2005 AMC 10A Problems/Problem 17
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 odd.
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
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. 
