form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence?
Solutions
Solution 1
(i.e., each number is counted twice). The sum will always be , so the arithmetic sequence has a sum of . The middle term must be the average of the five numbers, which is .
Solution 2
Let the terms in the arithmetic sequence be , , , , and . We seek the middle term .
These five terms are , , , , and , in some order. The numbers , , , , and are equal to 3, 5, 6, 7, and 9, in some order, so Hence, the sum of the five terms is But adding all five numbers, we also get , so Dividing both sides by 5, we get , which is the middle term. The answer is (D).
Solution 3
Not too bad with some logic and the awesome guess and check. Let . Then let and . Our arithmetic sequence is so our answer is .
Solution by franzliszt
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.