AMC 10 · 2005 · #17
Grade 6 geometry-2dIn 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 this order. What is the middle term of the arithmetic sequence?
Pick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: The numbers $3$, $5$, $6$, $7$, $9$ are placed one at each of the five points $A$, $B$, $C$, $D$, $E$ of a five-pointed star. Each of the five edges $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DE}$, $\overline{EA}$ gets a sum (its two endpoint numbers added), and those five sums, in some order, form an arithmetic sequence. Find the middle term of that sequence.
Givens: The numbers $3$, $5$, $6$, $7$, $9$ are placed one at each of the five points $A$, $B$, $C$, $D$, $E$; The five edge sums along $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DE}$, $\overline{EA}$ form an arithmetic sequence in some order; Answer choices: (A) $9$, (B) $10$, (C) $11$, (D) $12$, (E) $13$
Unknowns: The middle term of the arithmetic sequence formed by the five edge sums
Understand
Restated: The numbers $3$, $5$, $6$, $7$, $9$ are placed one at each of the five points $A$, $B$, $C$, $D$, $E$ of a five-pointed star. Each of the five edges $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DE}$, $\overline{EA}$ gets a sum (its two endpoint numbers added), and those five sums, in some order, form an arithmetic sequence. Find the middle term of that sequence.
Givens: The numbers $3$, $5$, $6$, $7$, $9$ are placed one at each of the five points $A$, $B$, $C$, $D$, $E$; The five edge sums along $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DE}$, $\overline{EA}$ form an arithmetic sequence in some order; Answer choices: (A) $9$, (B) $10$, (C) $11$, (D) $12$, (E) $13$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #5 Look for a Pattern, #4 Introduce a Variable
Placing $3$, $5$, $6$, $7$, $9$ correctly is a maze of $120$ orderings, and chasing the right one is slow. Tool #16 (Change Focus) sidesteps it: instead of finding the arrangement, add up all five edge sums at once. Tool #5 (Look for a Pattern) supplies the key structural fact — every point touches exactly two edges, so that grand total counts each number twice. Tool #4 (Introduce a Variable) names the middle term $m$ and turns the arithmetic-sequence fact into a single equation.
Execute — Answer: D
4.OA.A.3 Step 1 Stop arranging, start totaling
- There are $120$ ways to place the five numbers, and hunting for the right one is slow work.
- Change focus: don't ask what each individual edge sum is, ask what all five edge sums add up to together.
- That one grand total is enough to pin down the middle term.
💡 The middle term of an arithmetic sequence is fixed by the total of its terms, so the grand total is worth more than any single arrangement.
4.OA.A.3 Step 2 Each point is counted twice
- Every point of the star lies on exactly two edges, so when you add all five edge sums, each number is used exactly twice.
- The grand total is therefore twice the sum of the five numbers: $2(3+5+6+7+9)=2\cdot 30 = 60$.
💡 Adding every edge sum double-counts each corner, so the total is simply twice the sum of all the numbers.
6.SP.B.5 Step 3 Middle term equals the average
- In an arithmetic sequence with an odd number of terms, the terms are balanced around the center, so the middle term equals the average of all the terms.
- With five terms, the total is $5$ times that middle term.
- Call the middle term $m$, so $5m = 60$.
💡 Evenly spaced numbers balance around their center, so their average lands exactly on the middle term.
6.EE.B.7 Step 4 Solve for the middle term
- Divide both sides of $5m = 60$ by $5$ to get $m = 12$.
- So the middle term of the arithmetic sequence is $12$, which is choice (D).
💡 One total split into five equal shares — the middle term is just the grand total divided by five.
4.OA.A.3 There are $120$ ways to place the five numbers, and hunting for the right one is 4.OA.A.3 Every point of the star lies on exactly two edges, so when you add all five edge 6.SP.B.5 In an arithmetic sequence with an odd number of terms, the terms are balanced ar 6.EE.B.7 Divide both sides of $5m = 60$ by $5$ to get $m = 12$. So the middle term of the Review
Reasonableness: The middle term must be the average of the five edge sums, and that average never depends on how the numbers are arranged — it is always $60/5 = 12$. Check that a real arrangement exists so the problem is well-posed: place the numbers around the star in the cyclic order $9, 5, 6, 7, 3$. The edge sums are $9+5=14$, $5+6=11$, $6+7=13$, $7+3=10$, $3+9=12$, i.e. $\{10,11,12,13,14\}$ — an arithmetic sequence with common difference $1$ and middle term $12$. Also, $12$ is one of the answer choices and sits right in the middle of the list, both reassuring signs.
Alternative: Replace every number by the average of the five, which is $30/5 = 6$. Each edge sum then becomes $6+6 = 12$, so all five 'sums' equal $12$ — a constant arithmetic sequence whose middle term is $12$. Because the grand total of the edge sums depends only on the total of the numbers (each counted twice), this swap leaves the grand total unchanged, so the real middle term is also $12$.
CCSS standards used (min grade 6)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Adding the five numbers and doubling to get the grand total of the edge sums, $2\cdot 30 = 60$.)6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Using that the middle term of a $5$-term arithmetic sequence equals the average of the terms, so the total is $5$ times the middle term.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving the one-step equation $5m = 60$ to get $m = 12$.)
⭐ When the exact arrangement is hard to pin down, add everything up at once — the total often hands you the answer.
⭐ When the exact arrangement is hard to pin down, add everything up at once — the total often hands you the answer.
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