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2005 AMC 12A Problems/Problem 13: Difference between revisions

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== Problem ==
== Problem ==
The regular 5-point star <math>ABCDE</math> is drawn and in each [[vertex]], there is a number. Each <math>A,B,C,D,</math> and <math>E</math> are chosen such that all 5 of them came from set <math>\{3,5,6,7,9\}</math>. Each letter is a different number (so one possible way is <math>A = 3, B = 5, C = 6, D = 7, E = 9</math>). Let <math>AB</math> be the sum of the numbers on <math>A</math> and <math>B</math>, and so forth. If <math>AB, BC, CD, DE,</math> and <math>EA</math> form an [[arithmetic sequence]] (not necessarily in increasing order), find the value of <math>CD</math>.
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 at the ends of the line segments <math>\overline{AB}</math>, <math>\overline{BC}</math>, <math>\overline{CD}</math>, <math>\overline{CE}</math>, and <math>\overline{EA}</math> 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>


<math>
<math>

Revision as of 16:38, 24 July 2017

Problem

In the five-sided star shown, the letters $A$, $B$, $C$, $D$ and $E$ 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 $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{CE}$, and $\overline{EA}$ 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]

$(\mathrm {A}) \ 9 \qquad (\mathrm {B}) \ 10 \qquad (\mathrm {C})\ 11 \qquad (\mathrm {D}) \ 12 \qquad (\mathrm {E})\ 13$

Solution


$AB + BC + CD + DE + EA = 2(A+B+C+D+E)$. The sum $A + B + C + D + E$ will always be $3 + 5 + 6 + 7 + 9 = 30$, so the arithmetic sequence has a sum of $2 \cdot 30 = 60$. Since $CD$ is the middle term, it must be the average of the five numbers, of $\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}$.

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

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