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

Revision as of 22:46, 9 September 2024

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 $AB$ , $BC$ , $CD$ , $DE$ , and $EA$ form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence?

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

Solution 1

(meta-solving; answer choices imply solution exists)

Each corner $(A,B,C,D,E)$ goes to two sides/numbers. ( $A$ goes to $AE$ and $AB$ , $D$ goes to $DC$ and $DE$ ). The sum of every term is equal to $2(3+5+6+7+9)=60$

Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is $\frac{60}{5}=\boxed{\textbf{(D) }12}$

Solution 2 (Doesn't assume a solution exists)

We know that the smallest number in the arithmetic sequence must be $\geq 3 + 5 = 8$ , and the largest number must be $\leq 7 + 9 = 16$ .

Since there are $5$ terms in this sequence, the common difference $d \leq (16 - 8)/(5-1) = 2$ .

Since $d$ is an integer (difference of sums of integers), and since exactly 2 of the sums must be odd, $d$ must be odd. Therefore, $d=1$ .

The middle term must have the majority parity, so it must be even. The 2 terms adjacent to the middle are odd, $6+a$ and $6+b$ . $a-b = 2d = 2$ . $a$ and $b$ can't be the smallest (or largest) 2 odd numbers, because that would make it impossible to construct the smallest (or largest) sum from one of the remaining two numbers and one of the odd numbers. Therefore, $\{a,b\} = \{5,7\}$ . The middle sum must then be $(6+5) + 1 = (6+7)-1 = \boxed{12}$ . The remaining edges are $\{9,3\}$ (because $\{5,7\}$ can't be an edge, as that would make a triangle with 6), $\{3,7\}$ , and $\{5,9\}$ .

~oinava

Solution 3

We notice that the average of $3,5,6,7,$ and $9$ is 6. Now all we have to do is plug $6$ into each value. This results in all the terms being $12$ and since the term we are looking for is the middle term, we know that our answer is 12, D.

Video Solution by OmegaLearn

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

~ pi_is_3.14

@@ Line 27: / Line 27: @@
 ==Solution 3 ==
-We notice that the average of <math>3,5,6,7, and 9</math> is 6. Now all we have to do is plug <math>6</math> into each value. This results in all the terms being <math>12</math> and since the term we are looking for is the middle term, we know that our answer is 12, D.
+We notice that the average of <math>3,5,6,7,</math> and <math>9</math> is 6. Now all we have to do is plug <math>6</math> into each value. This results in all the terms being <math>12</math> and since the term we are looking for is the middle term, we know that our answer is 12, D.
 == Video Solution by OmegaLearn ==

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