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Difference between revisions of "2005 AMC 10A Problems/Problem 17"

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==Problem==
 
==Problem==
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In the five-sided star shown, the letters <math>A, B, C, D,</math> and <math>E</math> are replaced by the numbers <math>3, 5, 6, 7,</math> and <math>9</math>, although not necessarily in this order. The sums of the numbers at the ends of the line segments <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, and <math>EA</math> form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence?  
+
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 <math>3</math>, <math>5</math>, <math>6</math>, <math>7</math>, and <math>9</math>, although not necessarily in this 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{DE}</math>, and <math>\overline{EA}</math> form an arithmetic sequence, although not necessarily in this order. What is the middle term of the arithmetic sequence?  
  
−
[[Image:2005amc10a17.gif]]
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<asy>
 +
size(150);
 +
defaultpen(linewidth(0.8));
 +
string[] strng = {'A','D','B','E','C'};
 +
pair A=dir(90),B=dir(306),C=dir(162),D=dir(18),E=dir(234);
 +
draw(A--B--C--D--E--cycle);
 +
for(int i=0;i<=4;i=i+1)
 +
{
 +
path circ=circle(dir(90-72*i),0.125);
 +
unfill(circ);
 +
draw(circ);
 +
label("$"+strng[i]+"$",dir(90-72*i));
 +
}
 +
</asy>
  
−
<math> \textbf{(A) } 9\qquad \textbf{(B) } 10\qquad \textbf{(C) } 11\qquad \textbf{(D) } 12\qquad \textbf{(E) } 13 </math>
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<math>
 +
\textbf{(A) } 9\qquad \textbf{(B) } 10\qquad \textbf{(C) } 11\qquad \textbf{(D) } 12\qquad \textbf{(E) } 13
 +
</math>
  
−
==Solution==
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==Solution 1 (assuming a solution exists)==
−
Each corner <math>(A,B,C,D,E)</math> goes to two sides/numbers. (<math>A</math> goes to <math>AE</math> and <math>AB</math>, <math>D</math> goes to <math>DC</math> and <math>DE</math>). The sum of every term is equal to <math>2(3+5+6+7+9)=60</math>
+
Each of <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> forms part of exactly <math>2</math> sums along line segments (e.g. <math>A</math> forms part of the sums of <math>\overline{AB}</math> and <math>\overline{EA}</math>). Thus, the sum of all these line segments - i.e. the sum of the <math>5</math> terms of the arithmetic sequence - is exactly twice the sum of <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math>, and since those numbers are <math>3</math>, <math>5</math>, <math>6</math>, <math>7</math>, and <math>9</math> in some order, this reduces to <math>2 \cdot (3+5+6+7+9) = 60</math>.
  
−
Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is <math>\frac{60}{5}=\boxed{\textbf{(D) }12}</math>
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Now, since the middle term of an arithmetic sequence with an odd number of terms is the average of all the terms of the sequence, the middle number must be <math>\frac{60}{5} = \boxed{\textbf{(D) } 12}</math>.
  
−
== Video Solution ==
+
==Solution 2 (assuming a solution exists; quick)==
 +
We observe that the average of <math>3</math>, <math>5</math>, <math>6</math>, <math>7</math>, and <math>9</math> is <math>\frac{3+5+6+7+9}{5} = \frac{30}{5} = 6</math>. Accordingly, setting all <math>5</math> of <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> to <math>6</math> will give the same sum of sums along the line segments - i.e. the sum of the <math>5</math> terms of the arithmetic sequence (as in Solution 1) - as if we had constructed an actual solution. In particular, this results in all the terms of the arithmetic sequence being <math>6+6 = 12</math>, and again as in Solution 1, the middle term must be the average of the <math>5</math> terms, so it is necessarily <math>\boxed{\textbf{(D) } 12}</math>.
 +
 
 +
==Solution 3 (rigorous) ==
 +
As the numbers <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> are <math>3</math>, <math>5</math>, <math>6</math>, <math>7</math>, and <math>9</math> in some order, we deduce that the smallest term of the arithmetic sequence must be at least <math>3 + 5 = 8</math>, while the largest term must be at most <math>7 + 9 = 16</math>.
 +
 
 +
Thus, as this sequence has <math>5</math> terms, its common difference <math>d</math> is at most <math>\frac{16-8}{5-1} = 2</math>.
 +
 
 +
But as <math>d</math> is positive (taking the arithmetic sequence in increasing order without loss of generality) and an integer (since it is a difference of sums of integers), while exactly <math>2</math> of the sums must be odd (since exactly <math>4</math> of the numbers <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> are odd, while the <math>5</math>th number is even), it follows that <math>d</math> must be odd. Therefore, the only possibility is <math>d = 1</math>.
 +
 
 +
Next, again noting that the sequence has exactly <math>2</math> odd terms, and thus <math>5-2 = 3 > 2</math> even terms, we observe that the middle term must have the majority parity, i.e. must be even. This means the <math>2</math> terms adjacent to the middle term must be odd, and therefore must be <math>(6+a)</math> and <math>(6+b)</math> with <math>a,b \in \{3,5,7,9\}</math>. Moreover, <math>(6+a)-(6-b) = a-b</math> is exactly twice the common difference, i.e. <math>2d = 2 \cdot 1 = 2</math>.
 +
 
 +
The only other observation we need is that <math>a</math> and <math>b</math> can't be the smallest, or largest, <math>2</math> odd numbers, because that would make it impossible to construct the smallest, or (respectively) largest, sum from <math>1</math> of the remaining <math>2</math> even numbers and <math>1</math> of the odd numbers. Together with <math>a-b = 2</math> from above, this implies that <math>a = 5</math> and <math>b = 7</math>, so finally, the middle term of the sequence must be <math>(6+5) + 1 = (6+7) - 1 = \boxed{\textbf{(D) } 12}</math>.
 +
 
