Difference between revisions of "2005 AMC 12A Problems/Problem 13"
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+ | ===Solution 1=== | ||
<math>(A+B) + (B+C) + (C+D) + (D+E) + (E+A) = 2(A+B+C+D+E)</math> (i.e., each number is counted twice). The sum <math>A + B + C + D + E</math> will always be <math>3 + 5 + 6 + 7 + 9 = 30</math>, so the arithmetic sequence has a sum of <math>2 \cdot 30 = 60</math>. The middle term must be the average of the five numbers, which is <math>\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}</math>. | <math>(A+B) + (B+C) + (C+D) + (D+E) + (E+A) = 2(A+B+C+D+E)</math> (i.e., each number is counted twice). The sum <math>A + B + C + D + E</math> will always be <math>3 + 5 + 6 + 7 + 9 = 30</math>, so the arithmetic sequence has a sum of <math>2 \cdot 30 = 60</math>. The middle term must be the average of the five numbers, which is <math>\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}</math>. | ||
− | + | ===Solution 2=== | |
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== Solution 2 == | == Solution 2 == |
Revision as of 18:48, 11 July 2020
Problem
In the five-sided star shown, the letters ,
,
,
and
are replaced by the
numbers 3, 5, 6, 7 and 9, 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
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).
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.