Difference between revisions of "2005 AMC 10A Problems/Problem 17"
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− | We notice that the average of <math>3,5,6,7,</math> and <math>9</math> is 6. Now all we have to do is plug <math>6</math> into each value. This results in all the terms being <math>12</math> and since the term we are looking for is the middle term, we know that our answer is 12, D. | + | We notice that the average of <math>3,5,6,7,</math> and <math>9</math> is <math>6</math>. Now all we have to do is plug <math>6</math> into each value. This results in all the terms being <math>12</math> and since the term we are looking for is the middle term, we know that our answer is <math>12</math>, \boxed{D}. |
== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == |
Revision as of 23:47, 9 September 2024
Contents
Problem
In the five-sided star shown, the letters and
are replaced by the numbers
and
, although not necessarily in this 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?
Solution 1
(meta-solving; answer choices imply solution exists)
Each corner goes to two sides/numbers. (
goes to
and
,
goes to
and
). The sum of every term is equal to
Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is
Solution 2 (Doesn't assume a solution exists)
We know that the smallest number in the arithmetic sequence must be , and the largest number must be
.
Since there are terms in this sequence, the common difference
.
Since is an integer (difference of sums of integers), and since exactly 2 of the sums must be odd,
must be odd. Therefore,
.
The middle term must have the majority parity, so it must be even.
The 2 terms adjacent to the middle are odd, and
.
.
and
can't be the smallest (or largest) 2 odd numbers, because that would make it impossible to construct the smallest (or largest) sum from one of the remaining two numbers and one of the odd numbers. Therefore,
. The middle sum must then be
. The remaining edges are
(because
can't be an edge, as that would make a triangle with 6),
, and
.
~oinava
Solution 3
We notice that the average of and
is
. Now all we have to do is plug
into each value. This results in all the terms being
and since the term we are looking for is the middle term, we know that our answer is
, \boxed{D}.
Video Solution by OmegaLearn
https://youtu.be/tKsYSBdeVuw?t=544
~ pi_is_3.14
See Also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.