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

Revision as of 17:52, 11 July 2020

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

@@ Line 24: / Line 24: @@
 <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>.
 ===Solution 2===
-== Solution 2 ==
 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>.
@@ Line 35: / Line 33: @@
 <cmath>5a + 10d = 60.</cmath>
 Dividing both sides by 5, we get <math>a + 2d = \boxed{12}</math>, which is the middle term. The answer is (D).
+===Solution 3===
+Not too bad with some logic and the awesome guess and check. Let <math>A=6</math>. Then let <math>B=7,E=5</math> and <math>C=3,D=9</math>. Our arithmetic sequence is <math>10,11,12,13,14</math> so our answer is <math>12 \Longrightarrow \mathrm{(D)}</math>.
+Solution by franzliszt
 == 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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