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

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== Problem ==
 
== Problem ==
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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>.
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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
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numbers 3, 5, 6, 7 and 9, although not necessarily in that order. The sums of the
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numbers at the ends of the line segments <math>\overline{AB}</math>, <math>\overline{BC}</math>, <math>\overline{CD}</math>, <math>\overline{DE}</math>, and <math>\overline{EA}</math> form an
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arithmetic sequence, although not necessarily in that order. What is the middle
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term of the arithmetic sequence?
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<asy>
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draw((0,0)--(0.5,1.54)--(1,0)--(-0.31,0.95)--(1.31,0.95)--cycle);
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label("$A$",(0.5,1.54),N);
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label("$B$",(1,0),SE);
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label("$C$",(-0.31,0.95),W);
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label("$D$",(1.31,0.95),E);
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label("$E$",(0,0),SW);
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</asy>
  
 
<math>
 
<math>
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</math>
 
</math>
  
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== Solution ==
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== Solutions ==
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'''
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=== Solution 1 ===
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===Solution 1===
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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>.
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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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<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>. Since <math>CD</math> is the middle term, it must be the average of the five numbers, of <math>\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}</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).
  
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=== Solution 2 ===
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===Solution 3===
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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>.
  
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Assume AB is the smallest term (this assumption will drop out by the end). If AB is the smallest the sequence goes A+B,B+C,C+D,D+E,E+A. If (A+B)<(B+C) then A must be less than C. Using this method we eventually get A<C<E and B<D<A so we know B<D<A<C<E. Therefore D is 5 and C is 7 and CD is 12. If AB is the biggest it just works out to D being 7 and C being 5. '''D'''
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Solution by franzliszt
  
 
== See also ==
 
== See also ==

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

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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