# 2005 AMC 12A Problems/Problem 13

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## 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?

$[asy] draw((0,0)--(0.5,1.54)--(1,0)--(-0.31,0.95)--(1.31,0.95)--cycle); label("A",(0.5,1.54),N); label("B",(1,0),SE); label("C",(-0.31,0.95),W); label("D",(1.31,0.95),E); label("E",(0,0),SW); [/asy]$

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

## Solutions

### Solution 1

$(A+B) + (B+C) + (C+D) + (D+E) + (E+A) = 2(A+B+C+D+E)$ (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).

### Solution 3

Not too bad with some logic and the awesome guess and check. Let $A=6$. Then let $B=7,E=5$ and $C=3,D=9$. Our arithmetic sequence is $10,11,12,13,14$ so our answer is $12 \Longrightarrow \mathrm{(D)}$.

Solution by franzliszt