 +
(It is now easy to see that one possible solution is <math>A = 3</math>, <math>B = 9</math>, <math>C = 5</math>, <math>D = 6</math>, and <math>E = 7</math>; this gives the edge sums as <math>12</math>, <math>14</math>, <math>11</math>, <math>13</math>, and <math>10</math>, which indeed form the arithmetic sequence <math>10,11,12,13,14</math>.)
 +
 
 +
~oinava
 +
 
 +
==Video Solution 1==
 
https://youtu.be/tKsYSBdeVuw?t=544
 
https://youtu.be/tKsYSBdeVuw?t=544
 +
by OmegaLearn
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
==Video Solution 2==
 +
https://www.youtube.com/watch?v=RljLmFCF2tU
 +
 +
~By Ajeet Dubey (https://www.ioqm.in)
  
 
==See Also==
 
==See Also==
−
 
+
[[Category:Introductory Algebra Problems]]
 
{{AMC10 box|year=2005|ab=A|num-b=16|num-a=18}}
 
{{AMC10 box|year=2005|ab=A|num-b=16|num-a=18}}
  
−
[[Category:Introductory Geometry Problems]]
 
−
[[Category:Area Ratio Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:21, 2 July 2025

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 this 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 this order. What is the middle term of the arithmetic sequence?

[asy] size(150); defaultpen(linewidth(0.8)); string[] strng = {'A','D','B','E','C'}; pair A=dir(90),B=dir(306),C=dir(162),D=dir(18),E=dir(234); draw(A--B--C--D--E--cycle); for(int i=0;i<=4;i=i+1) { path circ=circle(dir(90-72*i),0.125); unfill(circ); draw(circ); label("$"+strng[i]+"$",dir(90-72*i)); } [/asy]

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

Solution 1 (assuming a solution exists)

Each of $A$, $B$, $C$, $D$, and $E$ forms part of exactly $2$ sums along line segments (e.g. $A$ forms part of the sums of $\overline{AB}$ and $\overline{EA}$). Thus, the sum of all these line segments - i.e. the sum of the $5$ terms of the arithmetic sequence - is exactly twice the sum of $A$, $B$, $C$, $D$, and $E$, and since those numbers are $3$, $5$, $6$, $7$, and $9$ in some order, this reduces to $2 \cdot (3+5+6+7+9) = 60$.

Now, since the middle term of an arithmetic sequence with an odd number of terms is the average of all the terms of the sequence, the middle number must be $\frac{60}{5} = \boxed{\textbf{(D) } 12}$.

Solution 2 (assuming a solution exists; quick)

We observe that the average of $3$, $5$, $6$, $7$, and $9$ is $\frac{3+5+6+7+9}{5} = \frac{30}{5} = 6$. Accordingly, setting all $5$ of $A$, $B$, $C$, $D$, and $E$ to $6$ will give the same sum of sums along the line segments - i.e. the sum of the $5$ terms of the arithmetic sequence (as in Solution 1) - as if we had constructed an actual solution. In particular, this results in all the terms of the arithmetic sequence being $6+6 = 12$, and again as in Solution 1, the middle term must be the average of the $5$ terms, so it is necessarily $\boxed{\textbf{(D) } 12}$.

Solution 3 (rigorous)

As the numbers $A$, $B$, $C$, $D$, and $E$ are $3$, $5$, $6$, $7$, and $9$ in some order, we deduce that the smallest term of the arithmetic sequence must be at least $3 + 5 = 8$, while the largest term must be at most $7 + 9 = 16$.

Thus, as this sequence has $5$ terms, its common difference $d$ is at most $\frac{16-8}{5-1} = 2$.

But as $d$ is positive (taking the arithmetic sequence in increasing order without loss of generality) and an integer (since it is a difference of sums of integers), while exactly $2$ of the sums must be odd (since exactly $4$ of the numbers $A$, $B$, $C$, $D$, and $E$ are odd, while the $5$th number is even), it follows that $d$ must be odd. Therefore, the only possibility is $d = 1$.

Next, again noting that the sequence has exactly $2$ odd terms, and thus $5-2 = 3 > 2$ even terms, we observe that the middle term must have the majority parity, i.e. must be even. This means the $2$ terms adjacent to the middle term must be odd, and therefore must be $(6+a)$ and $(6+b)$ with $a,b \in \{3,5,7,9\}$. Moreover, $(6+a)-(6-b) = a-b$ is exactly twice the common difference, i.e. $2d = 2 \cdot 1 = 2$.

The only other observation we need is that $a$ and $b$ can't be the smallest, or largest, $2$ odd numbers, because that would make it impossible to construct the smallest, or (respectively) largest, sum from $1$ of the remaining $2$ even numbers and $1$ of the odd numbers. Together with $a-b = 2$ from above, this implies that $a = 5$ and $b = 7$, so finally, the middle term of the sequence must be $(6+5) + 1 = (6+7) - 1 = \boxed{\textbf{(D) } 12}$.

(It is now easy to see that one possible solution is $A = 3$, $B = 9$, $C = 5$, $D = 6$, and $E = 7$; this gives the edge sums as $12$, $14$, $11$, $13$, and $10$, which indeed form the arithmetic sequence $10,11,12,13,14$.)

~oinava

Video Solution 1

https://youtu.be/tKsYSBdeVuw?t=544 by OmegaLearn

~ pi_is_3.14

Video Solution 2

https://www.youtube.com/watch?v=RljLmFCF2tU

~By Ajeet Dubey (https://www.ioqm.in)

See Also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 10 Problems and Solutions

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