Difference between revisions of "2005 AMC 12A Problems/Problem 13"
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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 3, 5, 6, 7 and 9, 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{ | + | 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 | ||
term of the arithmetic sequence? | term of the arithmetic sequence? | ||
Revision as of 17:45, 17 July 2018
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?
![[asy] draw((0,0)--(0.5,1.54)--(1,0)--(-0.31,0.95)--(1.31,0.95)--cycle); label("$A$",(0.5,1.54),N); label("$B$",(1,0),SE); label("$C$",(-0.31,0.95),W); label("$D$",(1.31,0.95),E); label("$E$",(0,0),SW); [/asy]](http://latex.artofproblemsolving.com/b/7/4/b742a9eaa62d04f46229732736080eeb97f05475.png)
Solution
(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 
.
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. 
