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

Latest revision as of 13:25, 13 October 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 $12$ , $\boxed{D}$ .

Video Solution by OmegaLearn

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

~ pi_is_3.14

@@ Line 24: / Line 24: @@
 ~oinava
+==Solution 3 ==
+We notice that the average of <math>3,5,6,7</math> and <math>9</math> is <math>6</math>. 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 <math>12</math>, <math>\boxed{D}</math>.
 == Video Solution by OmegaLearn ==
@@ Line 31: / Line 34: @@
 ==See Also==
+[[Category:Introductory Algebra Problems]]
 {{AMC10 box|year=2005|ab=A|num-b=16|num-a=18}}
-[[Category:Introductory Geometry Problems]]
-[[Category:Area Ratio Problems]]
 {{MAA Notice}}

2005 AMC 10A (Problems • Answer Key • Resources)
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