Difference between revisions of "2005 AMC 12A Problems/Problem 13"
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| − | <math>AB + BC + CD + DE + EA = 2(A+B+C+D+E)</math>. 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>. | + | <math>AB + BC + CD + DE + EA = 2(A+B+C+D+E)</math>. 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>. |
== See also == | == See also == | ||
Revision as of 16:40, 24 July 2017
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
. 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. 
