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Difference between revisions of "2005 AMC 12A Problems/Problem 13"

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<math>(A+B) + (B+C) + (C+D) + (D+E) + (E+A) = 2(A+B+C+D+E)</math> (i.e., each number is counted twice). 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>.
 
<math>(A+B) + (B+C) + (C+D) + (D+E) + (E+A) = 2(A+B+C+D+E)</math> (i.e., each number is counted twice). 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>.
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== Solution 2 ==
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Let the terms in the arithmetic sequence be <math>a</math>, <math>a + d</math>, <math>a + 2d</math>, <math>a + 3d</math>, and <math>a + 4d</math>. We seek the middle term <math>a + 2d</math>.
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These five terms are <math>A + B</math>, <math>B + C</math>, <math>C + D</math>, <math>D + E</math>, and <math>E + A</math>, in some order. The numbers <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> are equal to 3, 5, 6, 7, and 9, in some order, so
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<cmath>A + B + C + D + E = 3 + 5 + 6 + 7 + 9 = 30.</cmath>
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Hence, the sum of the five terms is
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<cmath>(A + B) + (B + C) + (C + D) + (D + E) + (E + A) = 2A + 2B + 2C + 2D + 2E = 60.</cmath>
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But adding all five numbers, we also get <math>a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 5a + 10d</math>, so
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<cmath>5a + 10d = 60.</cmath>
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Dividing both sides by 5, we get <math>a + 2d = \boxed{12}</math>, which is the middle term. The answer is (D).
  
 
== See also ==
 
== See also ==

Revision as of 09:11, 17 July 2019

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 $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DE}$, 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


$(A+B) + (B+C) + (C+D) + (D+E) + (E+A) = 2(A+B+C+D+E)$ (i.e., each number is counted twice). The sum $A + B + C + D + E$ will always be $3 + 5 + 6 + 7 + 9 = 30$, so the arithmetic sequence has a sum of $2 \cdot 30 = 60$. The middle term must be the average of the five numbers, which is $\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}$.


Solution 2

Let the terms in the arithmetic sequence be $a$, $a + d$, $a + 2d$, $a + 3d$, and $a + 4d$. We seek the middle term $a + 2d$.

These five terms are $A + B$, $B + C$, $C + D$, $D + E$, and $E + A$, in some order. The numbers $A$, $B$, $C$, $D$, and $E$ are equal to 3, 5, 6, 7, and 9, in some order, so \[A + B + C + D + E = 3 + 5 + 6 + 7 + 9 = 30.\] Hence, the sum of the five terms is \[(A + B) + (B + C) + (C + D) + (D + E) + (E + A) = 2A + 2B + 2C + 2D + 2E = 60.\] But adding all five numbers, we also get $a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 5a + 10d$, so \[5a + 10d = 60.\] Dividing both sides by 5, we get $a + 2d = \boxed{12}$, which is the middle term. The answer is (D).

See also

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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All AMC 12 Problems and Solutions

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